Explanation Given: The modulus of elasticity is, E = 200 GPa . The moment of inertia is, I = 150 × 10 6 mm 4 . The modulus of rigidity is, G = 75 GPa . The cross-sectional web area is, A = 2.5 × 10 3 mm 2 . Concept Used: Write the expression for the net force balance in the vertical direction of the beam. ∑ F Y = 0 ..... (I) Here, ∑ F Y is the sum of all the vertical forces. Write the expression for the net force balance in the horizontal direction in the beam. + ∑ F X = 0 ..... (II) Here, ∑ F X is the sum of all the horizontal forces. Write the expression for the net moment about end A of the beam. ∑ M A = 0 ..... (III) Here, ∑ M A is the sum of all the moments taken about the end A . Calculations: The free body diagram of beam is shown below. Figure (1) Here, the vertical support reaction at A is A y , the vertical support reaction at C is C y and the horizontal reaction at point A is A x . Consider the moment about C using Equation (III). − ( A y × 6 m ) + ( 16 × 6 × 6 2 ) kN-m = 0 A y = 288 kN-m 6 m A y = 48 kN Consider the forces in upward direction using Equation (I). C y + 48 kN = ( 16 × 6 ) kN C y = ( 96 − 48 ) kN C y = 48 kN Consider a section at a distance x 1 from the end A in the real beam The free body diagram of the two sections for real load is shown below. Figure (2) Calculate the moment at x 1 . M 1 − ( 48 × x 1 ) + ( 16 × x 1 × x 1 2 ) = 0 M 1 = 48 x 1 − 8 x 1 2 Consider the forces in vertical direction using Equation (I). V 1 + ( 16 × x 1 ) = 48 V 1 = 48 − 16 x 1 Here, V is the shear at appoint due to real loading. Consider a section at a distance x 2 from the end B in the real beam. Calculate the moment at x 2 . M 2 − ( 48 × x 2 ) + ( 16 × x 2 × x 2 2 ) = 0 M 2 = 48 x 2 − 8 x 2 2 Consider the forces in vertical direction using Equation (I). V 2 + 48 = 16 × x 2 V 2 = − 48 + 16 x 2 The free body diagram of the virtual beam to calculate the deflection at B is shown below. Figure (3) Consider the moment about point C in the virtual beam using Equation (III). − ( A y × 6 m ) + ( 1 × 4 ) kN-m A y = 4 kN-m 6 m A y = 0.667 kN Consider the forces in vertical direction in the virtual beam using Equation (I). 0.667 kN + C y = 1 kN C y = 0.33 kN The free body diagram of the two sections for virtual load is shown below. Figure (4) Consider a section at a distance x 1 from the end B in the virtual beam. Calculate the moment for the virtual beam at a distance x 1 . − ( 0.667 × x 1 ) + m 1 = 0 m 1 = 0.667 x 1 Consider the forces in vertical direction using Equation (I). v 1 − 0.667 = 0 v 1 = 0.667 kN Here, v is the shear due to virtual loading at a point. Calculate the moment for the virtual beam at a distance x 2 . ( 0.333 × x 2 ) − m 2 = 0 m 2 = 0.333 x 2 Consider the forces in vertical direction using Equation (I). v 2 + 0.33 kN = 0 v 2 = − 0.33 kN Write the equation for deflection in beam. Δ B = U b + U s Here, the strain energy due to bending at point B is U b and strain energy due to shear at point B is U s . Write the expression for strain energy due to bending at point B . U b = ∫ 0 L m M E I d x .....(IV) For the given beam, rewrite Equation (IV). U b = ∫ 0 2 m 1 M 1 E I d x 1 + ∫ 0 4 m 2 M 2 E I d x 2 .....(V) Substitute 0.667 x 1 for m 1 , ( 48 x 1 − 8 x 1 2 ) for M 1 , 0.333 x 1 for m 2 and ( 48 x 2 − x 2 2 ) for M 2 in Equation (V). U b = ∫ 0 2 0.667 x 1 × ( 48 x 1 − 8 x 1 2 ) E I d x 1 + ∫ 0 4 0.333 x 2 × ( 48 x 2 − x 2 2 ) E I d x 2 = 1 E I [ ∫ 0 2 0.667 x 1 × ( 48 x 1 − 8 x 1 2 ) d x 1 + ∫ 0 4 0.333 x 2 × ( 48 x 2 − x 2 2 ) d x 2 ] = 1 E I [ ∫ 0 2 ( 32 x 1 2 − 5.336 x 1 3 ) d x 1 + ∫ 0 4 ( 16 x 2 2 − 2.664 x 2 3 ) d x 2 ] Integrate and apply the limits. U b = 1 E I [ ( 32 × 2 3 3 ) − ( 5.336 × 2 4 4 ) + ( 16 × 4 3 3 ) − ( 2.666 × 4 4 4 ) ] = 1 E I ( 234.68 ) .....(VI) Substitute 200 GPa for E and 150 × 10 6 mm 4 for I in Equation (VI). U b = 1 ( 200 GPa ) ( 10 6 kN / m 2 1 GPa ) × ( 150 × 10 6 mm 4 ) ( 10 − 12 m 4 1 mm 4 ) ( 234

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ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 8 Solutions

Structural Analysis (10th Edition)

Ch. 8 - Determine the vertical displacement of joint A....Ch. 8 - Prob. 8.2P Ch. 8 - Prob. 8.3P Ch. 8 - Prob. 8.4P Ch. 8 - Prob. 8.5P Ch. 8 - Prob. 8.6P Ch. 8 - Prob. 8.7P Ch. 8 - Prob. 8.8P Ch. 8 - Prob. 8.9P Ch. 8 - Prob. 8.10P

Ch. 8 - Prob. 8.11P Ch. 8 - Prob. 8.12P Ch. 8 - Prob. 8.13P Ch. 8 - Prob. 8.14P Ch. 8 - Prob. 8.15P Ch. 8 - Prob. 8.16P Ch. 8 - Prob. 8.17P Ch. 8 - Prob. 8.18P Ch. 8 - Prob. 8.19P Ch. 8 - Prob. 8.20P Ch. 8 - Prob. 8.21P Ch. 8 - Prob. 8.22P Ch. 8 - Prob. 8.23P Ch. 8 - Prob. 8.24P Ch. 8 - Prob. 8.25P Ch. 8 - Prob. 8.26P Ch. 8 - Prob. 8.27P Ch. 8 - Prob. 8.28P Ch. 8 - Prob. 8.29P Ch. 8 - Prob. 8.30P Ch. 8 - Prob. 8.31P Ch. 8 - Prob. 8.32P Ch. 8 - Prob. 8.33P Ch. 8 - Prob. 8.34P Ch. 8 - Prob. 8.35P Ch. 8 - Prob. 8.36P Ch. 8 - Prob. 8.37P Ch. 8 - Prob. 8.38P Ch. 8 - Prob. 8.39P Ch. 8 - Prob. 8.40P Ch. 8 - Prob. 8.41P Ch. 8 - Prob. 8.42P Ch. 8 - Prob. 8.43P Ch. 8 - Prob. 8.44P Ch. 8 - Prob. 8.45P Ch. 8 - Prob. 8.46P Ch. 8 - Prob. 8.47P Ch. 8 - Prob. 8.48P Ch. 8 - Prob. 8.49P Ch. 8 - Prob. 8.50P Ch. 8 - Prob. 8.51P Ch. 8 - Prob. 8.52P Ch. 8 - Prob. 8.53P Ch. 8 - Prob. 8.54P Ch. 8 - Prob. 8.55P Ch. 8 - Prob. 8.56P Ch. 8 - Prob. 8.57P Ch. 8 - Prob. 8.58P Ch. 8 - Prob. 8.59P Ch. 8 - Prob. 8.60P Ch. 8 - Prob. 8.61P

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