Instructor's Solutions Design of Concrete Structures – Arthur H. Nilson – 15th Edition

Instructor's Solutions Manual
to accompany
Design of Concrete Structures, 15th Edition
Nilson/Darwin/Dolan
The authors welcome feedback on the problem solutions and on the text in
general. Please e-mail any comments to David Darwin at: daved@ku.edu
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Name: Bahzad B.Sabr
Chapter (1) - Chapter (9)
B
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Notes regarding reinforcement details.
In the following problems, the reinforcement details are compressed to clarify the problem solutions. The
reinforcement details are noted by the line:
Reinforcement details
Reinforcement is shown by bar size. Thus a No. 9 (No. 29) bar is As9. Expansion of the "Reinforcement
details" line gives all bar sizes and diameters, thus:
Area of Reinforcing Bars
As3 0.11in
2
 As7 0.60in
2

As11 1.56in
2

As4 0.20in
2
 As8 0.79in
2

As14 2.25in
2

As5 0.31in
2
 As9 1.00in
2

As18 4.00in
2

As6 0.44in
2
 As10 1.27in
2

Diameter of reinforcing bars
db3 0.375in
 db7 0.875in
 db11 1.410in

db4 0.500in
 db8 1.000in
 db14 1.693in

db5 0.625in
 db9 1.128in
 db18 2.257in

db6 0.750in
 db10 1.27in

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1/3
1.1 The building in figure P.1 is used for general office space. The slab is 8 in. thick on a beam
12 in. wide by 18 in. deep, the bay dimensions are 18'-6” in the x direction and 21'-0” in the y
direction and the superimposed service dead load is 12 psf. Calculate the slab service load in psf
and the beam service load in klf.. (Solution: qs = 162 psf, wu = 3.12 klf).
SOLUTION
From table 1.1 Office load ql 50psf

Concrete unit weight γc 150pcf

t 8in
 qd t γc
 100 psf


Slab load
Superimprosed dead load qsdl 12psf

Service load qs ql qd
 qsdl
 162 psf


The beam length is 21 feet and the tributary width is 18.5 ft. The beam is 12 x 18 in. of which 10
in is below the slab.
wbeam 18in t

( ) 12
 in γc
 125 plf


ws 18.5ft ql
 18.5ft qd qsdl

 

 wbeam
 3122 plf

 ws 3.12 klf

1.2 The building in figure P.1 is used for general office space. The slab is 8 in. thick on a 12
in. wide x 18 in. deep beam, the bay dimensions are 18’-6” in the x direction and 21’-0” in the
y direction and the superimposed service dead load is 12 psf. Calculate the factored column
load transferred to column C3 on the 3rd
floor. (Solution: Pu = 86.4 kips).
SOLUTION γc 150pcf
 ql 50psf
 qsdl 12psf

Slab load
t 8in
 qd t γc
 100 psf


Tributary area At 18.5ft 21
 ft 388.5 ft
2


wbeam 18in t

( ) 12
 in γc
 125 plf


Pu 1.6 At
 ql
 1.2 At
 qd qsdl

 

 1.2 wbeam
 21
 ft
 86.4 kip


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2/3
1.3 The building in figure P.1 is used for general office space. The slab is 8 in. thick on beams
12 in. wide x 18 in. deep, the bay dimensions are 18’- 6” in the x direction and 21’- 0” in the y
direction and the superimposed service dead load is 12 psf. Calculate the slab factored load in
psf and the beam factored load in klf. Comment on your solution in comparison with problem
1.1.
γc 150pcf
 ql 50psf
 qsdl 12psf

SOLUTION
Slab load t 8in
 qd t γc
 100 psf


qu 1.6 ql
 1.2 qd qsdl

 

 214.4 psf


From problem 1.1
qu
qs
1.323

wbeam 18in t

( ) 12
 in γc
 125 plf


The tributaty width is . l 18.5ft

wu 1.6 ql
 l
 1.2 qd qsdl

 
 l

 wbeam
 4.09 klf


Compare to problem 1.1 wu
ws
1.31

This ratio is between 1.2 and 1.6 and suggests that the majority of the load comes from long
term loadings.
1.4 A slab in figure P.1 is used for lobby space. The slab is 10 in. thick on a 14 in. wide x
24 in. deep beam, the bay dimensions are 21’- 0” in the x direction and 26’- 0” in the y
direction and the superimposed service dead load is 15 psf. Calculate the slab factored
load in psf and the beam factored load in klf.
SOLUTION
b 14in
 h 24in
 t 10in
 qsdl 15psf

