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PROBLEM-1

PROBLEM-1. 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:

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PROBLEM-1

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  1. PROBLEM-1 A composite A-36 steel bar shown in the figure has 2 segments, AB and BC. Cross-sectional area of AB is 500 mm2 and BC is 400 mm2. 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. A 1 m B 1.25 m C P

  2. PROBLEM-1 1. Displacement of point C relative to A: RA δAB + dBC δC/A= A 1 m Hence, PAB = PBC = P = 20 kN = 20x103 N EAB = EBC = E = 210 Gpa = 210x109 N/m2 B 1.25 m (20x103)(1) (500x10-6)(210x109) (20x103)(1.25) (400x10-6)(210x109) δC/A + = C = 1.904x10-4 + 2.976x10-4 m = 4.88x10-4 m = 0.488 mm P

  3. PROBLEM-1 2. Normal strain in each segment: RA dAB LAB P AAB 1.904x10-4 1 (20x103) (500x10-6) m/m eAB= = 1.904x10-4 = A = 40x106 N/m2 sAB= = = 40 MPa sAB= EeAB= (210x109)(1.904x10-4) N/m2 = 40 MPa 1 m dBC LBC P ABC (20x103) (400x10-6) 2.976x10-4 1.25 sBC= eBC= m/m = = 50x106 N/m2 = 50 MPa = 2.38x10-4 = B 3. Normal stress in each segment: sBC=EeBC= (210x109)(2.38x10-4) N/m2 = 50 MPa 1.25 m C Normal stress in each segment can be found using Hooke’s law: P

  4. PROBLEM-2 • A steel bar shown in the figure has 2 segments, AB and BC. Cross-sectional area of each segment is 500 mm2.The Young’s modulus of the steel bar is 210 GPa. The bar is subjected to two axial forces; P1 = 10 kN is acting at point A and P2 = 15kN 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.

  5. PROBLEM-2 PAB PBC= P2 – P1 = 5 kN (Compressive) PAB=P1 = 10 kN (Tensile) 1. Displacement of point Arelative to C: PBC δAB–dBC δA/C= (10x103)(1) (500x10-6)(210x109) (5x103)(0.5) (500x10-6)(210x109) δA/C – = = 0.0714x10-3 m = 0.0714 mm = 0.952x10-4–0.238x10-4 m

  6. PROBLEM-2 2. Normal strain in each segment: dAB LAB PAB A 0.952x10-4 1 (10x103) (500x10-6) m/m eAB= = 0.952x10-4 = = 20x106 N/m2 sAB= = = 20 MPa sAB= EeAB= (210x109)(0.952x10-4) N/m2 = 20 MPa dBC LBC PBC A (5x103) (500x10-6) –0.238x10-4 0.5 sBC= eBC= m/m = = 10x106 N/m2 = 10 MPa = –0.476x10-4 = 3. Normal stress in each segment: sBC=EeBC= (210x109)(–0.476x10-4) N/m2 = –10 MPa (Tensile) (Comprssv) Normal stress in each segment can be found using Hooke’s law:

  7. PROBLEM-3 Composite A-36 steel bar shown made from two segments AB and BD. Area AAB = 600 mm2 and ABD = 1200 mm2. Young’s modulus is 210 GPa. Determine: • Displacement of end A relative to D • Displacement of B relative to D. • Stress and strain in each segment.

  8. PAB=75 kN PAB=75 kN PAB=75 kN PAB=75 kN PAB=75 kN PAB PBC=35 kN PBC=35 kN PBC=35 kN PBC PCD=45 kN PCD 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.

  9. PAB=75 kN PBC=35 kN PCD=45 kN PROBLEM-3 1. Displacement A relative to D Vertical displacement of A relative to fixed support D is δAB+ dBC+ dCD δA/D= [75x103](1 ) (600x10-6)(210x109) = [35x103](0.75 ) (1200x10-6)(210x109) + [–45x103](0.5 ) (1200x10-6)(210x109) = (5.952x10-4+ 1.041x10-4– 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.

  10. PROBLEM-3 2. Displacement of B relative to D, dBC+ dCD δB/D= dAB LAB 5.952x10-4 1 eAB= = m/m = 5.952x10-4 = (1.041x10-4 – 0.893x10-4) m = 0.148x10-4 m = 0.015 mm 3. Strain in each segment: dBC LBC 1.041x10-4 0.75 eBC= = = 1.388x10-4 m/m dCD LCD –0.893x10-4 0.5 eCD= = m/m =–1.786x10-4

  11. PROBLEM-4 Stress in each segment: sAB= EeAB = (210x109)(5.952x10-4) = 125x106 N/m2 = 125 MPa (Tensile) sBC= EeBC = (210x109)(1.388x10-4) = 29.15x106 N/m2 = 29.15 MPa (Tensile) sCD= EeCD = (210x109)(–1.786x10-4) = –37.5x106 N/m2 = –37.5 MPa (Compression)

