INFORMATIO
strengthsadiq.pdf

R 2 0000018 m 2 0018 mm mm 100 mm 150 kn 150 kn 150 2

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R 2 =0.000018 m 2 =0.018 mm mm 100 mm 150 KN 150 KN 150 2 R 1 R
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7 Example 12: A steel bar of cross section 500 mm 2 is acted upon by the forces shown. Determine the total elongation of the bar. For steel, E=200 GPa . 15 KN 10 KN For portion AB 1 = AE PL = 9 6 3 3 10 200 10 500 10 500 10 50 =0.00025 m=0.25 mm For portion BC 2 = AE PL = 9 6 3 10 200 10 500 1 10 35 =0.00035 m=0.35 mm For portion CD 3 = AE PL = 9 6 3 10 200 10 500 5 . 1 10 45 =0.000675 m=0.675 mm T = 1+ 2 + 3 T =0.25+0.35+0.675=1.275 mm KN 50 KN 45 mm 500 m 1 m 5 . 1 A B C D A B KN 50 KN 50 KN 50 KN 15 KN 35 C B KN 45 KN 45 C D
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8 Example 13: Member AC shown is subjected to a vertical force of 3 KN . Determine the position x of this force so that the average compressive stress at C is equal to the average tensile stress in the tie rod AB . The rod has a cross-sectional area of 400 mm 2 and the contact area at C is 650 mm 2 . 0 y F F AB +F C -3000=0 F AB +F C =3000 ………………………(1) AB = C C C AB AB A F A F 6 6 10 650 10 400 C AB F F F AB =0.6153 F C …………………….(2) From equations (1) and (2) F C =1857.24 N F AB =1142.759 N 0 A M F C ×200×10 -3 -3000×x=0 3000 2 . 0 24 . 1857 x =0.123816 m=123.816 mm mm 200 KN 3 x A B C mm 200 KN 3 x A C AB F C F
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9 Example 14: The bar AB is considered to be absolutely rigid and is horizontal before the load of 200 KN is applied. The connection at A is a pin, and AB is supported by the steel rod EB and the copper rod CD . The length of CD is 1m , of EB is 2 m . The cross sectional area of CD is 500 mm 2 , the area of EB is 250 mm 2 . Determine the stress in each of the vertical rods and the elongation of the steel rod. Neglect the weight of AB . For copper E=120 GPa , for steel E=200 GPa . F Co ×1+F s ×2-200×10 3 ×1.5=0 F Co =300×10 3 -2 F s …………..(1) 1 2 Co s s =2 Co Co Co Co s s s E A L F E A L F 2 9 6 9 6 10 120 10 500 1 2 10 200 10 250 2 Co s F F F Co =1.2 F s …………………….(2) From equations (1) and (2) F s =93750 N F Co =112500 N s s s A F = 6 10 250 93750 =375000000 Pa s =375 MPa Co Co Co A F = 6 10 500 112500 =225000000 Pa Co =225 MPa E L s s = 9 6 10 200 2 10 375 =0.00375 m=3.75 mm A B C D E m 1 mm 500 mm 500 KN 200 KN 200 s F Co F y A x A A B C D E Co s 0 A M
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0 Thermal Stresses: A change in temperature can cause a material to change its dimensions. If the temperature increases, generally a material expands, whereas if the temperature decreases the material will contract. The deformation of a member having a length L can be calculated using the formula: T = × T×L T = AE FL = × T×L T =E× × T : Linear coefficient of thermal expansion. The units measure strain per degree of temperature. They are (1/ºF) in the foot-pound-second system and (1/ºC) or (1/ºK) in SI system.
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