R 977210 3 m the diameter of the bar d1954410 3 m b

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r =9.772×10 -3 m the diameter of the bar (d)=19.544×10 -3 m b) Double Shear A F 2 2 F A For brass material A= 6 3 10 50 2 10 30 =0.0003 m 2 0003 . 0 r =9.772×10 -3 m the diameter of the bar (d)=19.544×10 -3 m For steel material A= 6 3 10 100 2 10 30 =0.00015 m 2 00015 . 0 r =6.909×10 -3 m the diameter of the bar (d)=13.819×10 -3 m

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8 Example 3: The 80 kg lamp is supported by two rods AB and BC as shown. If AB has a diameter of 10 mm and BC has a diameter of 8 mm , determine the average normal stress in each rod. F BC =0.625F BA ……………..(1) F BC =1308-1.44337F BA ……….(2) 1308-1.44337F BA =0.625F BA F BA =632.38 N F BC =395.2375 N BA BA BA A F = 2 3 ) 10 5 ( 38 . 632 BA =8.051877×10 6 Pa BA =8.051877 MPa BC BC BC A F = 2 3 ) 10 4 ( 2375 . 395 BC =7.863149×10 6 Pa BC =7.863149 MPa A B C 5 4 3 60 BC F BA F N 8 . 784 81 . 9 80 0 60 cos 5 4 0 BA BC x F F F 0 8 . 784 60 sin 5 3 0 BA BC y F F F
9 Example 4: Shafts and pulleys are usually fastened together by means of a key, as shown. Consider a pulley subjected to a turning moment T of 1 KN.m keyed by a 10 mm×10 mm×75 mm key to the shaft. The shaft is 50 mm in diameter. Determine the shear stress on a horizontal plane through the key. 0 o M 0 025 . 0 10 1 3 F F=40000 N F=40 KN A F F F A is the shaded area = 3 3 3 10 75 10 10 10 40 =53.333×10 6 N/m 2 =53.333 MN/m 2 Example 5: Consider a steel bolt 10 mm in diameter and subjected to an axial tensile load of 10 KN as shown. Determine the average shearing stress in the bolt head, assuming shearing on a cylindrical surface of the same diameter as the bolt. A= dt A= ×10×10 -3 ×8×10 -3 =0.000251327 m 2 A F = 000251327 . 0 10 10 3 = 39.7888 ×10 6 N/m 2 = 39.7888 MN/m 2 mm 50 mm 10 mm 10 T F o KN 1 mm 75 mm 10 mm 8 KN 10

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0 Example 6: The bar shown has a square cross section for which the depth and thickness are 40 mm . If an axial force of 800 N is applied along the centroidal axis of the bar's cross sectional area, determine the average normal stress and average shear stress acting on the material along a) section plane a-a and b) section plane b-b . a) section plane a-a A P = 3 3 10 40 10 40 800 =500 KN/m 2 A F F=0 =0 b) section plane b-b d= 60 sin 40 =46.188 mm A F 2 = 3 3 10 40 10 188 . 46 60 sin 800 =375 KN/m 2 A F 1 = 3 3 10 40 10 188 . 46 60 cos 800 =216.50645 KN/m 2 a a b b 60 N 800 N 800 N 800 N 800 N 800 60 1 F 2 F d
1 Example 7: Determine the total increase of length of a bar of constant cross section hanging vertically and subject to its own weight as the only load. The bar is initially straight. : is the specific weight ( weight/unit volume ) A: is the cross-sectional area AE Aydy d L d 0 = L AE Aydy 0 = L ydy AE A 0 = L y AE A 0 2 2 1 = 2 2 L AE A = AE L AL 2 . W= AL = AE L W 2 . dy yA dy y L

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2 Example 8: A member is made from a material that has a specific weight and modulus of elasticity E . If its formed into a cone having the dimensions shown, determine how far its end is displaced due to gravity when its suspended in the vertical position.
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• Winter '15
• MAhmoudali

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