close

Вход

Log in using OpenID

embedDownload
College of Engineering
Mech. Eng. Dept.
Subject: Strength of Materials
Second Class
Lecturer: Sadiq Muhsin Almosawy
Strength of materials is a branch of applied mechanics that deals with the behavior of
solid bodies subjected to various types of loading. The solid bodies include axially-loaded
bars, shafts, beams, and columns. The objective of analysis will be the determination of the
stresses, strains, and deformations produced by the loads.
Simple Stress ():
If a cylindrical bar is subjected to a direct pull or push along its axis, then it is
said to be subjected to tension or compression.
P
P
P
Tension
P
Compression
In SI systems of units load is measured in Newton (N) or Kiloewton (KN) or
Meganewton (MN).
Normal stress () : is the intensity of normal force per unit area
Stress =
 
Load
Area
P
A
stress may thus be compressive or tensile depending on the nature of the load and will be
measured in units of Newton per square meter (N/m2). This unit, called Pascal
1 Pa=1 N/m2
1 KPa=1000 Pa=103 Pa
1 MPa=106 Pa
1 GPa=109 Pa
In the U.S. customary or foot-pound-second system of units, express stress in
pounds per square inch (Psi) or kilopound per square inch (Ksi)
1
Normal Strain ():
If a bar is subjected to a direct load, and hence a stress, the bar will change in
length. If the bar has an original length (L) and changes in length by an amount (L), the
strain produced is defined as follows:
Strain()=

change in length
original length
L
L
P
P
L
L
Strain is thus a measure of the deformation of the material and is non-dimensional,
i.e. it has no units. Tensile stresses and strains are considered positive sense. Compressive
stresses and strains are considered negative in sense.
Shear Stress () and Bearing Stress (b ):
Shearing
stress differs from both tensile and compressive stress in that it is
caused by forces acting along or parallel to the area resisting the forces, whereas tensile
and compressive stresses are caused by forces perpendicular to the areas on which they act.
For this reason, tensile and compressive stresses are called normal stresses, whereas a
shearing stress may be called a tangential stress.
A shearing stress is produced whenever the applied loads cause one section of a
body to tend to slide past its adjacent section.
Shear stress=
Shear load
Area resisting shear
Q
Q
Q
Q
 
Q
A
2
Area resisting shear is the shaded area as shown above.
P
P
P
A
P
Single shear stress
P

P
A
P
P/2
P
A
P/2
 
Double shear stress
P/2
A
Bearing stress is a normal stress that is produced by the compression of one
surface against another. The bearing area is defined as the projected area of the curved
bearing surface.
B
1
P
C
A
P
2
3
F
b  b
Ab
Consider the bolted connection shown above, this connection consists of a flat bar
A, a clevis C, and a bolt B that passes through holes in the bar and clevis. Consider the
bearing stresses labeled 1, the projected area Ab on which they act is rectangle having a
3
height equal to the thickness of the clevis and a width equal to the diameter of the bolt, the
bearing force Fb represented by the stresses labeled 1 is equal to P/2. The same area and the
same force apply to the stresses labeled 3. For the bearing stresses labeled 2, the bearing
area Ab is a rectangle with height equal to the thickness of the flat bar and width equal to
the bolt diameter. The corresponding bearing force Fb is equal to the load P.
Shear Strain (  ):
Shear strain is a measure of the distortion of the element due to shear. Shear
strain is measured in radians and hence is non-dimensional, i.e. it has no units.



Elastic Materials-Hook's Law:
A material is said to be elastic if it returns to its original, when load is
removed. In elastic material, stress is proportional to strain. Hook's law therefore states
that:
Stress ( )  strain (  )
stress
 constant
strain
Within the elastic limit, i.e. within the limits in which Hook's law applies, it has
been shown that:

E

This constant is given the symbol E and termed the modulus of elasticity or
Young's modulus.
Poisson's Ratio ( ):
Consider the rectangular bar shown below subjected to a tensile load. Under
the action of this load the bar will increase in length by an amount L giving a longitudinal
strain in the bar of:
L 
L
L
4
The bar will also exhibit a reduction in dimensions laterally i.e. its breadth and
depth will both reduce.
b
b
2
P
P
d
d
2
The associated lateral strains will both be equal, will be of opposite sense to the
longitudinal strain, and will be given by:
 lat  
d
b

d
b
Poisson's ratio is the ratio of the lateral and longitudinal strains and always
constant
Poisson's ratio=
 
Lateral Strain
Longitudin al Strain
d / d
L / L
Longitudinal Strain=
Lateral Strain= 

E

E
Modulus of Rigidity ( G ):
For materials within the elastic range the shear strain is proportional to the shear
stress producing it.
 
Shear Stress
=Constant
Shear Strain

=G

The constant G is termed the modulus of rigidity.
5
Example 1:A 25 mm square cross-section bar of length 300 mm carries an axial
compressive load of 50 KN. Determine the stress set up in the bar and its change of length
when the load is applied. For the bar material E=200 GN/m2.
25 mm
50 KN
300 mm
Cross-section area of the bar(A)=25×10-3×25×10-3=625×10-6 m2
 
P
A
50  10 3
=80000000 N/m2
6
625  10
=
=80 MN/ m2

=

E
80  10 6
=0.0004
200  10 9
L  L
L=0.0004×300×10-3=0.12×10-3m
L=0.12 mm
6
Example 2: Two circular bars, one of brass and the other of steel, are to be loaded by a
shear load of 30 KN. Determine the necessary diameter of the bars a) in single shear b) in
double shear, if the shear stress in the two materials must not exceed 50 MN/m2 and 100
MN/m2 respectively.
a) Single Shear
 
F
A
A
F

 For brass material
30  10 3
A=
=0.0006 m2
6
50  10
A=r2
r
A

r
0.0006

r=13.8197×10-3 m
the diameter of the bar (d)=27.639×10-3 m
 For steel material
30  10 3
=0.0003 m2
6
100  10
0.0003
r
=9.772×10-3 m

A=
the diameter of the bar (d)=19.544×10-3 m
b) Double Shear
 
F
2A
A
F
2
 For brass material
30  10 3
A=
=0.0003 m2
6
2  50  10
0.0003
r
=9.772×10-3 m

the diameter of the bar (d)=19.544×10-3 m
 For steel material
30  10 3
=0.00015 m2
6
2  100  10
0.00015
r
=6.909×10-3 m

A=
the diameter of the bar (d)=13.819×10-3 m
7
Example 3: The 80 kg lamp is supported by two rods AB and BC as shown. If AB has a
diameter of 10 mm and BC has a diameter of 8 mm, determine the average normal stress in
each rod.
FBC 
C
A
 Fx  0
5
4
 FBA  cos 60  0
5
60
B
FBA
FBC=0.625FBA ……………..(1)
F
y
FBC 
0
3
 FBA  sin 60  784 .8  0
5
FBC=1308-1.44337FBA
……….(2)
80  9.81  784 .8 N
1308-1.44337FBA=0.625FBA
FBA=632.38 N
FBC=395.2375 N
 BA 
FBA
632.38
=
ABA
 (5  10 3 ) 2
BA=8.051877×106 Pa
BA=8.051877 MPa
 BC 
3
4
FBC
395.2375
=
ABC  ( 4  10 3 ) 2
BC=7.863149×106 Pa
BC=7.863149 MPa
8
FBC
Example 4: Shafts and pulleys are usually fastened together by means of a key, as shown.
Consider a pulley subjected to a turning moment T of 1 KN.m keyed by a 10 mm×10
mm×75 mm key to the shaft. The shaft is 50 mm in diameter. Determine the shear stress on
a horizontal plane through the key.
T
1 KN
10 mm
F
10 mm
o
M
o
50 mm
0
1  10 3  F  0.025  0
75 mm
F=40000 N
F=40 KN
 
F
A
F
F
A is the shaded area
=
40  10 3
10  10 3  75  10 3
=53.333×106 N/m2
=53.333 MN/m2
Example 5: Consider a steel bolt 10 mm in diameter and subjected to an axial tensile load
of 10 KN as shown. Determine the average shearing stress in the bolt head, assuming
shearing on a cylindrical surface of the same diameter as the bolt.
A=dt
A=×10×10-3×8×10-3=0.000251327 m2
10 mm
F
 
A
10  10 3
=
0.000251327
8 mm
=39.7888×106 N/m2
=39.7888 MN/m2
9
10 KN
Example 6: The bar shown has a square cross section for which the depth and thickness are
40 mm. If an axial force of 800 N is applied along the centroidal axis of the bar's cross
sectional area, determine the average normal stress and average shear stress acting on the
material along a) section plane a-a and b) section plane b-b.
a) section plane a-a
 
P
A
a
b
800 N
800
=
3
40  10  40  10 3

60
F
A
 
b
a
=500 KN/m2
800 N
800 N
F=0
=0
b) section plane b-b
40
d=
=46.188 mm
sin 60
 
800 N
800 sin 60
=375 KN/m2
3
3
46.188  10  40  10

F1
A
60
F2
F2
A
=
=
F1
800 cos 60
=216.50645 KN/m2
3
3
46.188  10  40  10
10
d
800 N
Example 7: Determine the total increase of length of a bar of constant cross section
hanging vertically and subject to its own weight as the only load. The bar is initially
straight.
: is the specific weight ( weight/unit volume )
A: is the cross-sectional area
d 
Aydy
AE
dy
L
   d
L
dy
0
y
Aydy
= 
AE
0
L
A
ydy
AE 0
yA
L
=
A 1 2
y
=
AE 2
=
A 2
L
2 AE
=
AL.L
2 AE
L
0
W=AL
=
W .L
2 AE
11
Example 8: A member is made from a material that has a specific weight  and modulus of
elasticity E. If its formed into a cone having the dimensions shown, determine how far its
end is displaced due to gravity when its suspended in the vertical position.
r
x y
=
r L
x  r

3
P( y )
x
y
L
L
v= x 2 y
y
y

3
2 2
 r y
=   2 y
3
L
w(y)=v= x 2 y
w( y )
 r 3
y
3 L2
2
w(y)=
From equilibrium P(y)=w(y)
 r 3
P(y)=
y
3 L2
2
r
A(y)=x2=   2 y 2
L
2
 r 3

y dy
P( y )dy 3 L2
d=

2
A( y ) E
r 2
 2 y E
L

d= ydy
3E
L
L
ydy
   d  
3E
0
0
2
 
L2
6E
12
Example 9: A solid truncated conical bar of circular cross section tapers uniformly from a
diameter d at its small end to D at the large end. The length of the bar is L. Determine the
elongation due to an axial force P applied at each end as shown.
r
d
y
2
D d

y
P
2
2

x
L
D d x
y(  )
2 2 L
d
D d x
r  (  )
2
2 2 L
D
d
P
L
D d

2 2
y
A(x)=r2
r
P
P
2
d
D d x
A(x)=    (  ) 
2 2 L
2
Pdx
Pdx
d=
=
2
A( x) E
D d x
d
 (  )  E
2 2 L
2
L
Pdx
= 
0
x
dx
2
D d x
d
 (  )  E
2 2 L
2
P
D d x
d
=
(  ) 

 D d 2
2 2 L
E (  )
L 2 2
1
L
0
L
PL
PL
PL

=
=
D d d
D d d D d 
D d d
D d x
E (  )     E (  )
E (  )   (  )  0
2 2 2
2 2 2 2 2
2 2 2
2 2 L
PL
PL
4 PL 
1
1

=
=

 2

2
2
2 

E  D  dD Dd  d 
D
dD
Dd d
E (

) E (
 )
4
4
4
4
4 PL
=
dDE
13
Example 10: Determine the smallest dimensions of the circular shaft and circular end cop
if the load it is required to support is 150 KN. The allowable tensile stress, bearing stress,
and shear stress is (t)allow=175 MPa, is (b)allow=275 MPa, and allow=115 MPa.
(b)allow=
Fb
Ab
275×106=
150  10 3
Ab
P  150 KN
Ab=0.0005454 m2

4
4 Ab

d2=

d2
Ab= d 22
4  0.0005454

d2=0.026353 m=26.353 mm
P
A
150  10 3
175×106=
A
(t)allow=
t
A=0.0008571 m2
30 mm

A= [d12  (30  10 3 ) 2 ] =0.0008571
4
d1=0.04462 m=44.62 mm
allow=
d1
F
A
115×106=
150  10 3
A
A=0.0013043 m2
1. A=td
0.0013043= t××30×10-3
t=0.013839 m=13.839 mm
2. A=td2
0.0013043= t××26.353×10-3
t=0.01575 m=15.75 mm
14
Statically Indeterminate Members:
If
the values of all the external forces which act on a body can be
determined by the equations of static equilibrium alone, then the force system is statically
determinate.
P2
P1
P
R1
R2
R1
R3
R3
R2
In many cases the forces acting on a body cannot be determined by the equations of
static alone because there are more unknown forces than the equations of equilibrium. In
such case the force system is said to be statically indeterminate.
P
P
M1
R1
R4
R4
R1
R2
R2
R3
15
R3
Example 11: A square bar 50 mm on a side is held rigidly between the walls and loaded
by an axial force of 150 KN as shown. Determine the reactions at the end of the bar and the
extension of the right portion. Take E=200 GPa.
150 KN
150 KN
R2
R1
100 mm
R1+R2=150×103 ………………(1)
1=2
R1  100  10 3
R2  150  10 3

50  10 3  50  10 3  200  10 9 50  10 3  50  10 3  200  10 9
0.1R1=0.15R2
R1=1.5R2
…………………..(2)
From equations (1) and (2)
1.5R2 + R2=150×103
R2 =60000 N
R1=90000 N
2=
R2  150  10 3
60000  150  10 3

50  10 3  50  10 3  200  10 9 50  10 3  50  10 3  200  10 9
2=0.000018 m
2=0.018 mm
16
150 mm
Example 12: A steel bar of cross section 500 mm2 is acted upon by the forces shown.
Determine the total elongation of the bar. For steel, E=200 GPa.
50 KN
A
C
B
D
45 KN
15 KN 10 KN
1m
500 mm
50 KN
A
1 .5 m
B
50 KN
B
C
50 KN
15 KN
45 KN
C
D
 For portion AB
1=
PL 50  10 3  500  10 3
=
=0.00025 m=0.25 mm
AE 500  10 6  200  10 9
 For portion BC
35  10 3  1
PL
2=
=
=0.00035 m=0.35 mm
AE 500  10 6  200  10 9
 For portion CD
3=
45  10 3  1.5
PL
=
=0.000675 m=0.675 mm
AE 500  10 6  200  10 9
T=1+2+3
T=0.25+0.35+0.675=1.275 mm
17
45 KN
35 KN
Example 13: Member AC shown is subjected to a vertical force of 3 KN. Determine the
position x of this force so that the average compressive stress at C is equal to the average
tensile stress in the tie rod AB. The rod has a cross-sectional area of 400 mm2 and the
contact area at C is 650 mm2.
FAB
B
3 KN
x
3 KN
x
A
A
C
200 mm
F
y
FAB+FC-3000=0
FAB+FC=3000
200 mm
FC
0
………………………(1)
AB=C
FAB FC

AAB AC
FC
FAB

6
400  10
650  10 6
FAB=0.6153 FC
…………………….(2)
From equations (1) and (2)
FC=1857.24 N
FAB=1142.759 N
M
A
0
FC×200×10-3-3000×x=0
x
C
1857.24  0.2
=0.123816 m=123.816 mm
3000
18
Example 14: The bar AB is considered to be absolutely rigid and is horizontal before the
load of 200 KN is applied. The connection at A is a pin, and AB is supported by the steel
rod EB and the copper rod CD. The length of CD is 1m, of EB is 2 m. The cross sectional
area of CD is 500 mm2, the area of EB is 250 mm2. Determine the stress in each of the
vertical rods and the elongation of the steel rod. Neglect the weight of AB. For copper
E=120 GPa, for steel E=200 GPa.
M
E
A
0
C
A
3
D
FCo×1+Fs×2-200×10 ×1.5=0
FCo=300×103-2 Fs …………..(1)
Fs
Ax
 Fs  L 
F L

  2 Co

 As E  s
 ACo E  Co
F s 2
FCo  1
 2
6
9
250  10  200  10
500  10 6  120  10 9
Ay
200 KN
E
FCo=1.2 Fs …………………….(2)
C
From equations (1) and (2)
A
D
 Co
Fs=93750 N
FCo=112500 N
Fs
93750
=
=375000000 Pa
As 250  10 6
s=375 MPa
 Co 
200 KN
FCo
s=2Co
s 
500 mm 500 mm
1m
 s  Co

2
1
B
FCo
112500
=
=225000000 Pa
ACo 500  10 6
Co=225 MPa
 s L 375  10 6  2
s 
=
=0.00375 m=3.75 mm
E
200  10 9
19
B
s
Thermal Stresses:
A change in temperature can cause a material to change its dimensions. If the
temperature increases, generally a material expands, whereas if the temperature decreases
the material will contract.
The deformation of a member having a length L can be calculated using the
formula:
T=×T×L
T=
FL
=×T×L
AE
T=E××T
: Linear coefficient of thermal expansion. The units measure strain per degree of
temperature. They are (1/ºF) in the foot-pound-second system and (1/ºC) or (1/ºK) in SI
system.
T: Change in temperature of the member.
L: The original length of the member.
T: The change in length of the member.
Example 15: The A-36 steel bar shown is constrained to just fit between two fixed
supports when T1=60º F. If the temperature is raised to T2=120º F determine the average
normal thermal stress developed in the bar. For steel =6.6×10-6 1/ºF, E=29×103 Ksi.
0.5 in
F
T
F
y
0.5 in
FA
A
0
FA-FB=F
T-F=0
T=×T×L
T=E××T
=29×103 ×6.6×10-6 ×(120-60)
=11.484 Ksi
20 in
B
FB
20
Example 16: A 2014-T6 aluminum tube having a cross sectional area of 600 mm2 is used
as a sleeve for an A-36 steel bolt having a cross sectional area of 400 mm2. When the
temperature is T1=15º C, the nut hold the assembly in a snug position such that the axial
force in the bolt is negligible. If the temperature increases T2=80º C, determine the average
normal stress in the bolt and sleeve. For aluminum =23×10-6 1/ºC, E=73.1 GPa, for steel
=12×10-6 1/ºC, E=200 GPa.
F
y
150 mm
0
Fb
Fsl-Fb=0
Fsl=Fb=F
Initial Position
( b ) T
  ( b ) T  ( b ) F  ( sl ) T  ( sl ) F
[×T×L+
FL
FL
]b=[×T×L]sl
AE
AE
12×10-6×0.15×(80-15)+

( sl ) T
( sl ) F
F  0.15
F  0.15
= 23×10-6×0.15×(80-15)6
9
400  10  200  10
600  10 6  73.1  10 9
F=20255 N
20255
F
=
Ab 400  10 6
b=50.637655 MPa
sl=
( b ) F
Final Position
0.0052949×10-6F=0.00010725
b=
Fsl
20255
F
=
Asl 600  10 6
sl=33.758436 MPa
21
Example 17: The rigid bar shown is fixed to the top of the three posts made of A-36 steel
and 2014-T6 aluminum. The posts each have a length of 250 mm when no load is applied
to the bar, and the temperature is T1=20ºC. Determine the force supported by each posts if
the bar is subjected to a uniform distributed load of 150 KN/m and the temperature is
raised to T2=80ºC. For steel =12×10-6 1/ºC, E=200 GPa , for aluminum =23×10-6 1/ºC,
E=73.1 GPa.
300 mm
300 mm
Initial Position
150 KN / m
( al )T ( )
al F
( st ) T
( st ) F

60 mm
Final Position
250 mm
40 mm
Steel
F
y
40 mm
Aluminum
0
Steel
150  0.6  90 KN
2Fst+Fal=90000 …………….(1)
=(st)T-(st)F=(al)T-(al)F
Fst
[×T×L-
Fal
Fst
Fst L
F L
]st=[×T×L- al ]al
AE
AE
12×10-6×0.25×(80-20)- 
4
Fst  0.25
-6
=23×10
×0.25×(80-20)- 
( 40  10 3 ) 2  200  10 9
4
1.20956×10-9Fal-0.994718×10-9Fst=0.000165 ………………..(2)
From equations (1) and (2)
Fst=-16444.7 N
Fal=122888.8 N
22
Fal  0.25
(60  10 3 ) 2  73.1 109
Example 18: The rigid bar AD is pinned at A and attached to the bars BC and ED as
shown. The entire system is initially stress-free and the weights of all bars are negligible.
The temperature of bar BC is lowered 25ºK and that of the bar ED is raised 25ºK.
Neglecting any possibility of lateral buckling, find the normal stresses in bars BC and ED.
For BC, which is brass, assume E=90 GPa, =20×10-6 1/ºK and for ED, which is steel,
take =12×10-6 1/ºK, E=200 GPa. The cross-sectional area of BC is 500 mm2, of ED is
250 mm2.
M
A
E
0
250 mm
A
Pst×600×10-3-Pbr×250×10-3=0
B
D
300 mm
350 mm
250 mm
C
Pst=0.41666 Pbr ………..(1)
Pst
 br

 st
250 600
  L  T 
250
Ax
Pbr  L
Abr E br
  L  T 

Pst  L
Ast E st
Ay
Pbr
600
 br
( st )T
( st ) F
( br )T
( br ) F
Pbr  300  10 3
Pst  250  10 3
6
3
12

10

250

10

25

500  10 6  90  10 9 
250  10 6  200  10 9
250
600
20  10 6  300  10 3  25 
8.333×10-12 Pst+26.666×10-12 Pbr=475×10-9 …………..(2)
From equations (1) and (2)
Pbr=15760.5 N , Pst=6566.77 N
15760.5
 31.521 MPa
500  10 6
6566.77
Pst=
 26.267 MPa
250  10 6
br=
23
 st
Torsion:
Torque is a moment that tends to twist a member about its longitudinal axis.
When the torque is applied, the circles and longitudinal grid lines originally marked on the
shaft tend to distort into the pattern shown below.
T
T
Before deformation
After deformation
Twisting causes the circles to remain circles and each longitudinal grid line
deforms into a helix that intersects the circles at equal angles. Also, the cross sections at
the ends of the shaft remain flat that is, they do not warp or bulge in or out and radial lines
on these ends remain straight during the deformation.
The Torsion Formula:
Consider a uniform circular shaft is subjected to a torque it can be shown
that every section of the shaft is subjected to a state of pure shear.
T
 max


r
T
T
 
J
24
: The torsional shearing stress.
T: The resultant internal torque acting on the cross section.
: The distance from the centre (radial position).
J: The polar moment of inertia of the cross sectional area.
 max 
Tr
J
max: The maximum shear stress in the shaft, which occurs at the outer surface.
r: The outer radius of the shaft.
 4
r
2
 4
J
D
32
J
for a hollow shaft
 
Tr
J
 max 
J
r
ri
ro
Tro
J
 4

(ro  ri 4 ) 
( Do4  Di4 )
2
32
Angle of Twist ():
If a shaft of length L is subjected to a constant twisting moment along its
length, then the angle of twist  through which one end of the shaft will twist relative to the
other is:

TL
GJ
T
L
G: The shear modulus of elasticity or
modulus of rigidity.
: Angle of twist, measured in rad
r

A
25
T
If the shaft is subjected to several different torques or the cross sectional area or
shear modulus changes from one region to the next. The angle of twist of one end of the
shaft with respect to the other is then found from:
 
TL
GJ
In order to apply the above equation, we must develop a sign convention for the
internal torque and the angle of twist of one end of the shaft with respect to the other end.
To do this, we will use the right hand rule, whereby both the torque and angle of twist will
be positive, provided the thumb is directed outward from the shaft when the fingers curl to
give the tendency for rotation.
D
LCD
LBC
150 N.m
10 N.m
C
60 N.m
B
LAB
A
B
LAB
80 N.m
80 N.m
150 N.m
LBC
A
C
70 N.m
B
80 N.m
80 N.m
D
LAB
10 N.m
C
60 N.m
70 N.m
26
 A/ D 
80  L AB 70  LBC 10  LCD


GJ
GJ
GJ
Power Transmission (P):
Shaft and tubes having circular cross sections are often used to transmit
power developed by a machine.
P T
d
dt
,
d

dt
: The shaft's angular velocity (rad/s).
P  T 
In SI units power is expressed in (watts) when torque is measured in (N.m) and 
in (rad/s).
1 W=1 N.m/s
In the foot-pound-second or FPS system the units of power are (ft.lb/s); however
horsepower (hp) is often used in engineering practice where:
1 hp=550 ft.lb/s
For machinery the frequency of a shaft's rotation f is often reported. This is a
measured of the number of revolutions or cycles the shaft per second and is expressed in
hertz (1 Hz=1 cycle/s), 1 cycle=2 rad, then =2f
P=2fT
Example 19: If a twisting moment of 1 KN.m is impressed upon a 50 mm diameter shaft,
what is the maximum shearing stress developed? Also what is the angle of twist in a 1 m
length of the shaft? The material is steel, for which G=85 GPa.
 max 
Tr
J
J
 4 
4
D 
(50  10 3 ) 4  0.6135  10 6 m
32
32
 max
1  10 3  25  10 3

0.6135  10 6
max=40.74979 MPa
27
TL
GJ

1  10 3  1

85  10 9  0.6135  10 6
=0.01917 rad.
Example 20: The pipe shown has an inner diameter of 80 mm and an outer diameter of 100
mm. If its end is tightened against the support at A using a torque wrench at B, determine
the shear stress developed in the material at the inner and outer walls along the central
portion of the pipe when the 80 N forces are applied to the wrench.
T=80×200×10-3+80×300×10-3
T=40 N.m
Tr
J

J
( Do4  Di4 )
32
 
J


(100  10 3 ) 4  (80  10 3 ) 4
32

J=5.7962×10-6 m4
 Inner walls ri=40 mm
 
Tri 40  40  10 3
=
J
5.7962  10 6
=0.276042 MPa
 Outer walls ro=50 mm
Tro 40  50  10 3

=
J
5.7962  10 6
=0.345053 MPa
28
Example 21: The gear motor can developed 0.1 hp when it turns at 80 rev/min. If the
allowable shear stress for the shaft is allow=4 ksi, determine the smallest diameter of the
shaft that can be used.
 allow 
Tr
J
 4
r
2
J
 allow 
 allow 
r3
Tr
 4
r
2
2T
r 3
2T
 allow
P=T.
T
P

P=0.1×550=55 lb/s
=80×2/60=8.377 rad/s
T
55
=6.5655 lb.ft
8.377
T=6.5655×12=78.786 lb.in
r3
2  78.786
=0.2323 in
  4  10 3
d=0.4646 in
29
Example 22: The assembly consists of a solid 15 mm diameter rod connected to the inside
of a tube using a rigid disk at B. Determine the absolute maximum shear stress in the rod
and in the tube. The tube has an outer diameter of 30 mm and a wall thickness of 3 mm.
 The rod
T=50 N.m
r=7.5×10-3 m
 4
r
2

J  (75  10 3 ) 4
2
J
J=4.97009×10-9 m4
r 
Tr
J
r 
50  7.5  10 3
=75.4512 MPa
4.97009  10 9
 The tube
T=80 N.m
r=15×10-3 m
15 mm
 4
(ro  ri 4 )
2

-9
4
J  (15  10 3 ) 4  (12  10 3 ) 4 =46.9495×10 m
2
J

t 

Tro 80  15  10 3
=
=25.559 MPa
J
46.9495  10 9
30
12 mm
Example 23: The tapered shaft shown below is made of a material having a shear modulus
G. Determine the angle of twist of its end B when subjected to the torque.
x
d 2 d1

2
2
y
d1
d
x
d  d1 x
y d 2  d1

y 2
x
2L
2 L
x
d=d1+2y=d1+ (d 2  d1 )
L
 4 
x
J ( x)  d  [d1  (d 2  d1 ) ]4
32
32
L
L
L
L
T dx
T dx
32T
dx





x
x
G J ( x) 0
G 0
0
G
[ d1  ( d 2  d1 ) ]4
[ d1  ( d 2  d1 ) ]4
32
L
L
L




1
32TL
1
32TL
1







x
 3G (d 2  d1 ) 
 3G (d 2  d1 )  d 23 d13 
[ d 1  ( d 2  d 1 ) ]3 
L 0


 d13  d 23 
32TL


 3G (d 2  d1 )  d 23 .d13 

32TL  d12  d1d 2  d 22 


3G 
d 23 .d13

31
d2
Example 24: The gears attached to the fixed end steel shaft are subjected to the torques
shown. If the shear modulus of elasticity is G=80 GPa and the shaft has a diameter of 14
mm, determine the displacement of the tooth P on gear A.
 Segment AC
150-TAC=0
TAC=150 N.m
 Segment CD
150-280+TCD=0
TCD=130 N.m
 Segment DE
-130-40+TDE=0
TDE=170 N.m
 
=
=
1
GJ
TL
GJ
 TL
1
150  0.4  130  0.3  170  0.5

9
3 4
80  10  (7  10 )
2
=-0.21211 rad
32
A steel shaft ABC connecting three gears consists of a solid bar of diameter d
between gears A and B and a hollow bar of outside diameter 1.25d and inside diameter d
between gears B and C. Both bars have length 0.6 m. The gears transmit torques T1=240
N.m, T2=540 N.m, and T3=300 N.m acting in the directions shown in the figure. The shear
modulus of elasticity for the shaft is 80 GPa. a) what is the minimum permissible diameter
d if the allowable shear stress in the shaft is 80 MPa?. b) what is the minimum permissible

diameter d if the angle of twist between any two gears is limited to 4 ?.
Example 25:
TAB=240 N.m
TBC=300 N.m
a)
1. For solid bar AB
 4
d
32
T d / 2 T AB d / 2
  AB

 4
J
d
32
240  d / 2
80  10 6 
 4
d
32
16  240
d3 
  80  10 6
J=
d=0.02481 m
d=24.81 mm
2. For hollow bar BC


[ d o4  d i4 ]  [(1.25d ) 4  d 4 ]
32
32
TBC  1.25  d / 2
T  1.25  d / 2
 
 BC

J
[(1.25d ) 4  d 4 ]
32
300  1.25  d / 2
16  300  1.25
80  10 6 
d3 

  80  10 6  1.4414
[(1.25d ) 4  d 4 ]
32
J=
d=0.0255 m
d=25.5 mm
b)
Answer
1. For solid bar AB

TAB L
GJ
33
4


180
240  0.6

80  10 9  d 4
32
d=0.02263 m
d=22.63 mm
2. For hollow bar BC
TBC L
GJ

4

180

300  0.6

80  10 9  [(1.25d ) 4  d 4 ]
32
d=0.02184 m
d=21.84 mm
d=0.02263 m
d=22.63 mm
Answer
Example 26: The shaft is subjected to a distributed torque along its length of t=10x2
N.m/m, where x is in meters. If the maximum stress in the shaft is to remain constant at 80
MPa, determine the required variation of the radius r of the shaft for 0  x  3 m
T   tdx
  10 x 2 dx 
 max 
10 3
x
3
Tr
J
10 3
x r
20 x 3
80  10 6  3
 80  10 6 
 4
3r 3
r
2
34
r3
20 x 3
 0.002982 x
3  80  10 6
m
r=2.982 x mm
Example 26: The shaft has a radius 50 mm and is subjected to a torque per unit length of
100 N.m which is distributed uniformly over the shafts entire length 2 m. If it is fixed at its
far end A, determine the angle of twist of end B. The shear modulus is 73.1 GPa.
2m
100 N .m / m
T(x)=100x
2
2
T ( x)dx
 

GJ
0
0
100 xdx

73.1  10 9  (50  10 3 ) 4
2
2
 x2 
200

 
73.1  10 9   (50  10 3 ) 4  2  0

400
73.1  10   (50  10 3 ) 4
9
  2.786  10 4 rad
=0.01596o
35
Statically Indeterminate:
TA-T+TB=0
TA +TB=T
A/B=0
T A L AC TB LBC

0
GJ
GJ
If T1>T2
-TA-T2+T1-TB=0
T2
 TB L1 (TA  T2 ) L2 TA L3


0
GJ
GJ
GJ
L3
T1
L2
L1
T2
T2
T1
T2
36
Example 27: The solid steel shaft shown has a diameter of 20 mm. If it is subjected to the
two torques, determine the reactions at the fixed supports A and B.
-TB+800-500-TA=0
TB+TA=300 …………………………………....(1)
 TB  0.2 (TA  500)  1.5 TA  0.3


0
GJ
GJ
GJ
-0.2TB+1.5TA+750+0.3TA=0
1.8TA-0.2TB=-750 …………………………...…(2)
TA=-345 N.m
TB=645 N.m
37
Example 28: The shaft shown below is made from steel tube, which is bonded to a brass
core. If a torque of T=250 lb.ft is applied at its end, plot the shear stress distribution along a
radial line of its cross sectional area. Take Gst11.4×103 ksi, Gbr=5.2×103 ksi.
Tst+Tbr=250×12=3000 lb.in………….(1)
θ=θst=θbr
Tst L
Tbr L



11.4  10 6  [(1) 4  (0.5) 4 ] 5.2  10 6  [(1) 4  (0.5) 4 ]
2
2
Tst=32.88Tbr……………………….(2)
From (1) and (2)
Tst=2911 lb.in=242.6 lb.ft
Tbr=88.5 lb.in=7.38 lb.ft
( br ) max 
( st ) max 
( st ) min
 
88.5  0.5
 451 psi

(0.5) 4
2
2911  1
 1977 psi

4
4
[(1)  (0.5) ]
2
2911  0.5

 988 psi

4
4
[(1)  (0.5) ]
2

451
988


 0.0867  10 3 rad
6
G 5.2  10
11.4  10 6
38
Torsion of Solid Noncircular Shafts:
max
Shape of cross section
Square
a
a
a
Equilateral
triangle

4.81T
a3
7.1TL
a 4G
20T
a3
46TL
a 4G
a
b
b
2T
ab 2
Ellipse
a
a
(a 2  b 2 )TL
a 3 b 3 G
Example 28: The 2014-T6 aluminum strut is fixed between the two walls at A and B. If it
has a 2 in by 2 in square cross section and it is subjected to the torsional loading shown,
determine the reactions at the fixed supports. Also what is the angle of twist at C. Take
G=3.9×103 ksi.
TA-40-20+TB=0
TA+TB=60
……….(1)
A/B=0
 A/ B  
7.1TL
a 4G
7.1TA  12  2  12 7.1(TB  20 )  12  2  12 7.1TB  12  2  12


0
2 4  3.9  10 6
2 4  3.9  10 6
2 4  3.9  10 6
39
TA
TA-2TB=-20 ………………(2)
40 lb. ft
From equations (1) and (2)
TB=26.666 lb.ft
TA=33.333 lb.ft
C 
20 lb. ft
C
TB
D
7.1T A L
a 4G
C 
7.1  33.333  12  2  12
2 4  3.9  10 6
C=0.001092 rad
C=0.06258º
Thin walled tubes having closed cross sections:
Shear flow(q): is the product of the tube's thickness and the average shear stress. This value
is constant at all points along the tube's cross section. As a result, the largest average shear
stress on the cross section occurs where the tube's thickness is smallest.
The forces acting on the two faces are dFA=A(tAdx) , dFB=B(tBdx), these
forces are equal for equilibrium, so that:
AtA=BtB
q=avgt
40
Average shear stress(avg):
The average shear stress acting on the shaded
Area dA=tds
dF=avgdA=avgtds
dT=dF×h=avgtds×h
T=avgt  hds
1
2
Area of triangle dAm= hds
hds=2dAm
T=2avgt  dAm =2avgtAm
avg=
T
2tAm
avg: The average shear stress acting over the thickness of the tube.
T: The resultant internal torque at the cross section.
t: The thickness of the tube where avg is to be determined.
Am: The mean area enclosed within the boundary of the center line of the tube thickness.
q=avgt=
T
2 Am
Angle of Twist():

TL
ds
2

4 AmG t
41
Example 29: The tube is made of C86100 bronze and has a rectangular cross section as
shown below. If its subjected to the two torques, determine the average shear stress in the
tube at points A and B. Also, what is the angle of twist of end C? The tube is fixed at E.
Take G=38 GPa.
60-25-T=0
T=35 N.m
Am=(40×10-3-5×10-3)(60×10-3-3×10-3)
=0.001995 m2
A=
35
T
=
3
2tAm 2  5  10  0.001995
A=1.7543859 MPa
B=
35
T
=
3
2tAm 2  3  10  0.001995
B=2.9239766 MPa
 
=
TL
ds
2

4 Am G t
60  0.5
35  10 3
57  10 3
[
2


2

]
4  (0.001995 ) 2  3.8  10 9
3  10 3
5  10 3

35  1.5
35  10 3
57  10 3
[
2


2

]
4  (0.001995 ) 2  3.8  10 9
3  10 3
5  10 3
=0.0062912 rad
42
Example 30: A thin tube is made from three 5 mm thick A-36 steel plates such that it has a
cross section that is triangular as shown below. Determine the maximum torque T to which
it can be subjected, if the allowable shear stress is allow=90 MPa and the tube is restricted
to twist no more than =2×10-3 rad. Take G=75 GPa.
1
2
A= (200×10-3)× (200×10-3 sin60)=0.01732 m2
t=0.005 m
allow=
T
T
=
=90×106
3
2tAm 2  5  10  0.01732
T=15.588 KN.m

TL
ds
2

4 AmG t
2×10-3=
T 3
200  10 3
[
3

]
4  (0.01732) 2  75  10 9
5  10 3
T=500 N.m
43
Thin Walled Cylinder, Thin Walled Pressure Vessels:
Cylindrical or spherical vessels are commonly used in industry to serve as boilers
or tanks. When under pressure, the material of which they are made is subjected to a
loading from all directions. In general "thin wall" refers to a vessel having an inner radius
to wall thickness ratio of 10 or more (r/t  10)
1. Cylindrical Vessels:
Consider the cylindrical vessel having a wall thickness t and inner radius r as
shown below. A pressure p is developed within the vessel by a containing gas or fluid,
which is assumed to have negligible weight.
The stresses set up in the walls are:
a. Circumferential or hoop stress
2[1(tdy)]-p(2rdy)=0
1 
pr
t
b. Longitudinal or axial stress
2(2rt)-p(r2)=0
2 
pr
2t
c. Circumferential or hoop strain
1 
1
( 1   2 )
E
d. Longitudinal strain
2 
1
( 2   1 )
E
44
e. Change in length
The change in length of the cylinder may be determined from the
longitudinal strain.
Change in length=longitudinal strain×original length
1
E
L=ε2L= ( 2   1 ) L
L=
pr
(1  2 ) L
2tE
f. Change in diameter
The change in diameter may be found from the circumferential change.
Change in diameter=diametral strain×original diameter
Diametral strain=circumferential strain
1
E
d= ε1d= ( 1   2 ) d
d=
pr
(2   ) d
2tE
g. Change in internal volume
Volumetric strain=longitudinal strain+2diametral strain
1
E
1
E
εv= ε2+2 ε1= ( 2   1 ) +2 ( 1   2 )
1
E
1 pr
pr
pr
pr
= (   2  )
E 2t
t
t
t
pr
εv=
(5  4 )
2tE
εv= ( 2   1  2 1  2 2 )
diametral strain
longitudinal strain
diametral strain
change in internal volume=volumetric strain×original volume
v= εvv
v=
pr
(5  4 ) v
2tE
45
2. Spherical Vessels:
Because of the symmetry of the sphere the stresses set up owing to internal
pressure will be two mutually perpendicular hoop or circumferential stress of equal
value and a radial stress.
1(2rt)-p(r2)=0
1 
pr
2t
2=  1 
pr
2t
Change in internal volume
change in internal volume=volumetric strain×original volume
volumetric strain=3hoop strain
1
E
εv= ε1=3 ( 1   2 ) =
3 1
3 pr
(1   ) =
(1   )
E
2tE
v= εvv
v =
3 pr
(1   ) v
2tE
46
Cylindrical Vessels with Hemispherical Ends:
r=d/2
a) For the cylindrical portion
Pr
tc
Pr
2 
2t c
1 
hoop stress
longitudinal stress
1
1 Pr
Pr
( 1   2 ) = (  
)
E
E tc
2t c
pr
ε1=
hoop strain
(2   )
2t c E
1 
b) For the spherical ends
1 
Pr
2t s
hoop stress

1
( 1   2 ) = 1 (1   )
E
E
pr
ε1=
hoop strain
(1   )
2t s E
1 
Thus equating the two strains in order that there shall be no distortion of the
junction.
pr
pr
(1   ) =
(2   )
2t s E
2t c E
ts 1 

tc 2  
47
Example 31: A thin cylinder 75 mm internal diameter, 250 mm long with walls 2.5 mm
thick is subjected to an internal pressure of 7 MN/m2. Determine the change in internal
diameter and the change in length. If in addition to the internal pressure, the cylinder is
subjected to a torque of 200 N.m find the magnitude and nature of the stresses set up in the
cylinder.E=200 GN/m2, υ=0.3.
pr
(2   ) d
2tE
75
7  10 6   10 3
2
d=
[2  0.3]  75  10 3
3
9
2  2.5  10  200  10
d=
d=33.468×10-6 m=33.468 μm
pr
(1  2 ) L
2tE
75
7  10 6   10 3
2
L=
[1  2  0.3]  250  10 3
2  2.5  10 3  200  10 9
L=
L=26.25×10-6 m=26.25 μm
1 
pr
=
t
75
 10 3
2
2.5  10 3
7  10 6 
1=105×106 N/m2=105 MN/m2
pr
=
2 
2t
75
 10 3
2
2  2.5  10 3
7  10 6 
2=52.5×106 N/m2=52.5 MN/m2
200  40  10 3
Tr
Tr
=
=
 

J  4
4
[(40  10 3 ) 4  (37.5  10 3 ) 4 ]
[ro  ri ]
2
2
=8.743862 MN/m2
48
Example 32: A cylinder has an internal diameter of 230 mm, has walls 5 mm thick and is 1
m long. It is found to change in internal volume by 12×10-6 m3 when filled with a liquid at
a pressure p. If E=200 GN/m2 and υ=0.25, and assuming rigid end plates, determine a) the
values of hoop and longitudinal stresses b) the necessary change in pressure p to produce a
further increase in internal volume of 15%.
a)
pr
(5  4 ) v
2tE
230
p
 10 3
230
-6
2
[5  4  0.25]    (
 10 3 ) 2  1
12×10 =
3
9
2
2  5  10  200  10
v=
p=1.255763 MN/m2
pr
=
t
1 
1.255763  10 6 
230
 10 3
2
5  10 3
1=28.882549 MN/m2
2 
pr
=
2t
230
 10 3
2
2  5  10 3
1.255763  10 6 
2=14.4412745 MN/m2
b)
v=1.15×12×10-6=13.8×10-6 m3
v=
pr
(5  4 ) v
2tE
230
 10 3
230
2
13.8×10 =
[5  4  0.25]    (
 10 3 ) 2  1
3
9
2
2  5  10  200  10
-6
p
p=1.444128 MN/m2
Necessary increase=1.444128-1.255763=0.188365 MN/m2
49
Vessels Subjected to Fluid Pressure:
If a fluid is used as the pressurization medium the fluid itself will change in
volume as pressure is increased and this must be taken into account when calculating the
amount of fluid which must be pumped into the cylinder in order to raise the pressure by a
specific amount.
The bulk modulus of a fluid is defined as:
bulk modulus k=
Volumetric stress
Volumetric strain
volumetric stress=pressure p
volumetric strain=
k=
change in volume v
=
v
original volume
p pv
=
v v
v
change in volume of fluid under pressure=
pv
k
extra fluid required to raise cylinder pressure by p
=
pr
pv
(5  4 ) v+
2tE
k
extra fluid required to raise sphere pressure by p
=
3 pr
pv
(1   ) v+
2tE
k
50
Example 33: a) A sphere 1m internal diameter and 6 mm wall thickness is to be pressure
tested for safety purposes with water as the pressure medium. Assuming that the sphere is
initially filled with water at atmospheric pressure, what extra volume of water is required
to be pumped in to produce a pressure of 3 MN/m2 gauge? For water k=2.1 GN/m2
b) The sphere is now placed in service and filled with gas until there is a volume change of
72×10-6 m3. Determine the pressure exerted by the gas on the walls of the sphere. c) To
what value can the gas pressure be increased before failure occurs according to the
maximum principal stress theory of elastic failure? E=200 GPa, υ=0.3 and the yield stress
is simple tension=280 MPa.
a) extra volume of water=
3 pr
pv
(1   ) v+
2tE
k
4
3  10 6   (0.5) 3
3  3  10  0.5
4
3
=
(1  0.3)   (0.5) 3 
3
9
3
2  6  10  200  10
2.1  10 9
6
=0.001435221 m3
b)
v=
3 pr
(1   ) v
2tE
72×10-6=
3 p  0 .5
4
(1  0.3)   (0.5) 3
3
9
3
2  6  10  200  10
p=0.31430827 MN/m2
1 
pr
2t
280×106=
1=yield stress for maximum principal stress theory
p  0 .5
2  6  10 3
p=6.72 MN/m2
51
Shear and Moment Diagram:
Beams
are long straight members that carry loads perpendicular to their
longitudinal axis. They are classified according to the way they are supported, e.g. simply
supported, cantilevered, or overhanging.
P
P
Simply supported beam
overhanging beam
P
Cantilevered beam
Types of Loading:
Loads commonly applied to a beam may consist of concentrated forces(applied at
a point), uniformly distributed loads, in which case the magnitude is expressed as a certain
number of newtons per meter of length of the beam, or uniformly varying loads. A beam
may also be loaded by an applied couple.
100 N
(point load)
(concentrated force)
52
10 N / m
5 N / m
4m
5m
Uniformly distributed load
Uniformly varying load
Shearing force and bending moment diagrams show the variation of these
quantities along the length of a beam for any fixed loading condition. At every section in a
beam carrying transverse loads there will be resultant forces on either side of the section
which, for equilibrium, must be equal and opposite.
Shearing force at the section is defined as the algebraic sum of the forces taken on
one side of the section. The bending moment is defined as the algebraic sum of the
moments of the forces about the section, taken on either side of the section.
Sign Convention:
Forces upwards to the left of a section or downwards to the right of a section are
positive. Clockwise moments to the left and counter clockwise to the right are positive.
M

V

M
M
M
V -
V
V
-
Procedure of Analysis:
The shear and moment diagrams for a beam can be constructed using the
following procedure:1. Determine all the reactive forces and couple moments acting on the beam, and
resolve all the forces into components acting perpendicular and parallel to the beam's
axis.
2. Specify separate coordinates x having an origin at the beam's left end extending to
regions of the beam between concentrated forces and/or couple moments, or where
there is no discontinuity of distributed loading.
3. Section the beam perpendicular to its axis at each distance x, and draw the free body
diagram of one of the segments. Be sure V and M are shown acting in their positive
sense, in accordance with the sign convention given as above.
4. The shear is obtained by summing forces perpendicular to the beam's axis.
5. The moment is obtained by summing moment about the sectioned end of the
segment.
53
6. Plot the shear diagram(V versus x) and the moment diagram(M versus x). If
numerical values of the functions describing V and M are positive, the values are
plotted above the x-axis, whereas negative values are plotted below the axis.
Example 33: Draw the shear and moment diagrams for the beam shown below.
P
P
B
A
Ax
C
B
Cy
Ay
L/2
L/2
F
0
x
Ax=0
M
C
0
P×L/2-Ay×L=0
Ay=P/2
F
y
0
Cy+Ay-P=0
Cy=P/2
 Segment AB
F
y
0
V
P
2
P
-V=0
2
P
V=
2
M  0
M
x
P
×x=0
2
P
M= x
2
M-
P
 Segment BC
 Fy  0
V
M
P
2
x
54
P
-P-V=0
2
P
V=2
M  0
P
L
×x+P(x- )=0
2
2
P
M= (L-x)
2
M-
P
B
P
2
P
2
P
2
S.F. diagram

M max 
B.M. diagram
55
PL
4
P
2
Example 34: Draw the shear and moment diagrams for the beam shown below.
20 KN
10 KN
2m
30 KN
4m
B
A
C
2m
E
D
F
8m
4m
20 KN
F
y
0
B
A
y
C
2m
F
E
D
Fx
Fy
8m
4m
Ay
20 KN
10-10+20-20-30+Fy=0
Fy=30 KN
0 x2
 Segment AB
F
4m
0
-Ay×12+10×10-20×8+20×6
+30×2=0
Ay=10 KN
F
30 KN
2m
Fx=0
M
20 KN
10 KN
 Fx  0
0
10-V=0
V=10 KN
M  0
M-10×x=0
M=10x
 Segment BC
F
y
10 KN
2 x4
0
V
10-10-V=0
V=0
M  0
x
10 KN
M-10x+10(x-2)=0
M=20 KN.m
4 x6
 Segment CD
F
y
0
56
M
10 KN
10-10+20-V=0
V=20 KN
M  0
V
M-10x+10(x-2)-20(x-4)=0
M=20(x-3)
 Segment DE
F
y
M
20 KN
x
10 KN
6  x  10
0
10 KN
20 KN
10-10+20-20-V=0
V=0
M  0
M-10x+10(x-2)-20(x-4)+20(x-6)=0
M=60 KN.m
20 KN
x
10 KN
 Segment EF
F
y
M
V
10  x  12
0
10-10+20-20-30-V=0
V=-30 KN
M  0
M-10x+10(x-2)-20(x-4)+20(x-6)
+30(x-10)=0
M=30(12-x)
10 KN
20 KN
30 KN
V
20 KN
10 KN
57
x
M
20 KN
10 KN
2m
A
10 KN
30 KN
4m
C
B
F
E
D
8m
4m
2m
30 KN
20 KN
20 KN
10 KN
S.F Diagram
 30 KN
60 KN.m
20 KN.m
B.M. Diagram
58
Fx
Example 35: Draw the shear and moment diagrams for the beam shown below.
w
B
A
L
F
0
x
wL
Ax=0
M
B
0
L
wL -AyL=0
2
wL
Ay=
2
 Fy  0
A
B
Ax
L
Ay
wL
+By-wL=0
2
wL
By=
2
 Fy  0
By
wx
x/2
w
V
wL
-wx-V=0
2
L
V=-w(x- )
2
M  0
M
x
wL
2
wL
x
x+wx( )=0
2
2
w
M= (xL-x2)
2
M-
Maximum moment occur when
dM w
 ( L  2 x)  0
dx
2
L  2x  0
L
x
2
Location of maximum moment
59
dM
0
dx
w
B
A
wL
2
wL
2
L
wL
2
S.F Diagram
L/2
M max 
wL2
8
B.M Diagram
L/2
60
Example 36: Draw the shear and moment diagrams for the beam shown below.
wo
A
B
L
wo L
2
 Fx  0
Ax=0
F
y
0
MA
wo L
2
M A  0
Ax
Ay=
wo L 2
L =0
2 3
w L2
MA= o
3
MA-
Ay
2
L
3
1
L
3
wo x2
2L
F
y
w  wo
wo L2
3
0
V
2
wo L wo x
-V=0
2
2L
w
x2
V= o (L- )
2
L
x/3
wo L
2
w L
Vmax= o
2
x
Maximum shear force occur at
w x
dV
 o 0
dx
L
x=0
M  0
wo L2 wo L
wo x 2 1
M+
x+
x =0
2
3
2L 3
61
dV
0
dx
M
x
L
M=
wo
(3L2x-x3-2L3)
6L
2
Mmax=

wo L
3
wo
2
wo L
3
L
wo L
2
wo L
2
S.F Diagram

wo L2
3
62
Example 37: The horizontal beam AD is loaded by a uniform distributed load of 5 KN per
meter of length and is also subjected to the concentrated force of 10 KN applied as shown
below. Determine the shearing force and bending moment diagrams.
F
x
0
10 KN
Ax=0
5 KN / m
B
M A  0
A
D
Cy×3-30×2=0
Cy=20 KN
F
y
C
0
2m
Ay+20-30=0
Ay=10 KN
1m
1m
20 KN 10 KN
Ax
 Segment AB
F
0 x2
Ay
Cy
5x
y
0
x/2
5 KN / m
10-5x-V=0
V=10-5x
V
M  0
x
x
M-10x+5x =0
2
x
M=5x(2- )
2
 Segment BC
10 KN
2 x3
5x
F
y
10 KN
x/2
0
5 KN / m
10-5x-10-V=0
V=-5x
V
M  0
x
x
M-10x+5x +10(x-2)=0
2
5
M=20- x2
2
 Segment CD
M
10 KN
3 x  4
63
M
5x
F
10 KN
x/2
y
0
5 KN / m
V
10-5x-10+20-V=0
V=20-5x
M  0
x
20 KN
10 KN
x
M-10x+5x +10(x-2)-20(x-3)=0
2
5
M=-40+20x- x2
2
10 KN
5 KN / m
B
A
D
C
2m
1m
1m
20 KN
10 KN
10 KN
5 KN
S.F Diagram
 10 KN
 15 KN
10 KN .m
B.M Diagram
 2.5 KN .m
`
64
M
Example 38: A beam ABC is simply supported at A and B and has an overhang BC. The
beam is loaded by two forces P and a clockwise couple of moment Pa that act through the
arrangement shown. Draw the shear force and bending moment diagrams for beam ABC.
P
P
Pa
D
C
A
B
a
M
D
a
P
0
a
a
P
-Pa+RC(2a)-Pa=0
RC=P
F
y
0
RD+P-P-P=0
RD=P
M
Pa
A
P
RD
0
RC
RB(2a)-Pa-P(3a)=0
RB=2P
F
y
0
RA+2P-P-P=0
RA=0
0 xa
 Segment AD
F
y
RA
RB
0
V=0
M  0
V
M
M=0
x
P
 Segment DB
V
a  x  2a
x
65
P
F
y
0
-V-P=0
V=-P
M  0
M+P(x-a)=0
M=P(a-x)
 Segment DB
P
V
2a  x  3a
x
 Fy  0
2P
2P-P-V=0
V=P
M  0
M+P(x-a)-2P(x-2a)=0
M=P(x-3a)
P
P
2P
P
S.F. Diagram
 P
B.M. Diagram
 Pa
66
Graphical Method for Constructing Shear and Moment Diagram:
w(x )
dV
  w(x)
dx
Slope of shear diagram at each point=-distributed load intensity at each point.
dM
V
dx
Slope of moment diagram at each point=shear at each point.
 When the force acts downward on the beam, V is negative so the shear will jump
downward. Likewise, if the force acts upward, the jump will be upward.
 If moment Mo is applied clockwise on the beam, M is positive so the moment
diagram will jump upward. Likewise, when Mo acts counterclockwise, the jump will
be downward.
67
Example 39: Draw the shear and moment diagrams for the beam shown below.
P
 Fx  0
Ax=0
A
 Fy  0
Ay-P=0
Ay=P
M
A
L
0
M-PL=0
M=PL
P
M
Ax
Ay
At x=0 V=P
At x=L V=P
P
At x=0 M=-PL
At x=L M=0
 PL
68
Example 40: Draw the shear and moment diagrams for the beam shown below.
15 KN / m
M
A
0
10 KN.m
A
B
By×5.5-10-60×2=0
By=23.63 KN
F
y
0
4m
23.63+Ay-60=0
Ay=36.37 Kn
F
x
15 KN / m
0
Ax=0
0 .5 m
1m
10 KN.m
Ax
Ay
By
4m
0 .5 m
15 KN / m
1m
10 KN.m
23.63 KN
36.37 KN
4m
0 .5 m
1m
36.37 KN
4 x
x
4 x

36.37 23.63
23.63x=36.37×4-36.37x
x=2.4246 m
Maximum bending moment occur
when V=0, at x=2.4246 m
x
 23.63 KN
25.48 KN.m
44.092 KN.m
13.665 KN.m
69
 23.63 KN
23.665 KN.m
Example 41: Draw the shear and moment diagrams for the beam shown below.
48 KN / m
10 KN / m
M
C
0
A
B
80-RB×6+30×7.5+144×4=0
RB=146.83 KN
F
y
6m
3m
0
30 KN
RC+146.83-30-144=0
RC=27.17 KN
C
80 KN .m
144 KN
2m
1.5 m
80 KN.m
RB
RC
48 KN / m
10 KN / m
A
B
146 .83 KN
3m
C
80 KN .m
27.17 KN
6m
116 .83 KN
For segment 3  x  6
V=116.83+4(x-3)2-48(x-3)
Maximum bending moment occur when
V=0
0=116.83+4(x-3)2-48(x-3)
x2-18x+74.2075=0
x=6.39375 m from left end
3.39375 m
 30 KN
 45 KN .m
 125 KN .m
70
47.187 KN .m
 27.17 KN
Example 42: Draw the shear and moment diagrams for the beam shown below.
3 KN / m
F
x
0
A
C
B
Ax=0
3m
M A  0
3m
3m
By×6-9×7=0
By=10.5 KN
F
y
9 KN
0
Ay+10.5-9=0
Ay=-1.5 KN
2m
3 KN / m
Ax
By
Ay
3 KN / m
1.5 KN
10.5 KN
6.75 KN
 1.5 KN
 3.75 KN
 4.5 KN.m
 11.25 KN.m
71
Stresses in Beams:
Pure bending refers to flexure of a beam under a constant bending moment.
Therefore, pure bending occurs only in regions of a beam where the shear force is zero.
Nonuniform bending refers to flexure in the presence of shear forces, which means that the
bending moment changes as we move along the axis of the beam.
P
P
a
a
P
S.F. Diagram
P
Pa
B.M. Diagram
Nonuniform bending
Pure bending
Assumptions:
1.
2.
3.
4.
5.
6.
The beam is initially straight and unstressed.
The material of the beam is perfectly homogeneous.
The elastic limit is nowhere exceeded.
Young's modulus for the material is the same in tension and compression.
Plane cross-sections remain plane before and after bending.
Every cross-section of the beam is symmetrical about the plane of bending i.e. about
an axis perpendicular to the N.A.
7. There is no resultant force perpendicular to any cross-section.
If we now considered a beam initially unstressed and subjected to a constant bending
moment along its length, i.e. pure bending as would be obtained by applying equal couples
at each end, it will bend to a radius  as shown below.
72
As a result of this bending the top fibers of the beam will be subjected to
compression and the bottom to tension. Its reasonable to suppose, that somewhere between
the two there are points at which the stress is zero, these points is termed the neutral axis.
The neutral axis will always pass through the centre of area or centroid.
The length L1 of the line ef after bending takes place is:
L1=(-y)d
d=
dx

L1=(1-
y
)dx

The original length of line ef is dx
Strain(εx)=
L1  original length
=
original length
(1 
y
)dx  dx
y

=dx

εx=-ky
where k is the curvature.
The longitudinal normal strain will vary linearly with y from the neutral axis. A
contraction (-εx) will occur in fibers located above the neutral axis (+y), whereas
elongation (+εx) will occur in fibers located below the neutral axis (-y).
 max
εx
N.A
εx=-(
y
) εmax
c1
By using Hook's law x=Eεx
x=-Eky=-
E
y

