1.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 1 PROBLEM-1 The hanger assembly is used to support a distributed loading of w=16kN/m. Determine the average stress in the 12-mm-diameter bolt at A and the average tensile stress in rod AB, which has a diameter of 15 mm. If the yield shear stress for the bolt is  y = 180 MPa, and the yield tensile stress for the rod is  y = 275 MPa, determine factor of safety with respect to yielding in each case. 1 m 4/3 m2/3 m w A B C

2.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 2 V F AB =40 kN V Bolt A PROBLEM-1 Equation of equilibrium (F AB sin  )(4/3) – W(1) = 0 +  M C = 0 W 4/3 m2/3 m 1 m F AB  CxCx CyCy C A  = tan -1 (3/4) = 36.87 o F AB = 40 kN We get For bolt A Shear force: V = F AB /2 = 20 kN Shear stress:  = 176.8 N/mm 2

3.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 3 PROBLEM-1 Factor of safety for bolt A : C A F AB For rod AB = 226.4 N/mm 2 = 226.4 MPa Factor of safety for rod AB :

4.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 4 PROBLEM-2 Two bars are used to support a load P. When unloaded, AB is 125 mm long and AC is 200 mm long, and the ring at A has coordinates (0,0). If a load is applied to the ring at A, so that it moves it to the coordinate position (6.25 mm, -18.25 mm), determine the normal strain in each bar. A B C 200 mm 125 mm 60 o P

5.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 5 Initial Geometry A B C 200 mm 125 mm 60 o D BD = ABcos60 o = 62.5 mm AD = ABsin60 o = 108.25 mm PROBLEM-2 = 168.17 mm A’ B C y x 62.5 mm 6.25 mm 108.25 mm 18.25 mm 230.67 mm A’B = ?? Final Geometry = 143.975 mm = 205.47 mm

6.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 6 Average normal strain PROBLEM-2 = 0.1518 m/m A’ B C A A’B – AB AB  AB = 143.975 – 125 125 = = 0.0274 m/m A’C – AC AC  AC = 205.47 – 200 200 =

7.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 7 PROBLEM-3 The bar DBA is rigid and is originally held in the horizontal position. When the weight W is supported from D, it causes the end D to displace downward 0.5 mm. The wires are made of A-36 steel and have a cross-sectional area of 0.8 mm 2, and E st = 210 GPa. Determine: 1.The normal strain & stress in wire BE. 2.The reaction force F BE and weight W. 3.The normal stress & strain in wire CD.

8.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 8 PROBLEM-3 DD’ =  D = 0.5 mm Displacement: 1 m1.5 m D B A B’ D’ BB’ =  B  B = 0.3 mm Normal strain for wire BE : Applying Hook’s law for wire BE :  = E  = (210x10 9 )(2x10 -4 ) = 42x10 6 N/m 2 1) Normal strain & stress for wire BE :

9.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 9 PROBLEM-3 1 m1.5 m D B A Free-body diagram for the bar 2) Reaction force F BE & weight W  M A = 0; DyDy AyAy F BE Equations of equilibrium Applying uni-axial normal stress, where F BE is a tensile force: F BE = (42x10 6 )(0.8x10 -6 ) = 33.6 N Thus, D y = 0.6F BE = 20.16 N

10.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 10 The weight W Equilibrium at point C :  F y = 0; W = D y = 20.16 N 3) Normal stress and strain for wire CD PROBLEM-3 C DyDy W FBD of point C : Wire CD undergoes tensile force

11.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 11 PROBLEM-4 The rigid pipe is supported by a pin at C and an A-36 steel guy wire AB. If the wire has a diameter of 5 mm, determine how much it stretches when a load of P = 1500 N acts on the pipe. The material remains elastic. E = 210 GPa. A B C 2 m 60 o “how much it stretches” What does it mean?? stretch = change in length =   AB

12.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 12 PROBLEM-4 CyCy B C CxCx F BA P Free-body diagram CyCy B C CxCx PF BA ) X F BA ) Y F BA ) X = F BA cos 60 o Component of F BA : F BA ) Y = F BA sin 60 o A B C 2 m 60 o

13.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 13 PROBLEM-4 CyCy B C CxCx PF BA ) X F BA ) Y  M C = 0; Equations of equilibrium F BA ) X (2) – P(2) = 0 F BA ) X = P = 1500 N Applying uni-axial normal stress,

14.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 14 PROBLEM-4 Normal strain for wire BA : Change in length of wire BA :  AB =  BA L BA = (7.27x10 -4 )(2000/sin60 o ) mm = 1.67 mm F BA

15.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 15 PROBLEM-5 A shear spring is made from two blocks of rubber, each having a height h, width b, and thickness a. The blocks are bonded to the three plates as shown. If the plates are rigid and the shear modulus of the rubber is G, determine the displacement of plate A if a vertical load P is applied to this plate. Assume that the displacement is small so that  = a tan   a  aa h 

16.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 16 PROBLEM-5 Free-body diagram after deformation P P/2   V Shear force: V = P/2 Rubber cross section a b Shear stress: Shear strain: Displacement:  = a tan  a  aa h 

17.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN PROBLEM-6

18.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN PROBLEM-6

19.  2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 19