 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 1 PROBLEM-1 1 m P A composite A-36 steel bar shown in the figure has 2 segments, AB and BC. Cross-sectional area of AB is 500 mm 2 and BC is 400 mm 2. The Young’s modulus of the steel bar is 210 GPa. The bar is subjected to an axial force P = 20 kN. Determine: 1) Displacement of point C relative to A. 2) Normal strain in each segment. 3) Normal stress in each segment m A B C

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 2 PROBLEM -1 1 m P 1.25 m A B C δ C/A = 1. Displacement of point C relative to A: (20x10 3 )(1) (500x10 -6 )(210x10 9 ) = 4.88x10 -4 m = mm Hence, P AB = P BC = P = 20 kN = 20x10 3 N E AB = E BC = E = 210 Gpa = 210x10 9 N/m 2 + (20x10 3 )(1.25) (400x10 -6 )(210x10 9 ) RARA = δ AB +  BC = 1.904x x10 -4 m δ C/A

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 3 PROBLEM-1 1 m P 1.25 m A B C 2. Normal strain in each segment: RARA   =   L AB 1.904x = m/m = 1.904x10 -4  BC =  BC L BC 2.976x = m/m = 2.38x Normal stress in each segment:   = P A AB = (20x10 3 ) (500x10 -6 ) = 40x10 6 N/m 2 = 40 MPa  BC = P A BC = (20x10 3 ) (400x10 -6 ) = 50x10 6 N/m 2 = 50 MPa Normal stress in each segment can be found using Hooke’s law:   = E  AB  210x10 9 )(1.904x10 -4 ) N/m 2 = 40 MPa = 50 MPa  BC = E  BC  210x10 9 )(2.38x10 -4 ) N/m 2

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 4 PROBLEM-2 A steel bar shown in the figure has 2 segments, AB and BC. Cross- sectional area of each segment is 500 mm 2.The Young’s modulus of the steel bar is 210 GPa. The bar is subjected to two axial forces; P 1 = 10 kN is acting at point A and P 2 = 15 kN is acting at point B. Determine: 1) Displacement of point A relative to C. 2) Normal strain in each segment. 3) Normal stress in each segment.

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 5 PROBLEM -2 δA/C =δA/C = 1. Displacement of point A relative to C: (10x10 3 )(1) (500x10 -6 )(210x10 9 ) = x10 -3 m = mm – (5x10 3 )(0.5) (500x10 -6 )(210x10 9 ) = δ AB –  BC = 0.952x10 -4 – 0.238x10 -4 m δA/CδA/C P BC PABPAB P AB = P 1 = 10 kN P BC = P 2 – P 1 = 5 kN (Tensile) (Compressive)

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 6 PROBLEM-2 2. Normal strain in each segment:   =   L AB 0.952x = m/m = 0.952x10 -4  BC =  BC L BC –0.238x = m/m = –0.476x Normal stress in each segment:   = P AB A = (10x10 3 ) (500x10 -6 ) = 20x10 6 N/m 2 = 20 MPa  BC = P BC A = (5x10 3 ) (500x10 -6 ) = 10x10 6 N/m 2 = 10 MPa Normal stress in each segment can be found using Hooke’s law:   = E  AB  210x10 9 )(0.952x10 -4 ) N/m 2 = 20 MPa = –10 MPa  BC = E  BC  210x10 9 )(–0.476x10 -4 ) N/m 2 (Tensile) (Comprssv)

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 7 PROBLEM-3 Composite A-36 steel bar shown made from two segments AB and BD. Area A AB = 600 mm 2 and A BD = 1200 mm 2. Young’s modulus is 210 GPa. Determine: 1)Displacement of end A relative to D 2)Displacement of B relative to D. 3)Stress and strain in each segment.

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 8 PROBLEM-3

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 9 PROBLEM-3

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 10 PROBLEM-3

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 11 PROBLEM-3

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 12 PROBLEM-4 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

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 13 V F AB =40 kN V Bolt A PROBLEM-4 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) = o F AB = 40 kN We get For bolt A Shear force: V = F AB /2 = 20 kN Shear stress:  = N/mm 2

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 14 PROBLEM-4 Factor of safety for bolt A : C A F AB For rod AB = N/mm 2 = MPa Factor of safety for rod AB :

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 15 The assembly consists of three titanium rods and a rigid bar AC. The cross-sectional area of each rod is given in the figure. If a force of 30 kN is applied to the ring F, determine the horizontal displacement of point F. E = 121 GPa. PROBLEM-5

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 16 PROBLEM-5 Internal force in the rods 0.5 m 1.0 m F CD F AB F EF = 30 kN A E C  M A = 0, = 10 kN F CD = F EF (0.5) 1.5  F x = 0, F EF – F CD – F AB = 0 F AB = 20 kN

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 17 PROBLEM-5 Displacement AB,CD, and EF:  CD = F CD (L CD ) A CD E  AB = F AB (L AB ) A AB E  CD = mm 0.5 m F AB = 20 kN F CD = 10 kN  AB = mm  EF = F EF (L EF ) A EF E  EF = mm

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 18 PROBLEM-5 Free-body diagram for the displacement of point A, C, and E The rigid bar AC is displaced to A’C’ 0.5 m 1.0 m C A E A’ C’  CD  AB E’ Displacement of point F ?? F’  F = L EE’ +  EF L EE’ ??  L EE’ =  CD +   AB –  CD 1.5  1.0 =  = mm Thus,  F = (  CD +  ) +  EF = mm

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 19 PROBLEM-6 The assembly shown in the figure consists of an aluminum tube AB having a cross-sectional of 400 mm 2. A steel rod having a diameter of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement of the end C of the rod. Take E st = 200 GPa and E al = 70 GPa.

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 20 PROBLEM-6

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 21 PROBLEM-7 1 m Aluminum rod Steel rod The assembly shown in the figure consists of a steel rod BC and an aluminum rod CD. The cross-sectional of aluminum rod 900 mm 2 and for the steel rod is 400 mm 2. A force P = 20 kN is applied to the rod at C. If E st = 210 GPa and E al = 70 GPa, determine: 1.Reaction force at B and D, respectively 2.Displacement of point C. 3.Stress in each rod.

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 22 PROBLEM-7

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 23 PROBLEM-7

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 24 PROBLEM-8 Determine the maximum normal stress developed In the bar when it is subjected a tension of P = 10 kN.

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 25 PROBLEM-8 Stress concentration factor at fillet From the graph: K = 1.72 Average tensile stress at fillet: Maximum tensile stress at fillet:  max ) fillet = K  avg =(1.72)(160)= MPa

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 26 PROBLEM-8 Stress concentration factor at hole Referring to the graph: K = 2.45 Average tensile stress at hole: Maximum tensile stress at hole:  max ) hole = K  avg = (2.45)(177.78) = MPa

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 27 PROBLEM-8 Maximum tensile stress at hole:  max ) hole = MPa Maximum tensile stress at fillet:  max ) fillet = MPa Maximum tensile stress developed in the bar:  max ) bar =  max ) hole = MPa