Explanation The Figure (1) shows the free body diagram of the steel beam ABCD . Figure (1) Here, the reaction force at point A is R A , reaction force at point B is F B E , the applied load at point C is F and reaction force at point D is F D E . Refer to procedure 2 from Sec. 4–10. Write the expression for the net force in the beam R A + F B E + F D E = F (I) Write the expression for the net moment about point A in the beam. F D E l + F B E a = F b (II) Here, the length of the beam is l , length between point A B is a and the length between the point A C is b . The Figure (2) shows the beam at section (1). Figure (2) Write the expression for the bending moment at section (1). M = R A x + F B E 〈 x − a 〉 1 − F 〈 x − b 〉 1 (III) Here, the bending moment at section (1) is M and the distance from the left end is x . Write the expression for the bending moment in terms of elastic equation. E I d 2 y d x 2 = M (IV) Here, the modulus of elasticity is E , the deflection in the beam is y and the second area moment of beam is I Substitute R A x + F B E 〈 x − a 〉 1 − F 〈 x − b 〉 for M in Equation (IV). E I d 2 y d x 2 = R A x + F B E 〈 x − a 〉 1 − F 〈 x − b 〉 1 Integrate the above expression. E I d y d x = R A 2 x 2 + F B E 2 〈 x − a 〉 2 − F 2 〈 x − b 〉 2 + C 1 Further integrate the Equation (V). E I y = R A 6 x 3 + F B E 6 〈 x − a 〉 3 − F 6 〈 x − b 〉 3 + C 1 x + C 2 (V) Write the expression for the area of the cable. A = π 4 d 2 (VI) Here, the area of the cable is A , the diameter of the cable is d . Write the expression for elongation in the steel cable at point B. y B = − F B E h A E (VII) Here, the elongation is y B and the height of the cable is h . Write the expression for elongation in the steel cable at point D y D = − F D F h A E (VIII) Here, the elongation is y D . Applying Boundary conditions. (1) At x = 0 ; y = 0 . Substitute 0 in for x and 0 in for y in Equation (V). E I ( 0 ) = R A 6 ( 0 in ) 3 + 0 in − 0 in + C 1 ( 0 in ) + C 2 0 = 0 + 0 − 0 + 0 + C 2 C 2 = 0 (2) At x = 16 in ; y = y B . Substitute 16 in for x and y B for y in Equation (V). E I y B = [ R A 6 ( 16 in ) 3 + F B E 6 〈 16 in − a in 〉 3 − F 6 〈 16 in − b in 〉 3 + C 1 ( 16 in ) + C 2 ] (IX) (3) At x = 48 in ; y = y D . Substitute 48 in for x and y D for y in Equation (V). E I y D = [ R A 6 ( 48 in ) 3 + F B E 6 〈 48 in − a in 〉 3 − F 6 〈 48 in − b in 〉 3 + C 1 ( 48 in ) + C 2 ] (X) Write the expression for the normal stress in section BE. σ B E = F B E A (XI) Here, the normal stress is σ B E . Write the expression for the normal stress in section DF. σ D F = F D F A (XII) Here, the normal stress is σ D F . Conclusion: Refer to Table A-5 “Physical Constants of Materials” to obtain the modulus of elasticity for steel as 30 × 10 6 psi . Substitute 0.5 in for d in Equation (VI). A = π 4 ( 0.5 in ) 2 = 0.1963 in 2 Substitute 38 in for h , 30 × 10 6 psi for E and 0.1963 in 2 for A in Equation (VII). y B = − F B E × 38 in ( 30 × 10 6 psi ) ( 0.1963 in 2 ) = ( − 6.453 × 10 − 6 in / lbf ) F B E Substitute 38 in for h , 30 × 10 6 psi for E and 0.1963 in 2 for A in Equation (VIII). y D = − F D F × 38 in ( 30 × 10 6 psi ) ( 0.1963 in 2 ) = ( − 6.453 × 10 − 6 in / lbf ) F D F Substitute ( − 6.453 × 10 − 6 in / lbf ) F B E for y B , 30 × 10 6 psi for E , 1.2 in 4 for I 16 in for a , 32 in for b and 0 for C 2 in Equation (IX). [ ( 30 × 10 6 ) ( 0.1963 ) ( − 6.453 × 10 − 6 in / lbf ) F B E ] = R A 6 ( 16 in ) 3 + F B E 6 〈 16 in − 16 in 〉 3 − 0 38.0017 F B E = R A 6 ( 16 in ) 3 + 16 C 1 682.7 R A + 232.3 F B E + 16 C 1 = 0 (XIII) Substitute ( − 6.453 × 10 − 6 in / lbf ) F D F for y D , 30 × 10 6 psi for E , 1.2 in 4 for I 16 in for a , 5 kips for F , 32 in for b and 0 for C 2 in Equation (X). [ ( 30 × 10 6 ) ( 0.1963 ) × ( − 6.453 × 10 − 6 in / lbf ) F D F ] = [ R A 6 ( 48 in ) 3 + F B E 6 〈 48 in − 16 in 〉 3 − ( 5 kips × 1000 lbf 1 kips ) 6 〈 48 in − 32 in 〉 3 + C 1 ( 48 ) + 0 ] [ 18432 R A + 5461 F B E + 232

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

10th Edition

ISBN: 9780073398204

Author: Richard G Budynas, Keith J Nisbett

Publisher: McGraw-Hill Education

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Chapter 4, Problem 101P

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The stresses in the cable.

The deflection at point B.

The deflection at point C.

The deflection at point D.

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Chapter 4, Problem 100P

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Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

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The stresses in the cable. The deflection at point B . The deflection at point C . The deflection at point D .

Solution Summary: The author explains the stresses in the cable. The deflection at point B is -0.0255 in.

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The stresses in the cable.

The deflection at point B.

The deflection at point C.

The deflection at point D.

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Chapter 4 Solutions

The stresses in the cable. The deflection at point B . The deflection at point C . The deflection at point D .

Solution Summary: The author explains the stresses in the cable. The deflection at point B is -0.0255 in.

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The stresses in the cable. The deflection at point B. The deflection at point C. The deflection at point D.

Want to see the full answer?

Chapter 4 Solutions

The stresses in the cable.

The deflection at point B.

The deflection at point C.

The deflection at point D.