  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 ty = 180 MPa, and the yield tensile stress for the rod is sy = 275 MPa, determine factor of safety with respect to yielding in each case. B 1 m C A w 4/3 m 2/3 m

  13. FAB + MC = 0 Cx a A C Cy 4/3 m 2/3 m We get FAB = 40 kN 1 m 1 m W Shear force: V = FAB/2 = 20 kN V V Shear stress: Bolt A FAB=40 kN PROBLEM-4 Equation of equilibrium (FABsina)(4/3) – W(1) = 0 a = tan-1(3/4) = 36.87o For bolt A t = 176.8 N/mm2

  14. FAB C A PROBLEM-4 Factor of safety for bolt A: For rod AB = 226.4 N/mm2= 226.4 MPa Factor of safety for rod AB:

  15. PROBLEM-5 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.

  16. FEF(0.5) 1.5 C FCD= FCD 1.0 m FEF = 30 kN E 0.5 m FAB A PROBLEM-5 Internal force in the rods MA = 0, = 10 kN Fx = 0, FEF – FCD– FAB = 0 FAB = 20 kN

  17. FCD = 10 kN 0.5 m FAB = 20 kN dCD=0.2755 mm FCD(LCD) ACD E FAB(LAB) AAB E FEF(LEF) AEF E dCD= dAB= dEF= dAB=0.5510 mm dEF=0.1033 mm PROBLEM-4 Displacement AB,CD, and EF:

  18. dCD C’ C 1.0 m E F’ E’ d LEE’ = dCD + d 0.5 m A’ A d = 0.1837mm dAB = dAB – dCD 1.5 d 1.0 PROBLEM-4 Free-body diagram for the displacement of point A, C, and E The rigid bar AC is displaced to A’C’ Displacement of point F ?? dF = LEE’ + dEF LEE’ ?? = 0.5625 mm Thus, dF = (dCD + d) + dEF

  19. PROBLEM-6 The assembly shown in the figure consists of an aluminum tube AB having a cross-sectional of 400 mm2. 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 Est = 200 GPa and Eal = 70 GPa.

  20. PBCLBC AstEst PABLAB AalEal dC/B = = dB/A== PROBLEM-6 Displacement of steel rod, dC/B (80x103)(0.6) (0.0052)(200x109) = 3.056x10-3 m Force PBC makes point B moves to right. Displacement of aluminum tube, dB/A (– 80x103)(0.6) (400x10-6)(200x109) = –1.143x10-3 m Force PAB makes point B moves to right. Displacement of point C, dC Free-body diagram dC = dC/B + dB/A = 3.056x10-3 + 1.143x10-3 = 4.20x10-3 m

  21. PROBLEM-7 Aluminum rod Steel rod 1 m 1 m • The assembly shown in the figure consists of a steel rod BC and an aluminum rod CD. The cross-sectional of aluminum rod 900 mm2 and for the steel rod is 400 mm2. A force P = 20 kN is applied to the rod at C. If Est = 210 GPa and Eal = 70 GPa, determine: • Reaction force at B and D, respectively • Displacement of point C. • Stress in each rod.

  22. PROBLEM-7 1. Reaction force at B and D FB FD P − FB−FD = 0, FB = P −FD(a) 1 m 1 m FB FB FD FD B C C D FDLCD AalEal FDLCD AalEal FBLBC AstEst (P – FD)LBC AstEst δBC−δCD= 0 =0 =0 (b) − − Compatibility equation can bewritten as FD = 8.57 kN Substituting all values into the equation above, we get

  23. PROBLEM-7 From the Eq.(a) we get FB = P −FD = 20 – 8.57 = 11.43 kN 2. Displacement of point C, dC δC= δBC=δCD = =0.136x10-3m = 0.136 mm FBLBC AstEst 3. Stress in each rod, sBCandsCD FB Ast sBC= = 28.575 MPa FD Aal sCD= = 9.522 MPa

  24. PROBLEM-8 Determine the maximum normal stress developed In the bar when it is subjected a tension of P = 10 kN.

  25. PROBLEM-8 Stress concentration factor at fillet From the graph: K = 1.72 Average tensile stress at fillet: Maximum tensile stress at fillet: =(1.72)(160)= 275.2 MPa smax)fillet= K savg

  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: smax)hole= K savg = (2.45)(177.78) = 435.5 MPa

  27. PROBLEM-8 Maximum tensile stress at fillet: smax)fillet= 275.2 MPa Maximum tensile stress at hole: smax)hole= 435.5 MPa Maximum tensile stress developed in the bar: smax)bar = smax)hole = 435.5 MPa

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