73
 max
N.A
x=-(
y
) max
c1
Normal stress will vary linearly with y from the neutral axis. Stress will vary from
zero at the neutral axis to a maximum value max a distance c1 farthest from neutral axis.
N.A.
dF=xdA
M=  ydF =  ( x dA) y
A
A
y
c1
=  (  max ) ydA
A
M=
y
2
 max
c1
y
2
dA
A
dA =I moment of inertia
A
 max 
Mc1
I
max: The maximum normal stress in the member, which occurs at a point on the cross
sectional area farthest away from the neutral axis.
M: The resultant internal moment.
I: The moment of inertia of the cross sectional area computed about the neutral axis.
c1: The perpendicular distance from the neutral axis to a point farthest away from the
neutral axis, where max acts.
74
Mc1
I
M
1=S1
I
S1=
c1
Mc2
I
M
, 2=
S2
I
, S2=
c2
1=-
, 2=
The quantities S1 and S2 are known as the section moduli of the cross sectional
area.
Example 43: A simple beam AB of span length L=22 ft supports a uniform load of
intensity q=1.5 k/ft and a concentrated load P=12 k. The uniform load includes an
allowance for the weight of the beam. The concentrated load acts at a point 9 ft from the
left hand end of the beam. The beam is constructed of glued laminated wood and has a
cross section of width b=8.75 in and height h=27 in. Determine the maximum tensile and
compressive stresses in the beam due to bending.
9 ft
P  12 k
q  1.5 k / ft
B
A
22 ft
75
12 k
33 k
Ay
M
A
By
0
By×22-12×9-33×11=0
By=21.409 k
F
y
0
Ay+21.409-12-33=0
Ay=23.591 k
9 ft
P  12 k
q  1.5 k / ft
23.591 k
21.409 k
23.591 k
10.091 k
 1.909 k
Maximum bending moment
Mmax=151.569 k.ft
=151.569×12
=1818.828 ksi
S.F. Diagram
 21.409 k
151.569 k. ft
B.M. Diagram
76
c1=c2=13.5 in
Mc
1=  1
I
3
bh 8.75  (27) 3
I=
=
 14352.1875 in 4
12
12
1818.828  103  13.5
1= 
14352.1875
=-1710.8317 psi
1818.828  103  13.5
1=
14352.1875
=1710.8317 psi
c1
27 in
c2
8.75 in
Example 44: The simply supported beam has the cross sectional area shown below.
Determine the absolute maximum bending stress in the beam and draw the stress
distribution over the cross section at this location.
20 mm
5 KN / m
A
300 mm
20 mm
20 mm
B
6m
30 KN
250 mm
Ay
By
77
M
A
0
5 KN / m
By×6-30×3=0
By=15 KN
F
y
0
Ay+15-30=0
Ay=15 KN
15 KN
15 KN
15 KN
 15 KN
Maximum bending moment
Mmax=22.5 KN.m
22.5 KN.m
20 mm
1
2
300 mm
N.A
20 mm
c1=c2=170 mm
bh 3
I1 =
 Ad 2
12
3
250 mm
250  10 3  (20  10 3 ) 3
3
3
I1 =
 (250  10  20  10 )  (160  10 3 ) 2
12
I1=128.16667×10-6 m4
I3=I1=128.16667×10-6 m4
I2 =
bh 3 (20  10 3 )  (300  10 3 ) 3
=
=45×10-6 m4
12
12
I= I1+ I2+ I3=128.16667×10-6+128.16667×10-6+45×10-6
I=301.333×10-6 m4
max=
Mc1 22.5  10 3  170  10 3
=
I
301.333  10 6
=12.693598 MPa.
78
20 mm
B
My B
I
22.5  10 3  150  10 3
=
301.333  10 6
B=
=11.200233 MPa.
Example 45: The beam shown below has a cross section of channel shape with width
b=300 mm and height h=80 mm, the web thickness is t=12 mm. Determine the maximum
tensile and compressive stresses in the beam due to uniform load.
300 mm
3.2 KN / m
80 mm
A
B
12 mm
M
A
0
By×3-14.4×2.25=0
By=10.8 KN
F
y
x
Ax=0
14.4 KN
0
Ay+10.8-14.4=0
Ay=3.6 KN
F
1.5 m
3m
0
Ax
Ay
By
3.2 KN / m
79
10.8 KN
3.6 KN
4.8 KN
3.6 KN
 6 KN
M1=2.025 KN.m
M2=3.6 KN.m
2.025 KN.m
 3.6 KN.m
yc 
300 mm
 yA
A
2
3
1
80 mm
12 mm
No. of Area
1
2
3
yc 
A(m2)
960×10-6
3312×10-6
960×10-6
-6
 A =5232×10
321888  10 9
=61.52×10-3 m
6
5232  10
3
y (m)
yA (m )
40×10-3
74×10-3
40×10-3
38400×10-9
245088×10-9
38400×10-9
-9
 yA =321888×10
300 mm
yc=61.52 mm
c 2  18.48 mm
c1  61.52 mm
I1 =
12 mm
bh 3
 Ad 2
12
80
80 mm
I1 =
12  10 3 (80  10 3 ) 3
+960×10-6×(21.52×10-3)2=0.95658×10-6 m4
12
I3= I1=0.95658×10-6 m4
bh 3
 Ad 2
12
276  10 3 (12  10 3 ) 3
I2 =
+3312×10-6×(12.48×10-3)2=0.55558×10-6 m4
12
I2 =
I= I1+ I2+ I3=2.46874×10-6 m4
M 1c 2 2.025  10 3  61.52  10 3
( t )1 

 50.462179 MPa
I
2.46874  10 6
M c
3.6  10 3  18.48  10 3
( t ) 2  2 1 
 26.94815 MPa
I
2.46874  10 6
(t)max=50.462179 MPa
M 1c1
2.025  10 3  18.48  10 3

 15.158339 MPa
I
2.46874  10 6
M 2 c2
3.6  10 3  61.52  10 3
( c ) 2  

 89.71054 MPa
I
2.46874  10 6
( c )1  
(c)max=-89.71054 MPa
Composite Beams:
Composite beams are made from different materials in order to efficiently carry a
load.
Normal stress in material 1 is determined from =E1ε
Normal stress in material 2 is determined from =E2ε
dA=dydz
The force dF acting on the area dA of the beam is
dF=dA=( E1ε)dydz
If the material 1 is being transformed into material 2
b2=nb
81
dF   dA =( E2ε)ndydz
dF= dF
( E1ε)dydz=( E2ε)ndydz
n=
E1
E2
n: transformation factor (modular ratio).
If the material 2 is being transformed into material 1
b1= n b where n =
E2
E1
For the transformed material
=n 
82
Example 46: A composite beam is made of wood and reinforced with a steel strap located
on its bottom side. It has the cross sectional area shown below. If the beam is subjected to a
bending moment of M=2 KN.m determine the normal stress at point B and C. Take Ew=12
GPa and Est=200 GPa.
B
Ew
E st
12
n
 0.06
200
bst  n  bw
bst  0.06  150  9 mm
n
150 mm
20 mm
C
9 mm
150 mm
9 mm
B
2
150 mm
150 mm
20 mm
1
20 mm
C
150 mm
83
150 mm
No. of Area
1
2
A(m2)
3000×10-6
1350×10-6
-6
 A =4350×10
3
y (m)
yA (m )
10×10-3
95×10-3
30000×10-9
128250×10-9
-9
 yA =158250×10
9 mm
yc 
 y A
A
158250  10 9
yc 
 36.379  10 3 m
3
4350  10
 36.379 mm
c1  133.621 mm
150 mm
N. A
I=I1+I2
c 2  36.379 mm
bh 3
 Ad 2
12
150  10 3  (20  10 3 ) 3
I1 
 3000  10 6  (26.379  10 3 ) 2
12
I1 
I1=2.187554×10-6 m4
bh 3
 Ad 2
12
9  10 3  (150  10 3 ) 3
I2 
 1350  10 6  (58.621  10 3 ) 2
12
I2 
I2=7.170419×10-6 m4
I=9.35797×10-6 m4
My
I
2  10 3  133.621  10 3
 B  
 28.557689 MPa
9.35797  10 6
 B  n   B
 B  0.06  (28.557689)  1.71346134 MPa
My
B 
I
2  10 3  36.379  10 3
B 
 7.774976 MPa
9.35797  10 6
 B  
84
20 mm
150 mm
Shear Stresses in Beams
 
V
It
 ydA
A
 ydA  y A = Q
A
 
VQ
It
 :- the shear stress in the member at the point located a distance y  from the neutral axis.
V:-the internal resultant shear force.
I:-the moment of inertia of the entire cross sectional area computed about the neutral axis.
t:-the width of the members cross sectional area, measured at the point where  is to be
determined.
Q  y A , where A is the top (or bottom) portion of the members cross sectional area,
defined from the section where t is measured, and y  is the distance to the centroid of A ,
measured from the neutral axis.
85
Example 47: A metal beam with span L=3 ft is simply supported at points A and B. The
uniform load on the beam is q=160 lb/in. The cross section of the beam is rectangular with
width b=1 in and height h=4 in. Determine the normal stress and shear stress at point C,
which is located 1 in below the top of the beam and 8 in from the right hand support.
q  160 lb / in
M
A
0
C
A
B
By×3×12-5760×1.5×12=0
By=2880 lb
F
y
0
3 ft
Ay+2880-5760=0
Ay=2880 lb
q  160 lb / in
5760 lb
C
2880 lb
2880 lb
By
Ay
2880 lb
8 in
18 in
V
 2880 lb
M max  25.92 k .in
18 in
V
2880

10
18
V=1600 lb
86
At point C x=28 in from left end
from shear force diagram
A
1 in
y
N.A
4 in
1
1
 2880  18   1600  10
2
2
M=17.92 k.in
M 
I
bh 3 1  (4) 3

 5.3333 in 4
12
12
A =1×1=1 in
y  =1.5 in
1 in
2
Q  y A =1.5×1=1.5 in3
C  
My
17920  1

 3.36 ksi
I
5.3333
VQ
It
1600  1.5
 
 450 psi
5.3333  1
 
87
Example 48: Consider the cantilever beam subjected to the concentrated load shown
below. The cross section of the beam is of T-shape. Determine the maximum shearing
stress in the beam and also determine the shearing stress 25 mm from the top surface of the
beam of a section adjacent to the supporting wall.
50 mm
50 KN
2m
M
A
0
A
M-50×2=0
M=100 KN.m
F
y
B
0
Ay-50=0
Ay=50 KN
125 mm
50 KN
2m
M
50 mm
B
200 mm
Ay
50 KN
2m
B
100 KN.m
50 KN
50 KN
S.F. Diagram
B.M. Diagram
100 KN.m
50 mm
From shear and bending moment diagrams
V=50 KN
M=100 KN.m
125 mm
50 mm
2
1
200 mm
88
No. of Area
1
2
yc 
A(m2)
10000×10-6
6250×10-6
-6
 A =16250×10
3
y (m)
yA (m )
25×10-3
112.5×10-3
250000×10-9
703125×10-9
-9
 yA =953125×10
 y A
A
50 mm
953125  10 9
yc 
 58.65  10 3 m
3
16250  10
 58.65 mm
125 mm
c1  116.35 mm
2
N. A
c2  58.65 mm
1
50 mm
I=I1+I2
bh 3
I1 
 Ad 2
12
200  10 3  (50  10 3 ) 3
I1 
 10000  10 6  (33.65  10 3 ) 2
12
200 mm
I1=13.40655833×10-6 m4
bh 3
 Ad 2
12
50  10 3  (125  10 3 ) 3
I2 
 6250  10 6  (53.85  10 3 ) 2
12
I2 
I2=26.26191146×10-6 m4
I=39.6684×10-6 m4
50 mm
Q  y A
-3
-3
A =50×10 ×116.35×10
=0.0058175 m2
-3
y  =58.175×10 m
Q=0.000338433 m3
 max
 max
A
125 mm
c1  116.35 mm
2
y
N. A
50 mm
VQ

It
50  10 3  0.000338433

 8.5315553 MPa
39.6684  10 6  50  10 3
c2  58.65 mm
1
200 mm
89
50 mm
Q  y A
-3
-3
A =50×10 ×25×10
=0.00125 m2
-3
y  =103.85×10 m
Q=0.000129812 m3
 
A
25 mm
c1  116.35 mm
125 mm
y
N. A
50 mm
VQ
It
c2  58.65 mm
1
200 mm
50  10 3  0.000129812
 
 3.272441 MPa
39.6684  10 6  50  10 3
90
Curved Beams
Due to the curvature of the beam, the normal strain in the beam does not vary linearly with depth as in the
case of a straight beam .As result, the neutral axis does not pass through the centroid of the cross section.
C7
If we isolate a differential segment of the beam let a strip material located at r distance has an original length
r dθ .Due to the rotations δθ/2, the strip's total change in length is δθ)R-r(

 ( R  r )

Rr
,k 
   k(
)
rd
d
r
Strain is a nonlinear function of r, in fact it varies in a hyperbolic fashion .Hooke's law applies,
  Ek (
F
R
Rr
)
r
0
 dA  0
A
  Ek (
A
dA
  dA  0
r
A
A
Rr
) dA  0
r
R
91
R
A
dA
A r
R-:The location of the neutral axis, specified from the center of curvature 0' of the member.
A-:The cross -sectional area of the member.
r -:The arbitrary position of the area element dAon the cross section, specified from the center of
curvature 0' of the member.
 