From Table 1.1 the lobbly live load is ql 100psf
 γc 150pcf

qslab γc t
 125 psf


wbeam h t

( ) b
 γc
 204 plf


qu 1.2 qslab qsdl

 
 1.6 ql

 328 psf


The beam tributary width length is l 21ft

wu qu l
 1.2wbeam
 7.13 klf


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3/3
1.5 The building in figure P.1 is used for light storage space. The slab is 10 in. thick on a 16 in. wide x 20
in. deep beam, the bay dimensions are 20’- 0” in the x direction and 25’- 0” in the y direction and the
superimposed sprinkler dead load is 4 psf. Calculate the slab factored load in psf and the beamfactored
load in klf.
SOLUTION
b 16in
 h 20in
 t 10in
 qsdl 4psf
 γc 150pcf

From Table 1.1 the light storage live load is ql 125psf

qslab γc t
 125 psf


wbeam h t

( ) b
 γc
 167 plf


qu 1.2 qslab qsdl

 
 1.6 ql

 355 psf


The beam tributary width length is l 20ft

wu qu l
 1.2wbeam
 7.30 klf


1.6 The roof on the building in figure P.1 has a slab 7 in. thick on a 12 in. wide x 16 in. deep beam, the
bay dimensions are 19’- 0” in the x direction and 21’- 0” in the y direction and the superimposed service
dead load is 6 psf. Calculate the slab factored load in psf and the beamfactored load in klf.
SOLUTION
b 12in
 h 16in
 t 7in
 qsdl 6psf

From Table 1.1 the roof live load is ql 20psf
 γc 150pcf

qslab γc t
 87.5 psf


wbeam h t

( ) b
 γc
 112 plf


qu 1.2 qslab qsdl

 
 1.6 ql

 144 psf


The beam tributary width length is l 18.5ft

wu qu l
 1.2wbeam
 2.80 klf


B
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1/2
3.1. A 16 × 20 in. column is made of the same concrete and reinforced with the same six No.
9 (No. 29) bars as the column in Examples 3.1 and 3.2, except t hat a steel with yield strength
f y = 40 ksi is used. The stress-strain diagram of this reinforcing steel is shown in Fig. 2.15 for
fy = 40 ksi. For this column determine ( a ) the axial load that will stress the concrete to 1200
psi; ( b ) the load at which the steel starts yielding; ( c ) the maximum load; and ( d ) the share
of the total load carried by the reinforcement at these three stages of loading. Compare results
with those calculated in the examples for f y = 60 ksi, keeping in mind, in regard to relative
economy, that the price per pound for reinforcing steels with 40 and 60 ksi yield points is about
the same.
20
16
As 6.0in
2

Ag 16in 20
 in 320 in
2



Ac Ag As
 314 in
2



f'c 4000psi
 fy 40000psi
 fy1 60000psi

Ec 3600000psi
 Es 29000000psi

n
Es
Ec
8.1


Part a The solution is identical for grade 40 and grade 60 reinforcement
fc 1200psi

P fc Ac n As


 
 434800 lbf

 Ps fc n
 As
 58000 lbf


Ps
P
0.133
 The steel carries 13.3 percent of the load
Part b
εy
fy
Es
0.00138

 εy1
fy1
Es
0.00207


For slow loading fc 3000psi
 fc1 3300psi

P1 Ac fc1
 As fy1

 1396200 lbf


P Ac fc
 As fy

 1182000 lbf


Ps1 As fy1
 360000 lbf


Ps As fy
 240000 lbf


Ps1
P1
0.258

Ps
P
0.203

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2/2
Problem 3.1
Part c
fc 3400psi

Pu1 Ac fc
 As fy1

 1427600 lbf


Pu Ac fc
 As fy

 1307600 lbf


Ps1 As fy1
 360000 lbf


Ps As fy
 240000 lbf


Ps1
Pu
0.275

Ps
Pu
0.184

Comments
1. There is no difference at fc = 1200 psi and elastic assumptions are used
2. As the strain increases, the steel with fy = 60,000 psi contributes more to the total load and
the column has a higher total capacity
3. Grade 40 and Grade 40 have the same cost, therefore Grade 60 provides a 9% increase in
capacity for no increase in cost.
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1/2
3.2 The area of steel, expressed as a percentage of gross concrete area, for the column
of Problem 3.1 is lower than would often be used in practice. Recalculate
the comparisons of Problem 3.1, using f y of 40 ksi and 60 ksi as before, but
for a 16 × 20 in. column reinforced with eight No. 11 (No. 36) bars. Compare
your results with those of Problem 3.1.
20
16
4‐No. 11 (No. 36)
As=12.48in2
4‐No. 11 (No. 36)
As11 1.56in
2

As 8 As11
 12.48 in
2



Ag 16in 20
 in 320 in
2



Ac Ag As
 307.52 in
2



f'c 4000psi
 fy 40000psi
 fy1 60000psi

Ec 3600000psi
 Es 29000000psi

n
Es
Ec
8.1


Part a The solution is identical for grade 40 and grade 60 reinforcement
fc 1200psi

P fc Ac n As


 
 489664 lbf

 Ps fc n
 As
 120640 lbf


Ps
P
0.246
 The steel carries 25 percent of the load
Part b
εy
fy
Es
0.00138

 εy1
fy1
Es
0.00207


For slow loading fc 3000psi
 fc1 3300psi

P1 Ac fc1
 As fy1

 1763616 lbf


P Ac fc
 As fy

 1421760 lbf


Ps1 As fy1
 748800 lbf


Ps As fy
 499200 lbf


Ps1
P1
0.425

Ps
P
0.351

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2/2
Problem 3.2
Part c
fc 3400psi
 both cases
Pu1 Ac fc
 As fy1

 1794368 lbf


Pu Ac fc
 As fy

 1544768 lbf


Ps1 As fy1
 748800 lbf


Ps As fy
 499200 lbf


Ps1
Pu
0.485

Ps
Pu
0.323

Comments
1. There is no difference at fc = 1200 psi and elastic assumptions are used
2. There is a 16% capacity increase at nominal using Grade 60 reinforcement
3. The higher steel ratio produces a higher overall capacity compared to problem 3.1
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3.3. A square concrete column with dimensions 22 × 22 in. is reinforced with
a total of eight No. 10 (No. 32) bars arranged uniformly around the column

perimeter. Material strengths are f y = 60 ksi and f c = 4000 psi, with stressstrain
curves as given by curves a and c of Fig. 3.3 . Calculate the percentages
of total load carried by the concrete and by the steel as load is gradually
increased from 0 to failure, which is assumed to occur when the concrete strain
reaches a limit value of 0.0030. Determine the loads at strain increments of
0.0005 up to the failure strain, and graph your results, plotting load percentages
vs. strain. The modular ratio may be assumed at n = 8 for these materials.
Using Concrete data from Figure 3.3
As = 10.12 in2
Ac = 474 in3
fy = 60000 psi
f'c = 4000 psi
Strain fc (psi) Pc (kips) fs (psi) Ps (kips) Ptotal (kips) Pc/Ptotal Ps/Ptotal
0.0000 0 0 0 0 0 0.00% 0.00%
0.0005 1600 758 14500 147 905 83.8% 16.2%
0.0010 2600 1232 29000 293 1526 80.8% 19.2%
0.0015 3100 1469 43500 440 1910 76.9% 23.1%
0.0020 3300 1564 58000 587 2151 72.7% 27.3%
0.0025 3400 1612 60000 607 2219 72.6% 27.4%
0.0030 3400 1612 60000 607 2219 72.6% 27.4%
0.00%
10.00%
20.00%
30.00%
40.00%
50.00%
60.00%
70.00%
80.00%
90.00%
0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035
Pc/Ptotal
Ps/Ptotal
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1/2
3.4. A 20 × 24 in. column is made of the same concrete as used in Examples 3.1
and 3.2. It is reinforced with six No. 11 (No. 36) bars with f y = 60 ksi. For
this column section, determine ( a ) the axial load that the section will carry at
a concrete stress of 1400 psi; ( b ) the load on the section when the steel begins
to yield; ( c ) the maximum load if the section is loaded slowly; and ( d ) the
maximum load if the section is loaded rapidly. The area of one No. 11 (No. 36)
bar is 1.56 in 2 . Determine the percent of the load carried by the steel and the
concrete for each combination.
Reinforcement Areas
Given Properties
f'c 4000psi
 fy 60000psi
 fc 1400psi
 n 8
 Es 29000000psi

Column Properties
b 20in
 t 24in
 Ag b t
 480 in
2


Ast 6As11 9.36 in
2


Part (a) Compute the axial capacity of the section loaded below the elastic limit.
Solution: The axial capacity is based on the gross area of the column plus the effective area of
the steel. Since we count the holes where the steel is removed, the additional effective area of
the steel is (n-1)Ast.
Ac Ag Ast


Ag 480 in
2

 Ast 9.36 in
2

 Ac 471 in
2


P fc Ag n 1

( ) Ast



 


 P 764 kip

 Concrete and steel
contribution
Pc fc Ag Ast

 

 Pc 659 kip


100
Pc
P
 86.3

Ps fc n
 Ast

 Ps 105 kip


100
Ps
P
13.7

Part (b): Compute the capacity of the column when the steel begins to yield εy
fy
Es

εy 0.00207
 or 2/10 of one percent
Examining Figure 3.3, we are beyond the elastic portion of the concrete stress strain
curve, but we are at the elastic limit of the steel.
fs εy Es


fs 60000 psi


From Figure 3.3 fc 3100psi
 for slow loading
Since the problem is nonlinear, we must break out the concrete and
steel areas. We can no longer use the elastic equation.
B
a
h
z
a
d

2/2
P fc Ac
 fs Ast


 P 2021 kip


100
Pc
P
 32.6

Pc fc Ac


Pc 1459 kip


Ps fs Ast

 100
Ps
P
27.8

Ps 562 kip


Part (c): Compute the maximum load capacity of the section if loaded slowly
Examining Figure 3.3, we are beyond the elastic portion of the concrete stress strain curve
and we are in the plastic range of the steel.
fs fy

fs 60000 psi


steel areas. We can no longer use the elastic equation from 1.1.
P fc Ac
 fs Ast


 P 2162 kip


Pc fc Ac


Pc 1600 kip


100
Pc
P
 74

Ps fs Ast


Ps 562 kip

 100
Ps
P
26

Part (d): If we reexamine the problem with a fast loading, then the concrete stress would be
fc 4000psi

P fc Ac
 fs Ast



P 2444 kip


Pc fc Ac


Pc 1883 kip

 100
Pc
P
 77

Ps fs Ast


Ps 562 kip


100
Ps
P
23

1. As the concrete becomes non-linear, the steel picks up more load, but after the steel
yields, the load goes to the conrete.
2. The slow loading is approximately 88% of the fast load scenario - This is slightly
higher than the 0.85 given in eq. 3.6.
B
a
h
z
a
d

1/3
3.5 A 24 in. diameter column is made of the same concrete as used in Examples
3.1 and 3.2. The area of reinforcement equals 2.1 percent of the gross cross
section (that is, A s = 0.021 A g ) and f y = 60 ksi. For this column section, determine
( a ) the axial load the section will carry at a concrete stress of 1200 psi;
( b ) the load on the section when the steel begins to yield; ( c ) the maximum
load if the section is loaded slowly; ( d ) the maximum load if the section is
loaded rapidly; and ( e ) the maximum load if the reinforcement in the column is
raised to 6.5 percent of the gross cross section and the column is loaded slowly.
Comment on your answer, especially the percent of the load carried by the
steel and the concrete for each combination.
Reinforcement Properties
Given Properties
f'c 4000psi
 fy 60000psi
 fc 1200psi
 n 8
 Es 29000000psi

Column Properties
d 24in
 Ag π
d
2
4

 ρ 0.021
  is the reinforcement ratio or the fraction of
the section that is steel
Ast ρ Ag


The total area of steel Ast is Ast 9.5 in
2


Part (a) Compute the axial capacity of the section loaded below the elastic limit.
Solution: The axial capacity is based on the gross area of the column plus the effective area of
the steel. Since we count the holes where the steel is removed, the additional effective area of
the steel is (n-1)Ast.
Ac Ag Ast


Ag 452 in
2

 Ast 9.50 in
2

 Ac 443 in
2


P fc Ag n 1

( ) Ast



 


