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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. 1.25 m A B C
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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 = 0.488 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 = 1.904x10 -4 + 2.976x10 -4 m δ C/A
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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 1.904x10 -4 1 = m/m = 1.904x10 -4 BC = BC L BC 2.976x10 -4 1.25 = m/m = 2.38x10 -4 3. 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
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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.
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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 ) = 0.0714x10 -3 m = 0.0714 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)
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 6 PROBLEM-2 2. Normal strain in each segment: = L AB 0.952x10 -4 1 = m/m = 0.952x10 -4 BC = BC L BC –0.238x10 -4 0.5 = m/m = –0.476x10 -4 3. 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)
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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.
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 8 PROBLEM-3
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 9 PROBLEM-3
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 10 PROBLEM-3
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 11 PROBLEM-3
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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
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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) = 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
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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 = 226.4 N/mm 2 = 226.4 MPa Factor of safety for rod AB :
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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
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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
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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 = 0.2755 mm 0.5 m F AB = 20 kN F CD = 10 kN AB = 0.5510 mm EF = F EF (L EF ) A EF E EF = 0.1033 mm
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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 = = 0.1837mm Thus, F = ( CD + ) + EF = 0.5625 mm
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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.
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 20 PROBLEM-6
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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.
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 22 PROBLEM-7
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 23 PROBLEM-7
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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.
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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)= 275.2 MPa
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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) = 435.5 MPa
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2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 27 PROBLEM-8 Maximum tensile stress at hole: max ) hole = 435.5 MPa Maximum tensile stress at fillet: max ) fillet = 275.2 MPa Maximum tensile stress developed in the bar: max ) bar = max ) hole = 435.5 MPa
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