My
Ae( R  y )
y=R-r , e= r -R
σ -:The normal stress in the member.
M -:The internal moment, determined from the method of sections equations of equilibrium and computed about
the centroidal axis .
A -:The cross-sectional area of the member.
R -:The distance measured from the center of curvature to the neutralaxis.
r -:The distance measured from the center of curvature to the centroidof the cross-sectional area.
r -:The distance measured from the center of curvature to the point where the stress σis to be determined.
92
o 
M ( ro  R )
Aro ( r  R )
i 
M ( R  ri ) Normal stress at the bar's bottom.
Ari ( r  R )
Normal stress at the bar's top.
Example:-The curved bar has a cross-sectional area shown below .If it issubjected to bending moments of 4
kN • m, determine the maximum normal stress developed in the bar.
Area
A(mm2)
y-(mm)
y-A(mm3)
rectangle
2500
225
562500
triangle
750
260
195000
3250
r
R
757500
 yA  757500  233.076 mm
 A 3250
A
dA
A r
 A  3250 mm

r
br2
r
dA
 b ln 2 
ln 2  b
R
r1 (r2  r1 ) r1
= 50 ln
R
250
50  280
280

ln
 50  14.04389 mm
200 (280  250) 250
3250
 231.417 mm
14.04389
93
B  
M ( R  ri )
4  10 3 (231.417  10 3  200  10 3 )

 116.5373 MPa
Ari (r  R )
3250  10 6  200  10 3 (233.076  10 3  231.417  10 3 )
M (ro  R )
4  10 3 (280  10 3  231.417  10 3 )

 128.7231 MPa The
Aro (r  R ) 3250  10 6  280  10 3 (233.076  10 3  231.417  10 3 )
maximum stress at point A= 128.7231 MPa
A 
Example:-The frame of a punch press is shown below. Find the stresses at the inner and outer surface at
section x-x of the frame if W=5000 N.
W
x
x
25 mm
50 mm
W
18 mm 100 mm
6 mm
40 mm
A
bi  bo
h
2
18  6
A(
)  40  480 mm 2
2

bi
bo
h
b r  bo ri
r
dA
18  65  6  25
65
( i o
) ln o  (bi  bo )  (
) ln  (18  6)  12.365 mm
R
h
ri
40
25
R
A
480

 38.8175 mm
dA 12.365
A r
6 mm
18 mm
40 mm
65 mm
94
25 mm
Area(mm2)
A(mm2)
y-(mm)
y-A(mm3)
rectangle
720
45
32400
Triangle
-120
51.666
-6200
Triangle
-120
51.666
-6200
480
r
20000
 yA  20000  41.666 mm
 A 480
M=W×d=5000×(100×10-3+41.666×10-3)=708.33 N.m
M ( R  ri ) W
708.33(38.8175  10 3  25  10 3 )
5000
i 



6
3
3
3
Ari ( r  R ) A 480  10  25  10 ( 41.666  10  38.8175  10 ) 480  10 6
 296.747 MPa
o  
M ( ro  R ) W
708.33(65  10 3  38.8175  10 3 )
5000



6
3
3
3
Aro ( r  R ) A
480  10  65  10 ( 41.666  10  38.8175  10 ) 480  10 6
 198.26 MPa
95
Slop and Deflection in Beams
The elastic curve :-the deflection diagram of the longitudinal axis that passes
through the centroid of each cross sectional area of the beam.
96
x-axis extends positive to the right.
v-axis extends positive upward from the x-axis.
1



y


E
My
 
I
1 M

 EI
When M is positive,  extends above the beam, i.e.  in the positive v direction.
When M is negative,  extends below the beam, or in the negative v direction.
Integration Method
The elastic curve for a beam can be expressed mathematically as v=f(x)
1
d 2 v / dx 2

 [1  (dv / dx) 2 ]3 / 2
M
d 2 v / dx 2

EI [1  (dv / dx) 2 ]3 / 2
The slop of the elastic curve which is determined from dv/dx will be very small, and
its square will be negligible compared with unity.
1 d 2v

 dx 2
M d 2v

EI dx 2
97
V 
dM
dx
d
d 2v
( EI 2 )
dx
dx
dV
w
dx
d2
d 2v
 w( x)  2 ( EI 2 )
dx
dx
V ( x) 
EI always positive quantity
d 2v
M  EI ( 2 )
dx
d 3v
V  EI ( 3 )
dx
d 4v
 w  EI ( 4 )
dx
Sign convention and coordinates
Positive deflection v is upward, the positive slope  will be measured
counterclockwise from the x-axis when x is positive to the right.
If positive x is directed to the left, then  will be positive clockwise.

dv
dx
98
Boundary conditions
dv
 Mdx  C1
dx 
EIv   [  Mdx ]dx  C1 x  C 2
EI
Example: The cantilevered beam shown is subjected to a vertical load P at its end.
Determine the equation of elastic curve. EI is constant.
M  0
M+Px=0
M=-Px
d 2v
) =-Px
dx 2
dv
1
EI
  Px 2  C1 ……………………..(1)
dx
2
1
EIv   Px 3  C1 x  C 2 ………………………(2)
6
M  EI (
Boundary conditions
99
at x=L

dv
=0
dx
and
v =0
1
0   PL2  C1
2
1
1
0   PL3  PL3  C 2
6
2
C1 
1 2
PL
2
C2 
1
1
EI   Px 2  PL2
2
2

PL3
1 3 1 3
PL  PL = 
3
6
2
P
( L2  x 2 )
2 EI
1
PL2
PL3
EIv   Px 3 
x
6
2
3
v
P
( x 3  3L2 x  2 L3 )
6 EI
Example: The simply supported beam shown supports the triangular distributed loading.
Determine the maximum deflection. EI is constant.
Due to symmetry we take
M  0
0≤ x ≤
L
2
w L
w x3
L
x3
x  0
M  w ( x  )
4
3L
4
3L
2
3
d v
L
x
M  EI ( 2 ) = w ( x  )
4
3L
dx
4
w
L
w
x
dv
EI
  x 2    C1 ………………….(1)
dx
8
12 L
M
w L 3 w x 5
EIv 
x 
 C1 x  C 2 ……..………..(2)
24
60 L
100
Boundary conditions
at x=
L
2

dv
=0
dx
w L 2 w x 4
0
x 
 C1
8
12 L
at x=0 v=0
0=0-0+0+C2
5w L3
C1  
192
C2=0
w L 3 x 5
5 L3
( x 

x)
EI 24
60 L 192
L
Vmax at x=
2
v
v max 
w L4
w L4
L4
5 L4
(


) = 
EI 192 1920 384
120 EI
Discontinuity Method
101
xa

n
for x  a
0

n
( x  a)
x  a dx 
n
xa
for x  a
n 1
n 1
C
Example:- Find the moment expression using continuity equations.
3
2
1
6
M=2.75<x-0>1+1.5<x-3>0- <x-3>2- <x-3>3
3
2
1
6
=2.75x+1.5<x-3>0- <x-3>2- <x-3>3
Example:- Determine the equation of the elastic curve for the beam shown below. EI is
constant.
F
y
0
Ay-40-12=0
M
A
Ay=52 kN
0
MA-40×2.5-50-12×9=0
MA=258 kN.m
102
8
2
8
2
M=-258<x-0>0+52<x-0>1- <x-0>2+50<x-5>0+ <x-5>2
EI
d 2v
=-258+52x-4x2+50<x-5>0+4<x-5>2
2
dx
dv
4
4
=-258x+26x2- x3+50<x-5>1+ <x-5>3+C1………….….(1)
3
3
dx
26
1
1
EIv=-129x2+ x3- x4+25<x-5>2+ <x-5>4+C1x+C2………….(2)
3
3
3
EI
B.C
dv
=0 at x=0 in eq.(1)
dx
v=0 at x=0
C1=0
C2=0
v=
in eq.(2)
1
26
1
1
(-129x2+ x3- x4+25<x-5>2+ <x-5>4)
EI
3
3
3
Moment Area Method
B
M
dx
EI
A
B/ A  
The notation  B / A is referred to as the angle of the tangent at B measured with respect to
the tangent at A.
103
Theorem 1 The angle between the tangents at any two points on the elastic curve equals
the area under the M/EI diagram between these two points.
If the area under M/EI diagram is positive, the angle is measured counterclockwise
from the tangent A to tangent B.
If the area under M/EI diagram is negative, the angle  B / A is measured clockwise from
tangent A to tangent B.  B / A will measured in radians.
B
tA/ B   x
A
M
dx
EI
t A / B : the vertical deviation of the tangent at A with respect to the tangent at B.
 xdA  x  dA
B
M
 EI dx
represents the area under the M/EI diagram, we can also write:-
A
B
M
dx
EI
A
tA/ B  x
x is the distance from A to the centroid of the area under the M/EI diagram between A
and B.
104
Theorem 2 The vertical deviation of the tangent at a point A on the elastic curve with
respect to the tangent extended from another point B equals the moment of the area under
the M/EI diagram between these two points. This moment is computed about point A
where the vertical deviation t A / B is to be determined.
Example: Determine the slope of the beam shown at points B and C. EI is constant
 B = B / A
C =C / A
P
MA
M
A
0
Ay
MA-PL=0
MA=PL
F
y
0
Ay=P
B
M
dx Area under the M/EI diagram from A to B
EI
A
B/ A  
105
  PL  L  1   PL PL  L 
 B = B / A  

   
 
2 EI  2 
 2 EI  2  2  EI
 3PL2
=
rad clockwise
8 EI
2
  PL  1  PL
C =C / A = 
=


L
2 EI
 EI  2
rad
clockwise
Example: Determine the displacement of points B and C of the beam shown. EI is
constant.
Mo
vB= t B / A
vC= t C / A
F
0
y
Ay=0
M
A
0
M-Mo=0
M=Mo
B
M
dx
EI
A
vB= t B / A  x 
1  L   M o  L   M o L2
=  
  =
2  2  EI  2 
8EI
 M o L2
M
 L   M o 
dx =  
 L  =
EI
2 EI
 2  EI 
C
A
vC= t C / A  x 
106
Mo
Example: Determine the slope at point C for the steel beam shown. Take Est=200 GPa,
I=17×106 mm4
L
C
C =  A  C / A
tB / A
L
1  24   1  2  24   1 

t B / A =  2  (6)  (6)   (2) (2) 
3  EI   2  3  EI   2 

Since the angle is very small  A =tan  A =
320
EI
 8  1 8
 C / A =  ( 2)  
 EI   2  EI
320 8 32
C =
- =
8 EI EI EI
tB / A =
C =
32
=0.009411 rad
200  10  17  10 6
9
Castigliano’s Theorem Applied to Beams
L
  M(
0
M dx
)
P EI
Where:=displacement of the point caused by the real loads acting on the beam.
P=external force of variable magnitude applied to the beam in the direction of .
M=internal moment in the beam, expressed as a function of x and caused by both the force
P and the load on the beam.
L
  M(
0
M dx
)
M EI
 =the slope of the tangent at a point on the elastic curve.
107
M =an external couple moment acting at the point.
Example: Determine the displacement of point B on the beam shown below. EI is
constant.
M  0
x
M  wx ( )  Px  0
2
x2
M   w  Px
2
M
 x
P
When P=0
x 2 M
M  w
,
 x
2
P
 wx 2
( x)
L
L
L
M dx
w
B   M (
)
 2
dx 
x 3 dx

P EI 0
EI
2 EI 0
0
L
B 
w 4
wL4
x =
8EI
8EI
0
Example: Determine the displacement of point A of the steel beam shown below.
I=450 in4, Est=29×103 ksi.
P
Cx
M
By
B
0
Cy×20-60×10+15×
10
+P×10=0
3
Cy =27.5-0.5P
F
y
0
By+27.5-0.5P-60-15-P=0
108
Cy
3 2
x1
20
P
By=47.5+1.5P
F
x
0
Cx=0
V1
M  0
M1
x1
x
3
M1+ x12 ( 1 ) +Px1=0
20
3
3
x
M1=- 1 - Px1
20
M 1
=-x1
P
P
3x 2
15 kip
When P=0
M1=-
x13
20
V2 M 2
x2
M 1
=-x1
P
M  0
M2+3x2(
47.5  1.5 P
x2
10
)+15×( +x2)+P(10+x2)-( 47.5+1.5P)x2=0
2
3
3
2
M2=  x 22 +(32.5+0.5P)x2-10P-50
M 2
=0.5x2-10
P
When P=0
M 2
=0.5x2-10
P
L
M 1 dx1 L
M 2 dx 2
 A   M1(
)
 M2(
)
P EI 0
P EI
0
3
2
M2=  x 22 +32.5x2-50 ,
A 
10
20
x13
(12) 3
(12) 3
3

(

x
)
dx

( x 22  32.5 x 2  50)(0.5 x 2  10)dx 2
1
1


EI 0 20
EI 0 2
(12) 3
A 
EI
 x 5 10  3
1

x 24  10.4166 x 23  175 x 22  500 x 2

16
100 0
(12) 3
(1000  6667.2)
EI
 A  0.75 in
A 
109
20
0



Example: Determine the slope at point B of the A-36 steel beam shown below.I=70×106
mm4 and E=200 GPa.
2 KN / m
C
B
A
10 m
5m
10 KN
C
M
Ay
M
A
By
0
V1
By×10+ M -10×12.5=0
By=12.5-
F
y
x1
M
10
0
M
 2 .5
10
M
Ay+12.5- -10=0
10
M
Ay= -2.5
10
M  0
M
M1-( -2.5)x1=0
10
M
M1= ( -2.5)x1
10
M 1 1
 x1
M 10
When M =0
M 1 1
 x1
M1= -2.5x1 ,
M 10
M  0
M2+2x2(
M1
2x 2
M
V2 M 2
x2
M
 2 .5
10
x2
M
M
)-(12.5- )x2-( -2.5)(10+x2)+ M =0
2
10
10
110
12.5 
M
10
M
M
)x2+( -2.5)(10+x2)- x 22 - M
10
10
M 2
1
1
=  x1 +1+ x1 -1=0
M
10
10
M2=(12.5-
When M =0
M 2
=0
M
L
10
M dx 5
M dx
M dx
B   M (
)
=  M1( 1 ) 1   M 2 ( 2 ) 2
M EI 0
M EI 0
M EI
0
M2=- x 22 +10x2-25 ,
x
(2.5 x1 )( 1 )dx1
10
0
=
EI
0
10
=
10
1
70  10  10
6
12
 200  10
-3
  0.25x
2
1
6
dx1
0
3 10
1 0
=0.07142×10 (-0.0833 x
)
 B  0.00595 rad
=-0.3410
Statically Indeterminate Beams
1. Method of Integration:
Example: The beam is subjected to the distributed loading shown. Determine the reaction
at A. EI is constant.
111
M  0
wo x 3
0
6L
w x3
M  Ay x  o
6L
2
d v
M  EI ( 2 )
dx
wo x 3
d 2v
Ay x 
 EI ( 2 )
6L
dx
2
Ay x
w x4
dv
 o  C1
EI ( ) 
2
24 L
dx
3
5
Ay x
w x
 o  C1 x  C 2
EIv 
6
120 L
M  Ay x 
Boundary Conditions
at x=0 v=0 , at x=L
0=0-0+0+C2
2
Ay L
dv
 0 , at x=L v=0
dx
C2=0
3
wo L
 C1  0 …………………(1)
2
24
Ay L3 wo L4

 C1 L  0 ………………..(2)
6
120

From equations (1) and (2)
3
wo L
24
w L
Ay  o
10
C1= 
Example: The beam shown below is fixed supported at both ends and is subjected to the
uniform loading shown. Determine the reactions at the supports. Neglect the effect of axial
load.
M
112
F
y
0
wL
2
MA=MB= M 
M  0
VA=VB=
wx 2 wL

x0
2
2
wx 2 wL
M 

xM
2
2
d 2v
M  EI ( 2 )
dx
2
d v
wx 2 wL
EI ( 2 )  

xM
2
2
dx
wx 3 wL 2
dv

x  M x  C1
EI ( )  
6
4
dx
wx 4 wL 3 M  2
EIv  

x 
x  C1 x  C 2
24
12
2
M  M
Boundary Conditions
at x=0 v=0 , at x=0
0=-0+0-0+0+C2
0=-0+0-0+C1
dv
 0 , at x=L v=0
dx
C2=0
C1=0
wL4 wL4 M  2


L
24
12
2
wL2
M
12
0= 
113
Moment Area Method(Statically Indeterminate):
Since application of the moment area theorems requires calculation of both the area
under the M/EI diagram and the centroidal location of this area, it is often convenient to
use separate M/EI diagrams for each of the known loads and redundant rather than using
the resultant diagram to compute these geometric quantities.
114
Example: The beam is subjected to the concentrated loading shown. Determine the
reactions of the supports. EI is constant.
From the elastic curve  B  0 , tB/A=0
Using superposition method to draw the separate M/EI diagrams for the redundant reaction
By and the load P.
A
B
A
For load P
For redundant reaction By
1 By L
2
PL
L 1 PL
2
(
)( L)( L)  (
)( L)( )  (
)( L)( L)  0
2 EI
3
EI
2
2 EI
3
B y  2 .5 P
F
y
0
-Ay-P+2.5P=0
M
A
B
Ay=1.5P
0
MA=0.5PL
115
Example: The beam is subjected to the couple moment at it end C as shown below.
Determine the reaction at B. EI is constant.
From the elastic curve
tC / A t B / A

 t C / A  2t B / A
2L
L
Using superposition method to draw the separate M/EI diagrams for the redundant reaction
By and the load Mo.
M
EI
L
For redundant reaction By
2L
For load Mo
3
By L
M
M
1
1
1
1
L
t B / A  ( L)[ ( L)(
)]  ( L)[ ( L)( o )] 
(By  o )
3
2
2 EI
3
2
2 EI
12 EI
L
B
L
B
L
M
M
1
1
2
1
2
1
L3
y
y
t C / A  ( L  L)[ ( L)(
)]  ( L)[ ( L)(
)]  ( L)[ (2 L)( o )] 
(6 B y  8 o )
3
2
2 EI
3
2
2 EI
3
2
EI
12 EI
L
3
3
3 Mo
M
M
L
L
By 
(6 B y  8 o ) =2
( By  o )
2 L
12 EI
L
12 EI
L
M
5M o
Ay  o , C y 
4L
L
116
Combined Stresses
There are three types of loading: axial, torsional
Axial loading  a 
and flexural.
P
A
T.r
J
M .y
Flexural loading  f 
I
Torsional loading  
There are four possible combinations of these loadings:
1. Axial and flexural.
2. Axial and torsional.
3. Torsional and flexural.
4. Axial , torsional and flexural.
B
f
M .y
I
a
BB
P
A
A
a 
y
y
A
A
f 
Q
P
A
Q
a
A
For point A
P
yy
yy
BB
P
A
f
 A   a  f
for point B  A   a   f
117
yy
yy
P
Example: The bent steel bar shown is 200 mm square. Determine the normal stresses at A
and B.
500 kN
200
mm
4
3
A
4
B
3
M
C
0
250 mm
-500×200×10-3+R1×900×10-3=0
R1=111.111 kN
F
y
450
mm
100
mm
0
R2+111.111cos(53.1301)-500sin(53.1301)=0
R2=333.333kN
500 kN
4
3
750
mm
A
R1
C
4
3
R3
 Fx  0
150 mm
R2
R3+111.111sin(53.1301)-500cos(53.1301)=0
R3=388.888kN
A  
P
500

 12.5 MPa
A
(200  200  10 6 )
M=-500×200×10-3+111.111×700×10-3
M=-22.2223 kN.m
M .y
22.2223  100  10 3

 16.666 MPa
I
(200  10 3 )  (200  10 3 ) 3 / 12
 A   a   f  12.5  16.666  29.166 MPa
f 
 B   a   f  12.5  16.666  4.166 MPa
118
B
53.1301
Stresses at a Point
General State of Stress
x  (
y (
 x  y
2
 x  y
 x y  (
2
)(
)(
 x  y
2
 x  y
2
 x  y
2
Plane Stress
) cos 2   xy sin 2
……………………..(1)
) cos 2   xy sin 2
…………………….(2)
) sin 2   xy cos 2
…………………….(3)
The planes defining maximum or minimum normal stresses are found from:
tan 2 p 
2 xy
 x  y
………………(4)
The planes of maximum shearing stresses are defined by :
tan 2 s  
 x  y
2 xy
………………(5)
The planes of zero shearing stresses may be determined by setting τ equal to zero.
tan 2 
2 xy
 x  y
……………….(6)
Equation 6 and 4 show that maximum and minimum normal stresses occur on planes of
zero shearing stresses.
The maximum and minimum normal stresses are called the principal stresses.
119
Equation 5 is the negative reciprocal of equation 4. This means that the values of 2θ s from
equation 5 and equation 4 differ by 90ο. This means that the planes of maximum shearing
stress are at 45o with the planes of principal stress.
 max .  (
 x  y
2
min .
 max   (
 avg 
) (
 x  y
2
 x  y
) 2   xy
2
) 2   xy
2
2
…………………(7)
…………………..(8)
 x  y
………………….(9)
2
Example:The state of plane stress at a point is represented by the element shown.
Determine the state of stress at the point on another element oriented 30o clockwise from
the position shown.
 x  80 MPa
 y  50 MPa
 xy  25 MPa
50 MPa
θ=-30o
x  (
x
x
y
y
y
 x  y
xy
 x  y
80 MPa
) cos 2   xy sin 2
2
2
 80  50
 80  50
(
)(
) cos( 60)  25 sin( 60)
2
2
 25.849 MPa
 x  y
 x  y
(
)(
) cos 2   xy sin 2
2
2
 80  50
 80  50
(
)(
) cos( 60)  25 sin( 60)
2
2
 4.15 MPa
 x y  (
xy
)(
25 MPa
 x  y
) sin 2   xy cos 2
2
 80  50
 (
) sin( 60)  25 cos( 60)
2
 68.791 MPa
y
4.15 MPa
25 .849 MPa
x
68 .791 MPa
120
Example: When the torsional loading T is applied to the bar shown it produce a state of
pure shear stress in the material. Determine a) the maximum in plane shear stress and the
associated average normal stress. b) the principal stresses.
x  0
y 0
 xy  60 MPa
a)  max   (
 x  y
2
) 2   xy
2
60 MPa
00 2
)  ( 60) 2  60 MPa
2
 x  y 0  0
 avg 

0
2
2
 x  y
tan 2 s  
2 xy
 max   (
tan 2 s  
0
0
2  60
θs=0
b)  max .  (
 x  y
min .
 max .  (
min .
2
) (
 x  y
2
) 2   xy
2
00
00 2
) (
)  (60) 2  60 MPa
2
2
σmax=60 MPa
σmin=-60 MPa
tan 2 p 
2 xy
 x  y
2  60

0


o
2 p  
  p   45  or 135
2
4
 x  y
 x  y
x  (
)(
) cos 2   xy sin 2
2
2
0
0
 x  ( )  ( ) cos(90)  60 sin(90)  60 MPa
2
2
tan 2 p 
60 MPa
60 MPa
121
Example: The state of plane stress at a point on a body is shown on the element. Represent
this stress state in terms of the principal stresses.
 x  20 MPa
 y  90 MPa
 xMPa
 y
 x  y 2
  60
2
 maxxy.  (
) (
)   xy
2
2
min .
 20  90
 20  90 2
) (
)  (60) 2
2
2
 35  81.394
 max .  (
min .
 max .
min .
σmax=116.394 MPa
σmin=-46.394 MPa
tan 2 p 
2 xy
 x  y
=
2  60
 1.090909
 20  90
o
2 p  47.489 
 p  23.744 or 66.256
x  (
 x  y
 x  y
) cos 2   xy sin 2
2
2
 20  90
 20  90
x  (
)(
) cos( 47.489)  60 sin( 47.489)  46.349 MPa
2
2
)(
θp1=66.256o
θp2=-23.744o
122
Example: A sign of dimensions 2 m×1.2 m is supported by a hollow circular pole having
outer diameter 220 mm and inner diameter 180 mm as shown below. The sign is offset 0.5
m from the center line of the pole and its lower edge is 6 m above the ground. Determine
the principal stresses and maximum shear stresses at points A and B at the base of the pole
due to wind pressure of 20 kPa against the sign.
w=PA=2×(2×1.2)=4.8 kN
T=wr=4.8×(1+0.5)=7.2 kN.m
M=wd=4.8×(6+0.6)=31.68 kN.m
V=w=4.8 kN
I


[d 24  d 14 ]  [(220  10 3 ) 4  (180  10 3 ) 4 ]
64
64
I=63.46×10-6 m4
220  10 3
31.68 
M .y
2
A 

 54.91 MPa
I
36.46  10 6
220  10 3
7 .2 
T .r
2
1 

 6.24 MPa

J
3 4
3 4
[(220  10 )  (180  10 ) ]
32
VQ


2 
, t  2( r2  r1 ) , I  ( r24  r14 ) , A  ( r22  r12 )
It
4
2
4r 
4r 
( 2 )( r22 )  ( 1 )( r12 )
y1 A1  y 2 A2
4 r23  r13
3 2
y
 3 2

[
]
 2  2
A1  A2
3 r22  r12
r2  r1
2
2
4 r23  r13  2
2
Q  yA 
[ 2
]  (r2  r12 )  (r23  r13 )
2
3 r2  r1
2
3
2
V (r23  r13 )
4V (r22  r2 r1  r12 )
3
2 

 4
3 (r22  r12 )(r22  r12 )
4
(r2  r1 )  2(r2  r1 )
4
τ2=0.76 MPa
54.91 MPa
Point A
σ1=55.7 MPa
σ2=-0.7 MPa
τmax=28.2 MPa
Point B
σ1=7 MPa
σ2=-7 MPa
τmax=7 MPa
7 MPa
6.24 MPa
123
Mohr's Circle
For plane stresses transformation have a graphical solution that is often convenient to use
and easy to remember. Furthermore this approach will allow us to visualize how the
normal and shear stress components  x and  x y vary as the plane on which they act is
oriented in different directions. This graphical solution known as Mohr's circle.
x  (
 x  y
2
 x y  (
)(
 x  y
2
 x  y
2
) cos 2   xy sin 2
) sin 2   xy cos 2
The parameter θ can be eliminated by squaring each equation and adding the equations
together. The result is:
[ x  (
 x  y
Let c 
)]2   x y  (
2
 x  y
2
, R2=  (
2
2
[ x  c]   x y  R 2
2
 x  y
2
 x  y
2
) 2   xy
) 2   xy
2
2
this equation represents a circle having a radius R and center at point
(c,0).
Construction of the circle
1. Establish a coordinate system such that the abscissa represents the normal stress σ
with positive to the right and the ordinate represents the shear stress τ with positive
down ward.
2. Using the positive sign convention for σxσyτxy as shown:
124
Plot the center of the circle C which is located on the σ axis at a distance  avg 
 x  y
2
from the origin.
3. Plot the reference point A having coordinate A(  x ,  xy ). This point represents the
normal and shear stress components on the element's right hand vertical face, and
since the x axis coincides with the x axis, this represents θ=0.
4. Connect point A with the center C of the circle and determine CA by trigonometry.
This distance represents the radius R of the circle.
5. Once R has been determined , sketch the circle.
Principal Stresses
The principal stresses σ1 and σ2 (σ1≥σ2) are represented by the two points B and D where
the circle intersects the σ axis i.e where τ=0.
These stresses act on planes defined by angles θp1 and θp2. They are represented on the
circle by angles 2θp1 and 2θp2and are measured from the radial reference line CA to line CB
and CD respectively.
Using trigonometry only one of these angles needs to be calculated from the circle since θp1
and θp2 are 90o apart.
125
Maximum in Plane Shear Stress.
The average normal stress and maximum in plane shear stress components are determined
from the circle as the coordinates of either point E or F. In this case the angles θs1 and θs2
give the orientation of the planes that contain these components. The angle 2θ s1 can be
determined using trigonometry.
Stress on Arbitrary Plane.
The normal and shear stress components  x and  x y acting on a specified plane defined by
the angle θ can be obtained from the circle using trigonometry to determine the coordinates
of point P.
To locate P, the known angle θ for the plane must be measured on the circle in the same
direction 2θ from the radial reference line CA to the radial line CP.
Example: Due to the applied loading the element at point A on the solid cylinder is
subjected to the state of stress shown. Determine the principal stresses acting at this point.
 x  -12 ksi
y 0
 xy  -6 ksi
c
 x  y
R (

2
 x  y
2
 12  0
 6 ksi
2
) 2   xy
2
 12  0 2
)  (6) 2 =8.485 ksi
2
1  c  R
 1  -6+8.485=2.485 ksi
2  c  R
 2  -6-8.485=-14.485 ksi
6
o
2 p 2  tan 1 (
)  tan 1 (1)  45
12  6
o
 p 2  22.5
R (
126
Example: An element in plane stress at the surface of a large machine is subjected to
stresses shown below. Using Mohr's circle determine the following quantities a) the stress
acting on element inclined at an angle 40o b)the principal stresses and c) the maximum
shear stress.
 x  15ksi
5 ksi
 y  5 ksi
4 ksi
 xy  4ksi
15 ksi
c
 x  y
R (

2
 x  y
2
15  5
 10 ksi
2
) 2   xy
2
F
15  5 2
R (
)  (4) 2 =6.403 ksi
2
P
R  6.403 ksi
  41.34  B
D
A(15,4)
C
4
o
2 p1  sin (
)  38.66
6.403

 p1  19.33
1
1  c  R
 1  10+6.403=16.403 ksi
2  c  R
 2  10-8.485=3.597ksi
2 p1  38.66 
4 ksi
10 ksi
P
A
15 ksi
3.597 ksi
 p 2  109.33
16.403 ksi
 p1  19.33
 x  10  6.403 cos(41.34)  14.807 ksi
 y  10  6.403cos(41.34)  4.807 ksi
 x y  6.403sin(41.34)  4.23 ksi
4.23 ksi
4.807 ksi
14.807 ksi
  40 
127
 max  R  6.403 ksi
 avg  10 ksi
 avg  c  10 ksi
o
2 s1  38.66+90=128.66
o
 s1  64.33
counterclockwise
 s1  64.33
 max  6.403 ksi
Stresses Due to Axial Load and Torsion
P
A
Tc

J


P
T

 1 , 2
from Mohr’s circle or from stress transformation equations.
Example: An axial force of 900 N and a torque of 2.5 N.m are applied to the shaft as
shown. If the shaft has a diameter of 40 mm, determine the principal stresses at appoint P
on its surface.
T r 2.5  (20  10 3 )
=
=198.94367 kPa

J
3 4
(20  10 )
2
900
P
=716.19724 kPa
 =
A  (20  10 3 ) 2

 x  0 ,  y  716.19724 kPa ,   198.94367 kPa
128
c
 x  y
R (

2
 x  y
2
716.19724
 358.09862 kPa
2
) 2   xy  (
2
0  716.19724 2
)  (198.94367) 2  409.65 kPa
2
 1  c  R = 358 .09862 + 409 .65 =767.74862 kPa
 2  c  R = 358 .09862 - 409 .65 =-51.55138 kPa
198.94367
)  29.054
409.65
 14.527 clockwise
2 p 2  sin 1 (
 p2
Example: The beam shown below is subjected to the distributed loading of w=120 kN/m.
Determine the principal stresses in the beam at point P, which lies at the top of the web.
Neglect the size of the fillets and stress concentrations at this point. I=67.4×10-6 m4.
Ax=0
∑MB=0
Ay×2-240×1=0
By=120 kN
-V-36+120=0
V=84 kN
Ay=120 kN
129
M-120×0.3+36×0.15=0
M=30.6 kN.m
 
