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# Deflection and member deformation

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### Deflection and member deformation

1. 1. Problem Number (1) A 3-mm thick hollow polystyrene cylinder E = 3GPa and a rigid circular plate (only part of which is shown) are used to support a 250-mm long steel rod AB (E = 200 GPa) of 6-mm diameter. If a 3.2KB load P is applied at B, determine (a) the elongation of rod AB, (b) the deflection of point B, (c) the average normal stress in rod AB. Solution: βπΏ = πΉ Γ πΏ π΄ Γ πΈ = 3200 Γ 0.25 3.14 Γ 9 Γ 10β6 Γ 200 Γ 109 = 1.4 Γ 10β4 π Deflection of B = 3200 Γ 0.03 3.14((25)2 β (22)2) Γ 10β6 Γ 3 Γ 109 + 1.4 Γ 10β4 = 0.214 ππ π = πΉ π΄ = 3200 3.14 Γ9 Γ 10β6 = 113.2 πππ
2. 2. Problem Number (2) Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is made of steel E = 200GPa and rod BC of brass E = 105GPa. Determine (a) the total deformation of the composite rod ABC, (b) the deflection of point B. Solution: Assume that the force 40KN is directed to downward at point B βπΏ = 30 Γ 103 Γ0.25 3.14 Γ15 Γ15 Γ 10β6 Γ200 Γ 109 + 70 Γ 103 Γ0.3 3.14 Γ25 Γ25 Γ 10β6 Γ105 Γ 109 = 0.393 ππ Deflection of Point B = 70 Γ 103 Γ0.3 3.14 Γ25 Γ25 Γ 10β6 Γ105 Γ 109 = 0.102 mm Problem Number (3)
3. 3. Both portions of the rod ABC are made of an aluminum for which E = 70 GPa. Knowing that the magnitude of P is 4KN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B. Solution: βπΏ π΅πΆ = βπΏ π΄π΅ (π β 4000) Γ 0.5 3.14 Γ 0.03 Γ 0.03 Γ 70 Γ 109 = 4000 Γ 0.4 3.14 Γ 0.01 Γ 0.01 Γ 70 Γ 109 Then, Q = 32800 N Then, Deflection of B = (32800β4000) Γ0.5 3.14 Γ0.03 Γ0.03 Γ70 Γ 109 = 0.0728 mm Problem Number (4)
4. 4. The rod ABC is made of an aluminum for which E = 70GPa. Knowing that P = 6KN and Q = 42 KN, determine the deflection of (a) point A, (b) point B. Solution: Deflection of A = βπΏ π΄π΅ β βπΏ π΅πΆ = 6000 Γ0.4 3.14 Γ 0.01 Γ0.01 Γ70 Γ 109 β (42000β6000) Γ0.5 3.14 Γ0.03 Γ0.03 Γ70 Γ 109 = 0.01819 ππ Deflection of B = (42000β6000)Γ0.5 3.14 Γ0.03 Γ0.03 Γ70 Γ 109 = 0.091 ππ Problem Number (5)
5. 5. Each of the links AB and CD is made of steel (E = 200GPa) and has a uniform rectangular cross section of 6 * 24 mm. Determine the largest load which can be suspended from point E if the deflection of E is not to exceed 0.25 mm. Solution: βMB = P(375 + 250) β FDC (250) = 0 β΄ πΉ π·πΆ = 2.5π (ππππ πππ) βFy = FDC β FBA β P = 0 β΄ πΉ π΅π΄ = 1.5π (ππππ πππ) β΄ ΞCD = πΉ π·πΆ πΏ π·πΆ πΈ π·πΆ π΄ π·πΆ = 2.5π (200)(10)β3 (200)(10)9(6)(24)(10)β6 = 1.736π (10)β8 π (π·ππ€ππ€πππ) β΄ ΞBA = πΉ π΅π΄ πΏ π΄π΅ πΈπ΄π΅ π΄ π΄π΅ = 1.5π (200)(10)β3 (200)(10)9(6)(24)(10)β6 = 1.0416π (10)β8 π (πππ€πππ) From geometry of the deflected structure: β΄ Ξ πΈ = ( 250 + 375 250 ) ΞC β ( 375 250 )ΞB
6. 6. β΄ Ξ πΈ = (2.5)(β1.736π)(10)β8 β (1.5)(1.0416π)(10)β8 = β2.7776π(10)β8 π For maximum deflection |Ξ πΈ | = 0.25ππ β΄ 2.7776π(10)β8 = 0.25(10)β3 β΄P)max = 9.57 KN
7. 7. Problem Number (6) The length of the 2-mm diameter steel wire CD has been adjusted so that with no load applied, a gap of 1.5mm exists between the end B of the rigid beam ACB and a contact point E. knowing that E = 200 GPa, determine where a 20-kg block should be placed on the beam in order to cause contact between B and E. Solution: