 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 P AB 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 Internal force Due to external loadings, internal axial forces in segments AB, BC and CD are different. Apply method of sections and equation of vertical force equilibrium as shown. Variation is also plotted. P AB P AB =75 kN P BC P AB =75 kN P BC =35 kN P AB =75 kN P BC =35 kN P CD P BC =35 kN P AB =75 kN P CD =45 kN

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 9 PROBLEM-3 P AB =75 kN P BC =35 kN P CD =45 kN 1. Displacement A relative to D Vertical displacement of A relative to fixed support D is δ A/D = δ AB +  BC +  CD [75x10 3 ](1 ) (600x10 -6 )(210x10 9 ) = [35x10 3 ](0.75 ) (1200x10 -6 )(210x10 9 ) + [–45x10 3 ](0.5 ) (1200x10 -6 )(210x10 9 ) + = (5.952x x10 -4 – 0.893x10 -4 ) m = 6.1x10 -4 m = 0.61 mm Since the result is positive, the bar elongates and, therefore, the displacement at A is upward.

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 10 PROBLEM-3 δ B/D =  BC +  CD = (1.041x10-4 – 0.893x10 -4 ) m = 0.148x10 -4 m = mm 2. Displacement of B relative to D, 3. Strain in each segment:   =   L AB 5.952x = m/m = 5.952x10 -4  BC =  BC L BC 1.041x = m/m = 1.388x10 -4  CD =  CD L CD – 0.893x = m/m = –1.786x10 -4

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 11 PROBLEM-4 Stress in each segment:   = E  AB  210x10 9 )(5.952x10 -4 ) = 125x10 6 N/m 2 = 125 MPa (Tensile) = 29.15x10 6 N/m 2 = MPa  BC = E  BC  210x10 9 )(1.388x10 -4 ) = –37.5x10 6 N/m 2 = –37.5 MPa  CD = E  CD  210x10 9 )( – 1.786x10 -4 ) (Tensile) (Compression)

 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-4 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-4 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 Free-body diagram Displacement of steel rod,  C/B  C/B = = P BC L BC A st E st Force P BC makes point B moves to right. (80x10 3 )(0.6)  ( )(200x10 9 ) = 3.056x10 -3 m Displacement of aluminum tube,     = = P AB L AB A al E al (– 80x103)(0.6) (400x10 -6 )(200x10 9 ) = –1.143x10 -3 m Force P AB makes point B moves to right. Displacement of point C,  C  C =  C/B +  B/A = 3.056x x10 -3 = 4.20x10 -3 m

 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 1 m FBFB FDFD 22 PROBLEM-7 FBFB FBFB B C FDFD FDFD C D P − F B − F D = 0, F B = P − F D (a) δ BC − δ CD = 0 Compatibility equation can be written as = 0 F B L BC A st E st F D L CD A al E al − (b)(b) = 0 (P – F D )L BC A st E st F D L CD A al E al − Substituting all values into the equation above, we get F D = 8.57 kN 1. Reaction force at B and D

 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 23 PROBLEM-7 F B = P − F D = 20 – 8.57 = kN From the Eq.( a ) we get 2. Displacement of point C,  C δ C = δ BC = δ CD = = 0.136x10 -3 m = mm F B L BC A st E st 3. Stress in each rod,  BC and  CD   C  = = MPa F B A st  CD  = = MPa F D A al

 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