My
30.6  10 3  100  10 3

 45.4 MPa
I
67.4  10 6
175 mm
15 mm
y
P
VQ
It
-3
-3
-3
3
Q  y A =(107.5×10 )×(175×10 ×15×10 )=0.000282187 m
84  0.000282187
 
 35.168 MPa
67.4  10  6  10  10 3

 x  45.4 MPa  y  0   35.168 MPa
c
x  y
R (

2
 x  y
2
- 45.4
 22.7 MPa
2
) 2   xy  (
2
0  45.4 2
)  (35.168) 2  41.857 MPa
2
 1  c  R =  22.7 + 41.857 =19.157 MPa
 2  c  R =  22.7 - 41.857 =-64.557 MPa
35.168
)  57.16
41.857
 28.58 counterclockwise
2 p 2  sin 1 (
 p2
Strain at a Point
130
N. A
Plane Strain
x 
x y
y 
2
x y

x y
2
x y
cos 2 
 xy
2
 xy
sin 2

cos 2 
sin 2
2
2
2
xy
x y
 xy
 (
) sin 2 
cos 2
2
2
2
 xy
tan 2 p 
x y
 1, 2 
x y
2
tan 2 s  (
 (
x y
x y
 xy
2
)2  (
 xy
2
)2
Principal strain
)
x y 2
 xy
 max
 (
)  ( )2
2
2
2
x y
 avg 
2
Maximum in plane shear strain
Example:A differential element of material at a point is subjected to a state of plane strain
-6
-6
-6
 x =500×10 ,  y =-300×10 ,  xy =200×10 , which tends to distort the element as shown
below. Determine the equivalent strains acting on an element oriented at the point
clockwise 30o from the original position.
  30o
x y x y
 xy
x 

cos 2 
sin 2
2
2
2
131
500  10 6  300  10 6 500  10 6  300  10 6
200  10 6

cos(60) 
sin(60)
2
2
2
 213  10 6
x y x y
 xy


cos 2 
sin 2
2
2
2
500  10 6  300  10 6 500  10 6  300  10 6
200  10 6


cos(60) 
sin(60)
2
2
2
 13.4  10 6
x 
x
y
y
y
xy
x y
) sin 2 
 xy
cos 2
2
2
xy
500  10 6  300  10 6
200  10 6
 (
) sin( 60) 
cos(60)
2
2
2
 x y  793 106
2
 (
Mohr's Circle – Plane Strain
C=
x y
R= (
2
x y
2
Point A (  x ,
)2  (
 xy
2
 xy
2
)2
)
 The principal strain  1 and  2 are determined from the circle as the coordinates of
points B and D.
 The average normal strain and the maximum in plane shear strain are determined
from the circle as the coordinates of points E and F.
132
 The normal and shear strain components  x and  x y for a plane specified at angle 
can be obtained from the circle using trigonometry to determine the coordinates of
point P.
Example:The state of plane strain at a point is represented on an element having
components  x =-300×10-6 ,  y =-100×10-6,  xy =100×10-6. Determine the state of strain on
an element oriented 20o clockwise from this reported position.
x y
 300  10 6  100  10 6
 200  10 6
2
2
 

 300  10 6  100  10 6 2 100  10 6 2
R= ( x y ) 2  ( xy ) 2  (
) (
)
2
2
2
2
R  111.8 106
C=

Point A(-300×10-6,50×10-6)
50
)  26.56
111.8
  40  26.56  13.44
 x  (C  R cos )
  sin 1 (
 x  (200  10 6  111.8  10 6 cos(13.44))
 x  (309  10 6 )
xy
  R sin 
2
xy
 111.8  10 6 sin(13.44)
2
 x y  52  10 6
 y  (C  R cos )
 y  (200  10 6  111.8  10 6 cos(13.44))
 y  (91.3  10 6 )
133
Theories of Failure:1. Ductile Materials
a) Maximum Shear Stress Theory
1   Y
 2  Y
 1 , 2 have same signs. (Rankine)
1   2   Y
 1 , 2 have opposite signs.(Guest-Tresca)
b) Maximum Principal Strain Theory
 1   2   3   Y (Saint-Venant)
c) Maximum Shear Strain Energy Per Unit Volume (Distortion Energy
Theory)
For the case of triaxial stress


1
( 1   2 ) 2  ( 2   3 ) 2  ( 3   1 ) 2   Y2
2
(Maxwell-Huber-Von Mises)
For the case of plane or biaxial stress
 12   1 2   22   Y2
d) Total Strain Energy Per Unit Volume
 12   22   32  2 ( 1 2   2 3   3 1 )   Y2
(Haigh)
2. Brittle Materials
a) Maximum Normal Stress Theory
If the material is subjected to plane stress.
 1   ult
 2   ult
134
Example:- The steel pipe shown below has an inner diameter of 60 mm and an outer
diameter of 80 mm. If it is subjected to a torsional moment of 8 KN.m and a bending
moment of 3.5 KN.m, determine if these loadings cause failure as defined by the
maximum distortion energy theory. The yield stress for the steel found from a tension test
is σY=250 MPa.
 12   1 2   22 ?  Y2
 Point A
A 
A 
Tr
8  40  10 3

 116.41 MPa

J
3 4
3 4
[(40  10 )  (30  10 ) ]
2
My
3.5  40  10 3

 101.859 MPa

I
3 4
3 4
[(40  10 )  (30  10 ) ]
4
116 .41 MPa
σx=-101.859 MPa , σy=0 , xy=116.41 MPa
c=
 x  y
R= (
=
2
 x  y
2
101 .859 MPa
 101.859  0
 50.9295 MPa
2
) 2   xy2  (
A(-101.859,116.41)
 101.859  0 2
)  (116.41) 2 =127.063 MPa
2
Draw Mohr’s circle
 1  C  R =-50.9295+127.063=76.1335 MPa
 2  C  R =-50.9295-127.063
2
C
=-177.9925 MPa
 12   1 2   22   Y2
A
(76.1335)2-(76.1335)( -177.9925)+( -177.9925)2 ≤(250)2
51100≤62500 since 51100<62500 so these loadings will not cause failure.
135
1
Example:- The solid shaft shown below has a radius of 0.5 in. and is made of steel having
yield stress  Y  36 ksi. Determine if the loadings cause the shaft to fail according to the
maximum shear stress theory and the maximum distortion energy theory.
 
A 
P
15

 19.1 ksi
A  (0.5) 2
Tr 3.25  0.5

 16.55 ksi

J
4
(0.5)
2
 x  19.1 ksi ,  y  0 ,   16.55 ksi
 1, 2 
=
 x  y
2
± (
 x  y
2
) 2   xy2
 19.1  0 2
 19.1  0
)  (16.55) 2
± (
2
2
=-9.55±19.11
 1  9.56 ksi
 2  28.66 ksi
 Maximum shear stress theory
1   2   Y
9.56  28.66  36
38.2>36
So the failure will occur according to this theory.
 maximum distortion energy theory
 12   1 2   22   Y2
(9.56)2-(9.56)(-28.66)+(-28.66)2=(36)2
1186.515<1296
the failure will not occur according to this theory.
136
Example:- The solid cast iron shaft shown below is subjected to a torque of T=400 lb.ft.
Determine the smallest radius so that it does not fail according to the maximum normal
stress theory  ult =20 ksi.
 
Tr 400  12  r 3055.8



J
r3
4
(r )
2
x 0 , y  0,  
 1, 2 
 1, 2 
1 
 x  y
2
± (
3055.8
psi
r3
 x  y
2
) 2   xy2
0  0 2 3055.8 2
00
) ( 3 )
± (
2
r
2
3055.8
r3
,
2  
3055.8
r3
 1   ult
3055 .8
=20000
r3
r=0.535 in.
137
Columns
Columns are long slender members subjected to an axial compressive force. The force
may be large enough to cause the member to deflect laterally or sides way, this deflection
is called buckling.
Critical Load
The maximum axial load that a column support when it is on the verge of buckling is
called the critical load (Pcr).
Any additional loading will cause the column to buckle and therefore deflect laterally.
Ideal Column with Pin Supports
The column to beconsideredis an ideal column, meaning one that is perfectly straight
before loading, is made of homogeneous material, and upon which the load is
appliedthrough the centroid of the cross section. It is further assumed that the material
behaves in a linear-elastic manner and that the column buckles or bends in a single
plane.
138
In order to determine the critical load and the buckled shape of the column we will apply
the following equation:
d 2v
EI 2  M
dx
 M section  0
M+Pv=0
M=-Pv
d 2v
  Pv
dx 2
d 2v
P
 ( )v  0
2
EI
dx
EI
.......................(1)
The general solution of equation (1) is:
v  C1 sin(
P
P
x)  C 2 cos(
x) ......................(2)
EI
EI
C1 and C2 are determined from the boundary conditions :
v=0 at x=0
C2=0
v=0 at x=L
C1 sin(
P
L)  0
EI
C1≠0 therefore
sin(
P
L)  0
EI
P
L  n
EI
n 2 2 EI
P
L2
n  1,2,3,.............
The smallest value of P is obtained when n=1, so the critical load for the column is:
139
Pcr 
 2 EI
L2
This load is sometimes referred to as the Euler load, n represents the number of waves in
the deflected shape of the column; if n=2 two waves will appear in the buckled shape and
the column will support a critical load that is 4Pcr.
The corresponding buckled shape is:
v  C1 sin(
x
)
L
The constant C1 represent the maximum deflection vmax which occurs at the midpoint of the
column.
It is important to realize that the column will buckle about the principal axis of cross
section having the least moment of inertia(the weakest axis). For example a column having
a rectangular cross section as shown below will buckle about the a-a axis not the b-b axis.
As a result engineers usually try to achieve a balance keeping the moments of inertia the
same in all directions
Ix  Iy
Pcr 
 2 EI
L2
140
Pcr: critical or maximum axial load on the column just before it begins to buckle. This load
must not cause the stress in the column to exceed the proportional limit.
E: modulus of elasticity for the material.
I: least moment of inertia for the column's cross sectional area.
L: unsupported length of the column, whose ends are pinned.
I=Ar2
 cr 
 2E
L
( )2
r
cr :critical stress which is an average stress in the column just before the column buckles.
This stress is an elastic stress and therefore:
 cr   Y
r: smallest radius of gyration of the column r 
I
.
A
L/r: slenderness ratio, it’s a measure of the column flexibility.
Example: A 24 ft long A-36 steel tube having the cross section shown below is to be used
as a pin ended column. Determine the maximum allowable axial load the column can
support so that it does not buckle. Est=29×103ksi,  Y =36 ksi.
Pcr 
 EI

L2
2
 4
(3  2.75 4 )
4
( 24  12) 2
 2  29  10 3 
=64.52 kip.
 cr 
Pcr
64.52

2
A  (3  2.75 2 )
=14.28 ksi
Since  cr   Y
Pallow=64.52 kip.
Example: The A-36 steel W8×31 member shown below is to be used as a pin connected
column. Determine the largest axial load it can support before it either begins to buckle or
the steel yields. Est=29×103ksi,  Y =36 ksi. A=9.13 in2, Ix=110 in4, Iy=37.1 in4.
Buckling occurs about y-axis.
Pcr 
 2 EI  2  29  10 3  37.1

L2
(12  12) 2
=512 in4
Pcr 512

 56 ksi
A 9.13
 Y
 cr 
 cr
141
 y  36 
P
A

P
9.13
P=328.68 kip.
Columns Having Various Types of Supports
 Fixed-Free column
M
section
0
M-P(  -v)=0
M=P(  -v)
d 2v
EI 2  P( - v)
dx
2
d v P
P

v
 .......................(1)
2
EI
EI
dx
The solution of equation (1) consists of both a complementary and particular solution.
v  C1 sin(
P
P
x)  C2 cos(
x)   ......................(2)
EI
EI
C1 and C2 are determined from the boundary conditions :
v=0 at x=0
C2=- 
dv
=0 at x=0
dx
dv
P
P
P
P
 C1
cos(
x)  C 2
sin(
x)
dx
EI
EI
EI
EI
C1=0
v   [1  cos(
P
x)] .................(3)
EI
Since the deflection at the top of the column is  , that is at x=L v= 
 cos(
P
L)  0
EI
 ≠0
P
L)  0
EI
n 2 2 EI
P
4L2
cos(
or
P
n
L
EI
2
The smallest value of P is obtained when n=1, so the critical load for the column is:
Pcr 
 2 EI
4L2
142
Effective Length
The effective length (Le) is the distance between points of inflection (that is , points of zero
moment ) in its deflection curve, assuming that the curve is extended (if necessary) until
points of inflection are reached.
Le=KL
Pinned –Ends
K=1
Fixed-Free Ends K=2
Fixed-Ends
K=0.5
Pinned-Fixed Ends K=0.7
Euler's formula becomes:
Pcr 
 2 EI
(KL) 2
;  cr 
 2E
KL
( )2
r
KL/r: columns effective slenderness ratio.
For fixed-Free ends K=2
Pcr 
 2 EI
4L2
Example: A W6×15 steel column is 24 ft long and is fixed at is ends as shown below. Its
load carrying capacity is increased by bracing it about the y-y (weak) axis using strut that
are assumed to be pin connected to its midheight. Determine the load it can support so that
the column does not buckle nor the material exceed the yield stress. Take Est=29×103ksi
and  Y =60 ksi. A=4.43 in2, Ix=29.1 in4, Iy=9.32 in4.
( Pcr ) x 
( Pcr ) y 
 cr 
 2 EI x  2  29  10 3  29.1

 401.7 kip
( KL) 2x
(12  12) 2
 2 EI y
( KL) 2y
( Pcr ) y
 cr   Y
A


 2  29  10 3  9.32
 262.5 kip
(0.7  12  12) 2
262.5
 59.3 ksi
4.43
Pcr=262.5 kip.
143
Example: A viewing platform in a wild animal park is supported by a raw of aluminum
pipe columns having length 3.25 m and outer diameter 100 mm. The bases of the columns
are set in concrete footings and the tops of the columns are supported laterally by the
platform (pinned). The columns are being designed to support compressive loads 100 kN.
Determine the minimum required thickness t of the columns if a factor of safety n=3 is
required with respect to Euler buckling for aluminum use 72 GPa for the modulus of
elasticity and use 480 MPa for the proportional limit.
For fixed –pinned ends column
Pcr 
 2 EI
(0.7 L) 2
Pcr=nP=3×100=300 kN
300=
 2  72 10 6  I
(0.7  3.25) 2
I=2.185×10-6 m4
 4
(d o  d i4 )
64

2.185 10 6  [(100 10 3 ) 4  (100 10 3  2t ) 4 ]
64
I=
t=6.846×10-3 m
t=6.846 mm
di=86.308 mm

4

4
A= (d o2  d i2 ) = [(100 103 ) 2  (86.308 103 ) 2 ]
A=2.0034×10-3 m2
Pcr
300

 149.738 MPa
A 2.0034 10 3
 cr   Y
 cr 
t=6.846 mm
144
1/--pages
Пожаловаться на содержимое документа