Explanation Given: The supports F and E are fixed and the points B and C are fixed connected. The each girder has thickness of 2 ft . Concept Used: Write the expression for fixed end moments. F E M = coefficients × w L 2 .......(I) Here, uniform distributed load is w and length of the span is L . Write the expression for absolute stiffness factor for member A B . K A B = k A B E I L .......(II) Here stiffness factor for member A B is k A B moment of inertia is I , modulus of elasticity is E and length of the span is L . Write the expression for absolute stiffness factor for member B A . K B A = k B A E I L .......(III) Here, stiffness factor for member B A is k B A . Write the expression for absolute stiffness factor for member B C . K B C = k B C E I L .......(IV) Here, stiffness factor for member B C is k B C . Write the expression for absolute stiffness factor for member C B . K C B = k C B E I L .......(V) Here, stiffness factor for member C B is k C B . Write the expression for absolute stiffness factor for member C D . K C D = k C D E I L .......(VI) Here, stiffness factor for member C D is k C D . Write the expression for absolute stiffness factor for member D C . K D C = k D C E I L .......(VII) Here, stiffness factor for member D C is k C B . Write the expression for absolute stiffness factor for member C E . K C E = 4 E I L .......(VIII) Here, stiffness factor for member C E is k C E . Write the expression for absolute stiffness factor for member B F . K B F = 4 E I L .......(IX) Here, stiffness factor for member B F is k B F . Write the expression for slope deflection equation. M N = K N ( θ N + C N θ F − ψ ( 1 + C N ) ) + ( F E M ) N .......(X) Here, internal moment at the near end of the span is M N , absolute stiffness of the near end is K N , near end slope of the span at the support is θ N , far end slope of the span at the support is θ F and the span cord rotation due to a linear displacement. Calculation: The diagram of the beam is shown below. Figure (1) Consider Figure (1). Length of span B C is 40 ft . Length of span A B is 30 ft . Length of span C D is 30 ft . Length of haunch at end C is 6 ft . Length of haunch at end B is 9 ft . Calculate ratio of length of haunch at end A to the length of the span A B . a A = 6 ft 30 ft = 0.2 Calculate ratio of length of haunch at end B to the length of the span A B . a B = 9 ft 30 ft = 0.3 Calculate ratio for rectangular cross-sectional area at end A for span A B . r A = 4 ft − 2 ft 2 ft = 1 Calculate ratio for rectangular cross-sectional area at end B for span A B . r B = 4 ft − 2 ft 2 ft = 1 Calculate ratio of length of haunch at end B to the length of the span B C . a B = 8 ft 40 ft = 0.2 Calculate ratio of length of haunch at end C to the length of the span B C . a C = 8 ft 40 ft = 0.2 Calculate ratio for rectangular cross-sectional area at end B for span B C . r B = 4 ft − 2 ft 2 ft = 1 Calculate ratio for rectangular cross-sectional area at end C for span B C . r C = 4 ft − 2 ft 2 ft = 1 Calculate ratio of length of haunch at end C to the length of the span C D . a C = 6 ft 30 ft = 0.2 Calculate ratio of length of haunch at end D to the length of the span C D . a D = 9 ft 30 ft = 0.3 Calculate ratio for rectangular cross-sectional area at end C for span C D . r C = 4 ft − 2 ft 2 ft = 1 Calculate ratio for rectangular cross-sectional area at end D for span C D . r D = 4 ft − 2 ft 2 ft = 1 Determine carry over factor of member A B at end A from parabolic haunches-constant width table and ratios calculated. C A B = 0.683 Determine carry over factor of member A B at end B from parabolic haunches-constant width table and ratios calculated. C B A = 0.598 Determine carry over factor of member B C at end B from parabolic haunches-constant width table and ratios calculated. C B C = 0.619 Determine carry over factor of member B C at end C from parabolic haunches-constant width table and ratios calculated. C C B = 0.619 Determine carry over factor of member C D at end C from parabolic haunches-constant width table and ratios calculated. C C D = 0.683 Determine carry over factor of member C D at end D from parabolic haunches-constant width table and ratios calculated. C D C = 0.598 Determine carry over factor of member B F at end B from parabolic haunches-constant width table and ratios calculated. C B F = 0.5 Determine carry over factor of member C E at end C from parabolic haunches-constant width table and ratios calculated. C C E = 0.5 Determine stiffness factor at end A for span A B from parabolic haunches-constant width table and ratios calculated. k A B = 6.73 Determine stiffness factor at end B for span A B from parabolic haunches-constant width table and ratios calculated. k B A = 7.68 Determine stiffness factor at end B for span B C from parabolic haunches-constant width table and ratios calculated. k B C = 6.41 Determine stiffness factor at end C for span B C from parabolic haunches-constant width table and ratios calculated. k C B = 6.41 Determine stiffness factor at end C for span C D from parabolic haunches-constant width table and ratios calculated. k C D = 6.73 Determine stiffness factor at end D for span C D from parabolic haunches-constant width table and ratios calculated. k D C = 7.68 Calculate absolute stiffness factor for member A B . Substitute 6.73 for k A B and 30 ft for L in Equation (II). K A B = ( 6.73 ) E I 30 ft = 0.2243 E I Calculate absolute stiffness factor for member B A . Substitute 7.68 for k B A and 30 ft for L in Equation (III). K B A = ( 7.68 ) E I 30 ft = 0.256 E I K ′ B A = 0.2565 E I [ 1 − ( 0.683 ) ( 0.598 ) ] = 0.15144 E I Calculate absolute stiffness factor for member B C . Substitute 6.41 for k B C and 40 ft for L in Equation (IV). K B C = ( 6.41 ) E I 40 ft = 0.16025 E I Calculate absolute stiffness factor for member C B . Substitute 6.41 for k C B and 40 ft for L in Equation (V). K C B = ( 6.41 ) E I 40 ft = 0.16025 E I Calculate absolute stiffness factor for member C D . Substitute 6.73 for k C D and 30 ft for L in Equation (VI). K C D = ( 6.73 ) E I 30 ft = 0.2243 E I Calculate absolute stiffness factor for member D C . Substitute 7.68 for k D C and 30 ft for L in Equation (VII). K D C = ( 7.68 ) E I 30 ft = 0.2565 E I K ′ D C = 0.2565 E I [ 1 − ( 0.683 ) ( 0.598 ) ] = 0.15144 E I Calculate absolute stiffness factor for member B F . Substitute 25 ft for L in Equation (VIII). K B F = 4 E I 25 ft = 0.16 E I Calculate absolute stiffness factor for member C E . Substitute 25 ft for L in Equation (IX). K C E = 4 E I 25 ft = 0.16 E I Calculate fixed end moment for member A B . Determine coefficients from straight haunches-constant width table. Substitute 4 k/ft for w , − 0.0911 for coefficient and 30 ft for L in Equation (I). ( F E M ) A B = ( − 0.0911 ) × ( 4 k/ft ) ( 30 ft ) 2 = − 327.96 k-ft Calculate fixed end moment for member B A . Determine coefficients from straight haunches-constant width table. Substitute 4 k/ft for w , 0.1042 for coefficient and 30 ft for L in Equation (I). ( F E M ) B A = ( 0.1042 ) × ( 4 k/ft ) ( 30 ft ) 2 = 375.12 k-ft Calculate fixed end moment for member B C . Determine coefficients from straight haunches-constant width table. Substitute 4 k/ft for w , − 0.0956 for coefficient and 40 ft for L in Equation (I). ( F E M ) B C = ( − 0.0956 ) × ( 4 k/ft ) ( 40 ft ) 2 = − 6.1184 k-ft Calculate fixed end moment for member C B . Determine coefficients from straight haunches-constant width table. Substitute 4 k/ft for w , 0.0956 for coefficient and 40 ft for L in Equation (I). ( F E M ) C B = ( 0.0956 ) × ( 4 k/ft ) ( 40 ft ) 2 = 6.1184 k-ft Calculate fixed end moment for member C D . Determine coefficients from straight haunches-constant width table. Substitute 4 k/ft for w , − 0.1042 for coefficient and 30 ft for L in Equation (I). ( F E M ) C D = ( − 0.1042 ) × ( 4 k/ft ) ( 30 ft ) 2 = − 375.12 k-ft Calculate fixed end moment for member D C . Determine coefficients from straight haunches-constant width table. Substitute 4 k/ft for w , 0.0911 for coefficient and 30 ft for L in Equation (I). ( F E M ) D C = ( 0.0911 ) × ( 4 k/ft ) ( 30 ft ) 2 = 327.96 k-ft Calculate fixed end moment for member B F . ( F E M ) B F = 0 Calculate fixed end moment for member F B . ( F E M ) F B = 0 Calculate fixed end moment for member C E . ( F E M ) C E = 0 Calculate fixed end moment for member E C . ( F E M ) E C = 0 Calculate internal moment at the near end of the span A B . Substitute 0.683 for C A B , − 327

Structural Analysis, Student Value Edition

10th Edition

ISBN: 9780134622088

Author: HIBBELER, Russell C.

Publisher: PEARSON

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Chapter 13 Solutions

Structural Analysis, Student Value Edition

Ch. 13 - Prob. 13.1P Ch. 13 - Prob. 13.2P Ch. 13 - Prob. 13.3P Ch. 13 - Prob. 13.4P Ch. 13 - Prob. 13.5P Ch. 13 - Prob. 13.6P Ch. 13 - Prob. 13.7P Ch. 13 - Prob. 13.8P Ch. 13 - Prob. 13.9P Ch. 13 - Prob. 13.10P

Ch. 13 - Prob. 13.11P Ch. 13 - Prob. 13.12P

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