Please show all of your work, Thank you for your help! Problem 7.29 Calculations show that 2.24 in. 22.24 in.² of top or negative moment steel is required for the beamshown. Three #8 bars have been selected. Are the 4 ft. 6 in. embedment lengths shown satisfactory iff = 4000 psi f = 4000 psi and f₁ = 60,000 psify = 60,000 psi? Bars are spaced 3 in. o.c. with 3-in.3-in.side and top cover measured from c.g. of bars. Use Ktr OK 0. (Ans. No; ld = 58.8 in. > 4 ft 6 in.€₁ = 58.8 in. > 4 ft 6 in., not adequate)3 #8==tr3 #83 in.21 in.4 ft 6 in.4 ft 6 in.4 ft 6 in.4 ft 6 in.

Given Data:Development Length CalculationConclusionThe calculated development length of 58.8 in.…

Answered: Problem 7.29 Calculations show that…

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

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter3: Tension Members

Section: Chapter Questions

Problem 3.8.3P

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Similar questions

Compute the nominal shear strength of an M107.5 of A572 Grad 65 steel.
Use an elastic analysis and determine the maximum load per inch of weld.
Use an elastic analysis and compute the extra load in the weld (in kips per inch of length) caused by the eccentricity.
Compute the nominal shear strength of an M1211.8 of A572 Grade 65 steel.
Use an elastic analysis and determine the maximum load in the weld (in kips per inch of length).
Determine the smallest value of yield stress Fy, for which a W-, M-, or S-shape from Part 1 of the Manual will become slender. To which shapes does this value apply? What conclusion can you draw from your answer?
Select an American Standard Channel shape for the following tensile loads: dead load = 54 kips, live load = 80 kips, and wind load = 75 kips. The connection will be with longitudinal welds. Use an estimated shear lag factor of U = 0.85. (In a practical design, once the member was selected and the connection designed, the value of U would be computed and the member design could be revised if necessary.) The length is 17.5 ft. Use Fy=50 ksi and Fu=65 ksi. a. Use LRFD. b. Use ASD.
Use an elastic analysis and compute the extra load in the weld (in kips per inch of length) caused by the eccentricity.
A W1422 acts compositely with a 4-inch-thick floor slab whose effective width b is 90 inches. The beams are spaced at 7 feet 6 inches, and the span length is 30 feet. The superimposed loads are as follows: construction load = 20 psf, partition load = 10 psf, weight of ceiling and light fixtures = 5 psf, and live load = 60 psf, A992 steel is used, and fc=4 ksi. Determine whether the flexural strength is adequate. a. Use LRFD. b. Use ASD.

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Steel Design (Activate Learning with these NEW ti…

Civil Engineering

ISBN:

9781337094740

Author:

Segui, William T.

Publisher:

Cengage Learning