Tutorial 02- Statics (Chapter 3.0 – 5.3) MECH 101 Spring 2009 F b 16th Feb-16 Tutor: LEO Contents Reviews Free Body Diagram (FBD) Truss structure Practices Review -Force vector, Moment, Equilibrium of Forces, 2D/3D Vectors (position, force) Adding/Subtracting vectors (Parallelogram law/trigonal) Resolution of a vector using Cartesian vector notation (CVN) Additi using Addition i CVN Multiplication of vectors (dot product) Moment / torque (2D/3D) MO = r × F Cross product Typical supports Review -Force vector, Moment, Equilibrium of Forces The equivalent system Moving a force from point A to O requires creating an additional couple moment. Since this new couple moment is a “free” vector, it can be applied at any point P on the body. When several forces and couple p moments act on a body, y, you y can move each force and its associated couple moment to a common point O. Afterwards, you can add all the forces and couple moments together and find one resultant force-couple force couple moment pair. Review -- GROUP PROBLEM SOLVING Given: Handle forces F1 and F2 are applied li d to t the th electric l t i drill. d ill Find: An equivalent q resultant force and couple moment at point O. Plan: a) Find FRO = Σ Fi b) Find MRO = Σ MC + Σ ( ri × Fi ) Where Where, Fi are the individual forces in Cartesian vector notation (CVN). MC are any free f couple l moments t in i CVN (none ( in i this thi example). l ) Ri are the position vectors from the point O to any point on the line of action of Fi . SOLUTION F1 = {6 i – 3 j – 10 k} N F2 = {{0 i + 2 j – 4 k}} N FRO = {6 i – 1 j – 14 k} N r1 = {0.15 {0 15 i + 0.3 0 3 k} m r2 = {-0.25 j + 0.3 k} m MRO = r1 × F1 + r2 × F2 MRO i j = { 0.15 0 6 -33 k 0.3 -10 10 + i j k 0 - 0.25 0.3 0 2 -44 } N·m = {0.9 {0 9 i + 3.3 3 3 j – 0.45 0 45 k + 0.4 0 4 i + 0 j + 0 k} N·m = {1.3 i + 3.3 j – 0.45 k} N·m Rigid Body Equilibrium THE PROCESS OF SOLVING RIGID BODY EQUILIBRIUM PROBLEMS For analyzing an actual physical system, first we need to create an idealized model. Then we need to draw a free-body diagram showing all the external (active and reactive) forces. Finally, we need to apply the equations of equilibrium to solve for any unknowns. Rigid body: y The distance between any two points within the body is constant. ∑ Fx = 0 ∑ Fy = 0 ∑ MO = 0 Where point O is any arbitrary point. Note: Newton’s 3rd LAW F3 F4 F1 x O F2 Rigid Body Equilibrium IMPORTANT NOTES 1 If we hhave more unknowns 1. k than h the h number b off iindependent d d equations, then we have a statically indeterminate situation. W cannott solve We l these th problems bl using i just j t statics. t ti 2. The order in which we apply equations may affect the simplicity of the solution. solution For example, example if we have two unknown vertical forces and one unknown horizontal force, then solving ∑ FX = O first allows us to find the horizontal unknown quickly. g number,, 3. If the answer for an unknown comes out as negative then the sense (direction) of the unknown force is opposite to that assumed when startingg the pproblem. 4. Two force members; Three force members. Rigid Body Equilibrium Free Body Diagram – FBD PROCEDURE FOR DRAWING A FREE BODY G DIAGRAM 1. Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape. 2. Show all the external forces and couple moments. These typically include: a) applied loads, b) support reactions, and, c) the weight of the body (if applicable) . 3. Label loads and dimensions: All known forces and couple moments should be labeled with their magnitudes and directions. For the unknown forces and couple moments, use letters like Ax, Ay, MA, etc.. Indicate any necessary dimensions. dimensions Force direction in free body diagram Whether a force positive or negative depends on your own d fi iti definition. y x =0 ⎧ FAx ⎪ A ⇒⎨ 1 ⎪⎩ FAy = 2 P =0 ⎧ FBx ⎪ B ⎨F = 1 P ⎪⎩ By 2 ⎧ F Ax = 0 ⎪ ⇒⎨ 1 F = P A Ay ⎪⎩ 2 ⎧ FBx = 0 ⎪ ⎨F = 1 P B ⎪⎩ By 2 ⎧ FAx = 0 ⎪ ⇒⎨ 1 − F = P Ay ⎪⎩ 2 ⎧ FBx = 0 ⎪ ⎨F = − 1 P ⎪⎩ By 2 Moment direction in free body y diagram g ∑ M = 0 ⇒ −M O + M1 − M 2 + M 3 = 0 ∑M = 0⇒ M O − M1 + M 2 − M 3 = 0 ∑M = 0 ⇒ M 0 + M1 − M 2 + M 3 = 0 ∑ M = 0 ⇒ −M 0 − M1 + M 2 − M 3 = 0 A General Case Aim: Find the reaction force on point A and B. FAx FAy FBy Equations of Equilibrium ∑F =0 ⇒ FAx = 0 ∑F Y =0 ⇒ FAy − qa + FBy − F = 0 B (F ) = 0 5 ⇒ qa a − 2aFAy − M − Fa = 0 2 X ∑M 1 M 5 FAy = − ( F + − qa ) 2 a 2 1 M 1 FAx = 0, FBy = (3F + − qa) 2 a 2 Force Analysis in a Truss Structure Zero‐force Members: In Engineering mechanics, mechanics a zero force member refers to a member in a truss which, given a specific load, is at rest ‐‐ neither in tension nor in compression. p In a truss a zero force member is often found at pins (connections within the truss) where no external load is applied and three or fewer truss members meet. (Wikipedia: zero-force member) How to identify a zero‐force member? ¾Two principles: 1) The geometry of a truss structure is fixed. ) Th f i fi d a) Rigid body b) Triangles 2) Inspect those potential two‐force members first, then the others. Force Analysis y in a Truss Structure Zero-force Members in a Truss Structure ¾ Two principles: 1) The geometry of a truss structure is fixed. a) Rigid body b) Triangles 2) Inspect those potential two‐force members first, then the others. Example: Indicate all zero-force zero force members. members Answer: AB,BC,CD,DE,HI, and GI JJoint Method + G ⊗ E Dy D Ay A B Ax 1m C 1m 0 + 1m 100 kN ¾ Simplified Si lifi d the h structure to a frame f to solve l out the h support reactions. i Force balance in X-direction: Force balance in Y-direction: Moment balance at C: Ax=0 Ay + Dy -100=0 Ay= 33.33 kN, Dy=66.67 kN 2Ay -1Dy=0 G E Ay=33.33 Dy=66.67 B A Ax=0 C 1m 1m Joint A: 0 D 1m 100 kN Definition: The forces are applied pp to