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For Problems 4.25 and 4.26, design interior one-way slabs for the situations shown. Concrete weight = 150 lb/ft³, Concrete weight = 150 lb/ft³, fy = 60,000 psify = 60,000 psi, and f = 4000 psi f = 4000 psi. Do not use the ACI Code's minimum thickness for deflections (Table 4.1). Reinforcement ratios are given in the figures. The only dead load is the weight of the slab. Problem 4.25 (One ans. 5 in. slab with #6 at 10 in. main reinf.) WL = 100 psf 16 ft- 0.18f' Use p = fv Problem 4.26 ト WL = 150 psf p = 1/1√ Pb 24 ft

For Problems 4.25 and 4.26, design interior one-way slabs for the situations shown. Concrete weight = 150 lb/ft³, Concrete weight = 150 lb/ft³, fy = 60,000 psify = 60,000 psi, and f = 4000 psi f = 4000 psi. Do not use the ACI Code's minimum thickness for deflections (Table 4.1). Reinforcement ratios are given in the figures. The only dead load is the weight of the slab. Problem 4.25 (One ans. 5 in. slab with #6 at 10 in. main reinf.) WL = 100 psf 16 ft- 0.18f' Use p = fv Problem 4.26 ト WL = 150 psf p = 1/1√ Pb 24 ft

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Please solve problem 4.26. Please show all of your work and solve the problem in the picture. Show your answer in pictures please do not type it please. 

For Problems 4.25 and 4.26, design interior one-way slabs for the situations shown. Concrete weight = 150 lb/ft³, Concrete weight = 150 lb/ft³, fy = 60,000 psify = 60,000 psi, and f = 4000 psi f = 4000 psi. Do not use the ACI Code's minimum thickness for deflections (Table 4.1). Reinforcement ratios are given in the figures. The only dead load is the weight of the slab. Problem 4.25 (One ans. 5 in. slab with #6 at 10 in. main reinf.) WL = 100 psf 16 ft- 0.18f' Use p = fv Problem 4.26 ト WL = 150 psf p = 1/1√ Pb 24 ft
Transcribed Image Text:For Problems 4.25 and 4.26, design interior one-way slabs for the situations shown. Concrete weight = 150 lb/ft³, Concrete weight = 150 lb/ft³, fy = 60,000 psify = 60,000 psi, and f = 4000 psi f = 4000 psi. Do not use the ACI Code's minimum thickness for deflections (Table 4.1). Reinforcement ratios are given in the figures. The only dead load is the weight of the slab. Problem 4.25 (One ans. 5 in. slab with #6 at 10 in. main reinf.) WL = 100 psf 16 ft- 0.18f' Use p = fv Problem 4.26 ト WL = 150 psf p = 1/1√ Pb 24 ft
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