PROBLEMS4.1.Compare the cracking moment based on the gross section properties and thetransformed section properties with four No. 11 (No. 36) bars in Figure P4.1abased on a concrete tensile capacity of 7.5 Vf..Compare the cracking moment based on the gross section properties and thetransformed section properties with two No. 10 (No. 32) bars in Figure P4.1bbased on a concrete tensile capacity of 7.5 Vf..Compare the cracking moment based on the gross section properties and thetransformed section properties with four No. 9 (No. 29) bars in Figure P4.1cbased on a concrete tensile capacity of 7.5 Vf..Compare the cracking moment based on the gross sectiotransformed section properties with two No. 11 (No. 36) bars in Figure P4.1dbased on a concrete tensile capacity of 7.5 Vf.4.5.4.2.4.3.4.4.properties and theDetermine the cracking moment based on the gross section properties inFigure P4.1d if the section is prestressed such that there is a 300 psi compressionstress in the extreme tension zone and the concrete tensile capacity is 7.5 Vf.4.6. Determine the service level moment capacity of the section in Figure P4.la ifthe allowable stress for concrete is 0.45 f and the allowable stress for the rein-forcement is 30,000 psi. Use the areas of reinforcement from Problem 4.1.28"36"-F12"-36"*8"-2-2"typ24"36"36"2-2"typ8"(a) = 4000 psi(b) = 5000 psi(c) = 6000 psi(d) %= 3000 psiFIGURE P4.1 4.7. Determine the service level moment capacity of the section in Figure P4.1bif the allowable stress for concrete is 0.45 f! and the allowable stress for thereinforcement is 30,000 psi. Use the areas of reinforcement from Problem 4.2.

Answered: Determine the service level moment…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Concept explainers

Question

PROBLEMS
4.1.
Compare the cracking moment based on the gross section properties and the
transformed section properties with four No. 11 (No. 36) bars in Figure P4.1a
based on a concrete tensile capacity of 7.5 Vf..
Compare the cracking moment based on the gross section properties and the
transformed section properties with two No. 10 (No. 32) bars in Figure P4.1b
based on a concrete tensile capacity of 7.5 Vf..
Compare the cracking moment based on the gross section properties and the
transformed section properties with four No. 9 (No. 29) bars in Figure P4.1c
based on a concrete tensile capacity of 7.5 Vf..
Compare the cracking moment based on the gross sectio
transformed section properties with two No. 11 (No. 36) bars in Figure P4.1d
based on a concrete tensile capacity of 7.5 Vf.
4.5.
4.2.
4.3.
4.4.
properties and the
Determine the cracking moment based on the gross section properties in
Figure P4.1d if the section is prestressed such that there is a 300 psi compression
stress in the extreme tension zone and the concrete tensile capacity is 7.5 Vf.
4.6. Determine the service level moment capacity of the section in Figure P4.la if
the allowable stress for concrete is 0.45 f and the allowable stress for the rein-
forcement is 30,000 psi. Use the areas of reinforcement from Problem 4.1.
28"
36"-
F12"-
36"
*8"-
2-2"
typ
24"
36"
36"
2-2"
typ
8"
(a) = 4000 psi
(b) = 5000 psi
(c) = 6000 psi
(d) %
= 3000 psi
FIGURE P4.1

4.7. Determine the service level moment capacity of the section in Figure P4.1b
if the allowable stress for concrete is 0.45 f! and the allowable stress for the
reinforcement is 30,000 psi. Use the areas of reinforcement from Problem 4.2.

Expert Solution

Step 1 Solution of problem 4.7

$s t r e s s i n c o n c r e t e f_{c} = 5000 p s i = 34.47379 M P a a l l o w a b l e s r e s s i n c o n c r e t e = 0.45 f_{c} s t r e s s i n s t e e l f_{y} = 30000 p s i = 206.84272 M P a A s s u m e a l l o w a b l e s t r e s s i n s t e e l = 0.87 f_{y} w i d t h o f b e a m b = 12 i n e f f e c t i v e d e p t h o f b e a m d = 36 - 2.5 = 33.5 i n a r e a o f s t e e l o f N o . 10 = 1.27 i n^{2} A_{s t} = 2 \times 1.27 = 2.54 i n^{2} \Rightarrow h e r e c r i t i c a l d e p t h o f n e u t r a l a x i s X_{u l i m i t} = \frac{700}{1100 + 0.87 f_{y}} \times d = \frac{700}{1100 + 0.87 \times 206.84} \times 33.5 = 20.87 i n c h \Rightarrow D e p t h o f a c t u a l n e u t r a l A x i s X_{u} = \frac{0.87 f_{y} A_{s t}}{0.36 f_{c} b} = \frac{0.87 \times 30000 \times 2.54}{0.36 \times 5000 \times 12} = 3.07 i n c h \Rightarrow X_{u} < X_{u l i m i t} s o s e c t i o n i s u n d e r r e i n f o r c e m e n t$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions