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=td 0.0013043= t××30×10-3 t=0.013839 m=13.839 mm 2. A=td2 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=2Co 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 =2f P=2fT 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 r3 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 r3 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 3G (d 2 d1 ) 3G (d 2 d1 ) d 23 d13 [ d 1 ( d 2 d 1 ) ]3 L 0 d13 d 23 32TL 3G (d 2 d1 ) d 23 .d13 32TL d12 d1d 2 d 22 3G 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 3r 3 r 2 34 r3 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=2avgt dAm =2avgtAm 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(2rt)-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(2rt)-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 x2 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 x4 0 V 10-10-V=0 V=0 M 0 x 10 KN M-10x+10(x-2)=0 M=20 KN.m 4 x6 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 x2 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 x3 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 xa 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 ) Rr ,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 Rr ) r 0 dA 0 A Ek ( A dA dA 0 r A A Rr ) 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 xa n for x a 0 n ( x a) x a dx n xa 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 x0 2 2 wx 2 wL M xM 2 2 d 2v M EI ( 2 ) dx 2 d v wx 2 wL EI ( 2 ) xM 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 00 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 00 00 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 793 106 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 111.8 106 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 00 ) ( 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 ) = [(100 103 ) 2 (86.308 103 ) 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

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