n to the beam weight) that will cause the crack if they are used for 28-ft simple spa =75√, and reinforced concrete weight = oblem 2.6 447
n to the beam weight) that will cause the crack if they are used for 28-ft simple spa =75√, and reinforced concrete weight = oblem 2.6 447
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![For Problems 2.6 and 2.7, calculate the uniform load (in addi-
tion to the beam weight) that will cause the sections to begin
to crack if they are used for 28-ft simple spans. f = 4000 psi,
f = 7.5√, and reinforced concrete weight = 150 lb/ft².
Problem 2.6
447
Problem 2.7 (Ans. 0.343 k/ft)
349
12 in
21 in
-14 in.
4 in.
24 in
22 in 30 in
Transformed-Area Method
For Problems 2.8 to 2.14, assume the sections have cracked and
use the transformed-area method to compute their flexural stresses
for the loads or moments given.
Problem 2.8
17 in
20 in
3 in.
M = 60 ft-k
n=8
Problem 2.9 Repeat Problem 2.8 if four #6 bars are used.
(Ans. f = 1356 psi, f, = 26,494 psi)
Problem 2.10
849
6 #9
21 in.
-14 in
27 in
3 in.
Problem 2.11 (Ans. f = 1258 psi, f,= 14,037 psi in
bottom layer, f, = 12,889 psi at steel centroid)
18 in.
3 in..
M = 120-k
24 in.
3 in.
M-110-k](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa85fabed-e218-4822-9051-c34bf359d546%2F118851a6-6915-4d4d-b62a-c1209c75a9ca%2Fmk3mo_processed.png&w=3840&q=75)
Transcribed Image Text:For Problems 2.6 and 2.7, calculate the uniform load (in addi-
tion to the beam weight) that will cause the sections to begin
to crack if they are used for 28-ft simple spans. f = 4000 psi,
f = 7.5√, and reinforced concrete weight = 150 lb/ft².
Problem 2.6
447
Problem 2.7 (Ans. 0.343 k/ft)
349
12 in
21 in
-14 in.
4 in.
24 in
22 in 30 in
Transformed-Area Method
For Problems 2.8 to 2.14, assume the sections have cracked and
use the transformed-area method to compute their flexural stresses
for the loads or moments given.
Problem 2.8
17 in
20 in
3 in.
M = 60 ft-k
n=8
Problem 2.9 Repeat Problem 2.8 if four #6 bars are used.
(Ans. f = 1356 psi, f, = 26,494 psi)
Problem 2.10
849
6 #9
21 in.
-14 in
27 in
3 in.
Problem 2.11 (Ans. f = 1258 psi, f,= 14,037 psi in
bottom layer, f, = 12,889 psi at steel centroid)
18 in.
3 in..
M = 120-k
24 in.
3 in.
M-110-k
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