PROBLEM 4.51
4 in.
24 in.
Knowing that the bending moment in the reinforced concrete beam is
100 kip  ft and that the modulus of elasticity is 3.625  106 psi for the
concrete and 29  106 psi for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
20 in.
1-in.
diameter
2.5 in.
12 in.
SOLUTION
n
Es
29  106

 8.0
Ec
3.625  106

 
As  (4)   (1)2  3.1416 in 2
4
nAs  25.133 in
2
Locate the neutral axis.
 x
(24)(4)( x  2)  (12 x)    (25.133)(17.5  4  x)  0
2
96 x  192  6 x 2  339.3  25.133x  0
x
Solve for x.
or
6 x 2  121.133x  147.3  0
121.133  (121.133)2  (4)(6)(147.3)
 1.150 in.
(2)(6)
d3  17.5  4  x  12.350 in.
1
1
b1h13  A1d12 
(24)(4)3  (24)(4)(3.150) 2  1080.6 in 4
12
12
1 3 1
I 2  b2 x  (12)(1.150)3  6.1 in 4
3
3
I1 
I 3  nA3d32  (25.133)(12.350) 2  3833.3 in 4
I  I1  I 2  I 3  4920 in 4
 
(a)
Steel:
nMy
I
where M  100 kip  ft  1200 kip  in.
n  8.0
y  12.350 in.
s  
(b)
Concrete:
n  1.0,
c  
(8.0)(1200)(12.350)
4920
 s  24.1 ksi 
y  4  1.150  5.150 in.
(1.0)(1200)(5.150)
4920
 c  1.256 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
500
PROBLEM 4.52
7
8
16 in.
-in. diameter
2 in.
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 3  106 psi for the concrete and 29  106 psi for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
8 in.
SOLUTION
n
Es 29  10 6

 9.67
Ec
3  106
As  3
2

   7 
d 2  (3)     1.8040 in 2
4
 4  8 
nAs  17.438 in 2
x
 (17.438)(14  x)  0
2
4 x 2  17.438 x  244.14  0
8x
Locate the neutral axis:
Solve for x.
x
17.438  17.4382  (4)(4)(244.14)
 5.6326 in.
(2)(4)
14  x  8.3674 in.
I 
1 3
1
8 x  nAs (14  x)2  (8)(5.6326)3  (17.438)(8.3674)2  1697.45 in 4
3
3
 
nMy
I
 M
I
ny
n  1.0,
Concrete:
M 
M 
Choose the smaller value.
  1350 psi
(1350)(1697.45)
 406.835  103 lb  in.  407 kip  in.
(1.0)(5.6326)
n  9.67,
Steel:
y  5.6326 in.,
y  8.3674 in.,   20  103 psi
(20  103 )(1697.45)
 419.72 lb  in.  420 kip  in.
(9.67)(8.3674)
M  407 kip  in.
M  33.9 kip  ft 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
on a website, in whole or part.
501