the jjoints by y the beams. ∑F ∑F x = 0 ⇒ FAB − FAG cos 45° = 0 y = 0 ⇒ 33.33 − FAG sin 45° = 0 FAB = 33.3kN Joint B: , FAG = 47.1kN (C) (T) FBG = 0 , FBC = 33.3 kN (T) G E Ay=33.33 Dy=66.67 B A Ax=0 1m 1m Joint D: C 0 ∑F =0 ∑F =0 1m 100 kN x ⇒ − FDC + FDE cos 45° = 0 y ⇒ 66.67 − FDE sin 45° = 0 FDC = 66.67 kN=66.7 kN FDE = 94.29 kN=94.3 kN Joint E: D ∑F ∑F (T) (C) ⇒ FEG − 94.29sin 45° = 0 94 29 cos 45° = 0 = 0 ⇒ − FEC + 94.29 =0 x y FEC = 66.67 kN=66.7 kN FEG = 66.67 kN=66.7 kN G ((T)) (C) E Ay=33.33 Dy=66.67 B A Ax=0 1m 1m 0 Joint C: ∑F y C D 1m 100 kN 66 67 − 100 = 0 = 0 ⇒ FCG cos 45° + 66.67 FCG = 47.1 47 1 kN ((T)) Joint Method 1. Joint Method ∑M A + = 0 ⇒ 20 × 1 + 30 × 2 − RD cos 30°× 3 = 0 RD = 30.79kN ∑F y =0 ⇒ Ay − 10 − 20 − 30 + 30.79 cos 30° = 0 Ay = 33.34kN ∑F x =0 ⇒ Ax − 30.79sin 30 79sin 30° = 0 Ax = 15.4kN ⊗ + Joint Method + ⊗ + 15.4 kN 30.79 kN 33.34 kN Joint A: ∑F y = 0 ⇒ 33.34 − 10 − FAB = 33kN (C) ∑ Fx = 0 ⇒ 15.4 + FAF − FAF = 7.93kN (T) 1 FAB = 0 2 1 × 33 = 0 2 Joint Method + ⊗ + 15.4 kN 33.34 kN Joint B: 1 × 33 − FBF = 0 y 2 FBF = 23.33kN=23.3kN (T) ∑F =0 ⇒ 1 × 33 − FBC = 0 2 = 23.33kN=23.3kN (C) ∑ Fx = 0 FBC ⇒ 30.79 kN + Joint F: ⊗ + ⊗ + 1 FFC − 20 + 23.33 = 0 2 FFC = 4.714kN=4.71kN (C) 1 ∑ Fx = 0 ⇒ FFE − 7.93 − 2 × 4.714 = 0 FFE = 11.26kN 11 26kN = 11.3kN 11 3kN (T) ∑ Fy = 0 ⇒ − Joint E: ∑F = 0 ⇒ FEC = 30kN (T) ∑F = 0 ⇒ FED = 11.26kN=11.3kN (T) y x + Joint C: ∑ Fx = 0 ⇒ 1 1 × 4.714 + 23.33 − FCD = 0 2 2 ⇒ FCD = 37.71kN = 37.7kN (C) ∑ Fy = 0 ⇒ 1 1 × 4.714 − 30 + 37.71 = 0 (check) 2 2 JJoint Method The Approach of Joint Method: 1) FBD ‐‐ Free body diagram. Remove the boundaries; add back support reactions. back support reactions 2) Use EoE equilibrium equations (forces & moments) to solve out the support reactions. solve out the support reactions 3) Inspect the force condition in each joints. Figure out the forces. forces 4) Indicate the members are in tension or compression. Section Method 1. Support Reactions: ∑M + + = 0 ⇒ Ey × 9 − 8 × 6 − 6 × 3 = 0 E y = 7.333kN 7 333kN Moment balance at C ∑ MC = 0 ⇒ 7.33 × 4.5 − 8 × 1.5 − FFG × 3 × sin 60° = 0 FFG = 8.08kN (T) Moment balance at F ∑ M F = 0 ⇒ 7.33 × 3 − FCD × 3 × sin 60° = 0 Y A FCD = 8.47kN X (C) Force balance along Y direction ∑ Fy = 0 ⇒ FCF sin 60° + 7.333 − 8 = 0 FCF = 0.770kN (T) Stress condition 2. + Y + X Force balance along g Y direction 3 4 ⇒ F − × 1500 = 0 F = 0 DF ∑ y 5 5 FDF = 2000kN (C) Moment balance at D ∑M D =0 4 3 ⇒ × 1500 × 12 + × 1500 × 3 − FGF × 3 = 0 5 5 FGF = 5700kN (C) Moment balance at F 4 ⇒ × 1500 × 16 − FDE × 3 = 0 M = 0 ∑ F 5 FDE = 6400kN (T) Section Method Notes when using section method: 1. Imaginary cutting shall through the cross‐section of the member you want to analyze. y y 2. The cutoff rule is not necessarily straight. 3. The section must be continuous to split the structure into Th ti t b ti t lit th t t i t TWO parts.

1/--pages