Explanation Given data: Percentage of dead load is 25 % . Percentage of live load is 75 % . Grade of steel for beam and column is A992 . Grade of steel for the end plate is A36 . Service load is 53 kips . Moment is 134 ft-kips . Calculation: Write the expression to calculate the dead load. D = 25 % of V ..... (I) Here, the dead load is D and the given load is V . Substitute 53 kips for V in Equation (I). D = 25 100 × 53 kips = 13 .25 kips Write the expression to calculate the live load. L = 75 % of V ..... (II) Here, the live load is L and the given load is V . Substitute 53 kips for V in Equation (II). L = 75 100 × 53 kips = 39.75 kips Write the expression to calculate the moment due to dead load. M D = 25 % of M ..... (III) Here, moment due to dead load is M D and the given moment is M . Substitute 134 ft-kips for M in Equation (III) M D = 25 100 × 134 ft-kips = 33 .5 ft-kips Write the expression to calculate moment due to live load. M L = 75 % of M ..... (IV) Here, moment due to live load is M L and given moment is M . Substitute 134 ft-kips for M in Equation (IV). M L = 75 100 × 134 ft-kips = 100.5 ft-kips Write the expression to calculate the factored moment. M u = M D + M L ..... (V) Here, M u is the factored moment. Substitute 33.5 ft-kips for M D and 100.5 ft-kips for M L in Equation (V). M u = 1.2 × 33.5 ft-kips + 1.6 × 100.5 ft-kips = 40.2 ft-kips + 160.8 ft-kips = 201 ft-kips Consider table 1-1 of ASIC manual for W 18 × 35 beam. f y = 50 ksi Z x = 66.5 in 3 Write the expression to calculate the plastic moment strength of the beam. M p = f y × Z x ...... (VI) Here, the plastic moment is M p , yield strength of steel is f e and section modulus about z axis is Z x . Substitute 50 ksi for f y and 66.5 in 2 for Z x in Equation (VI). M p = 50 ksi × 66.5 in 2 = 3325 in-kips ( 1 ft 12 in ) = 277.08 ft-kips Write the expression to calculate the available moment strength of the beam. ϕ b M n = ϕ b M p ..... (VII) Here, resistance factor for bending is ϕ b , nominal strength is M n and plastic moment strength is M p . Substitute 0.90 for ϕ b and 277.08 ft-kips for M p in Equation (VII). ϕ b M n = 0.90 × 277.08 ft-kips = 249.372 ft-kips Calculate the 60 % of available moment strength. 60 % ( ϕ b M n ) = 60 100 × 249.372 ft-kips = 149.62 ft-kips Apply the check for moment. M u > 60 % ( ϕ b M n ) 201 ft-kips > 149.62 ft-kips Hence, ok Check for tensile strength of bolts: Consider table 1-1 of ASIC manual for W 18 × 35 beam. D = 17.7 in t f = 0.425 in Write the expression to calculate the force developed in the beam. F = M u D − t f ..... (VIII) Here, the force developed in the beam is F , the overall depth of the beam is D and the thickness is t f . Substitute 201 ft-kips for M u , 17.7 in for D and 0.425 in for t f in Equation (VIII) To determine (b) The adequacy of the bolts in the given end connection using allowable strength design (ASD) method.

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN: 9781337094740

Author: Segui, William T.

Publisher: Cengage Learning

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Question

Chapter 8, Problem 8.8.2P

To determine

(a)

The adequacy of the bolts in the given end connection using load and resistance factor design (LRFD) method.

To determine

(b)

The adequacy of the bolts in the given end connection using allowable strength design (ASD) method.

Expert Solution & Answer

Chapter 8 Solutions

Steel Design (Activate Learning with these NEW titles from Engineering!)

Ch. 8 - Prob. 8.2.1P Ch. 8 - Prob. 8.2.2P Ch. 8 - A plate is used as a bracket and is attached to a...Ch. 8 - Prob. 8.2.4P Ch. 8 - Prob. 8.2.5P Ch. 8 - Prob. 8.2.6P Ch. 8 - Prob. 8.2.7P Ch. 8 - Prob. 8.2.8P Ch. 8 - Prob. 8.2.9P Ch. 8 - Prob. 8.2.10P

Ch. 8 - Prob. 8.2.11P Ch. 8 - Prob. 8.2.12P Ch. 8 - Prob. 8.2.13P Ch. 8 - Prob. 8.3.1P Ch. 8 - Prob. 8.3.2P Ch. 8 - Prob. 8.3.3P Ch. 8 - Prob. 8.3.4P Ch. 8 - Prob. 8.3.5P Ch. 8 - Prob. 8.3.6P Ch. 8 - Prob. 8.3.7P Ch. 8 - Prob. 8.3.8P Ch. 8 - Prob. 8.3.9P Ch. 8 - Prob. 8.3.10P Ch. 8 - Use an elastic analysis and determine the maximum...Ch. 8 - Use an elastic analysis and determine the maximum...Ch. 8 - Use an elastic analysis and determine the maximum...Ch. 8 - Prob. 8.4.4P Ch. 8 - Prob. 8.4.5P Ch. 8 - Prob. 8.4.6P Ch. 8 - Use an elastic analysis and compute the extra load...Ch. 8 - Use an elastic analysis and compute the extra load...Ch. 8 - Prob. 8.4.9P Ch. 8 - Prob. 8.4.10P Ch. 8 - Prob. 8.4.11P Ch. 8 - Prob. 8.4.12P Ch. 8 - Prob. 8.4.13P Ch. 8 - Prob. 8.4.14P Ch. 8 - Prob. 8.4.15P Ch. 8 - Prob. 8.4.16P Ch. 8 - Prob. 8.4.17P Ch. 8 - Prob. 8.4.18P Ch. 8 - a. Use LRFD and design a welded connection for the...Ch. 8 - Prob. 8.4.20P Ch. 8 - Prob. 8.5.1P Ch. 8 - Prob. 8.5.2P Ch. 8 - Prob. 8.5.3P Ch. 8 - Prob. 8.5.4P Ch. 8 - Prob. 8.5.5P Ch. 8 - Prob. 8.6.1P Ch. 8 - Prob. 8.6.2P Ch. 8 - Prob. 8.6.3P Ch. 8 - Prob. 8.6.4P Ch. 8 - Prob. 8.7.1P Ch. 8 - Prob. 8.7.2P Ch. 8 - Prob. 8.7.3P Ch. 8 - Prob. 8.8.1P Ch. 8 - Prob. 8.8.2P Ch. 8 - Prob. 8.8.3P Ch. 8 - Prob. 8.8.4P

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(a) The adequacy of the bolts in the given end connection using load and resistance factor design (LRFD) method.

Solution Summary: The author analyzes the adequacy of the bolts in the given end connection using load and resistance factor design (LRFD) method.

(a) The adequacy of the bolts in the given end connection using load and resistance factor design (LRFD) method.

(b) The adequacy of the bolts in the given end connection using allowable strength design (ASD) method.

Chapter 8 Solutions

(a)

The adequacy of the bolts in the given end connection using load and resistance factor design (LRFD) method.

(b)

The adequacy of the bolts in the given end connection using allowable strength design (ASD) method.