 P 623 kip

 Concrete and steel
contribution
Pc fc Ag Ast

 

 Pc 531 kip


100
Pc
P
 85.4

Ps fc n
 Ast

 Ps 91 kip


100
Ps
P
14.6

Part (b): Compute the capacity of the column when the steel begins to yield εy
fy
Es

εy 0.00207
 or 2/10 of one percent
Examining Figure 1.16, we are beyond the elastic portion of the concrete stress strain
curve, and we are at the elastic limit of the steel.
fs εy Es


fs 60000 psi


 for slow
loading
B
a
h
z
a
d

2/3
P fc Ac
 fs Ast


 P 1943 kip


100
Pc
P
 27.4

Pc fc Ac


Pc 1373 kip


Ps fs Ast

 100
Ps
P
29.3

Ps 570 kip


Part (c): Compute the maximum load capacity of the section if loaded slowly
Examining Figure 1.16, we are beyond the elastic portion of the concrete stress strain curve
and we are in the plastic range of the steel.
fs fy

fs 60000 psi


P fc Ac
 fs Ast


 P 2076 kip


Pc fc Ac


Pc 1506 kip

 100
Pc
P
 72.5

Ps fs Ast


Ps 570 kip

 100
Ps
P
27.5

Part (d): If we reexamine the problem with a fast loading as would occur in a building, then the
concrete stress would be
fc 4000psi

P fc Ac
 fs Ast



P 2342 kip


Pc fc Ac


Pc 1772 kip


100
Pc
P
 75.7

Ps fs Ast


Ps 570 kip


Note: the total increase is in the concrte contribution. 100
Ps
P
24.3

Part (e): Determine the capacity for a slow loaded column with the steel changed to 6.5%
Ast 0.065 Ag

 Ast 29.4 in
2


fs fy

fs 60000 psi


P fc Ac
 fs Ast


 P 3270 kip


Pc fc Ac


Pc 1506 kip


100
Pc
P
 46

Ps fs Ast

 Ps 1764 kip

 100
Ps
P
54

B
a
h
z
a
d

3/3
Comments
1. As the concrete becomes non-linear, the steel picks up more load, but after the steel
yields, the load goes to the conrete.
2. The slow loading is approximately 88% of the fast load scenario - This is slightly
higher than the 85 percent given earlier.
B
a
h
z
a
d

4.1 Compare the cracking moment based on the gross section properties and the transformed
section properties with 4‐No. 11 (No. 36) bars in Figure P4.1a based on a concrete tensile capacity of
7.5vf’c.
f'c 4000psi
 Es 29000000
psi

24“
36
2‐½”
typ
a) f’c = 4000 psi
Ec 57000 f'c psi
 3.605 10
6
 psi


b 36in
 h 24in
 Ig
b h
3

12
41472in
4



Mcr
7.5 f'c psi

 Ig

h
2
1639 in kip




Calculation of transformed section
n
Es
Ec
 As 4 As11
 6.24 in
2


 n As
 50.2 in
2

 ys 2.5in

yb
b h
2

2
n As
 ys


b h
 n As


11.48in



Ig
b h
3

12
b h

h
2
yb







2

 n As
 yb ys

 2

 45754in
4



Mcrt
7.5 f'c psi

 Ig

yb
1891 in kip




Reinforcement results in a 15% increase in cracking moment.
Mcrt
Mcr
1.153

B
a
h
z
a
d

section properties with 2‐No. 10 (No. 32) bars in Figure P4.1b based on a concrete tensile capacity of
7.5vf’c.
3 6
1 2 f'c 5000psi
 Ec 57000 f'c psi
 4.031 10
6
 psi


Es 29000000
psi

b 12in
 h 36in

Ig
b h
3

12
46656in
4



Mcr
7.5 f'c psi
 Ig

h
2
1375 in kip




Calculate transformed section properties
n
Es
Ec
7.2

 As 2 As10
 2.54 in
2


 n As
 18.3 in
2

 ys 2.5in

yb
b h
2

2
n As
 ys


b h
 n As


17.37in



Ig
b h
3

12
b h

h
2
yb







2

 n As
 yb ys

 2

 50869in
4



Mcrt
7.5 f'c psi

 Ig

yb
1553 in kip




As
b h

0.0059

Mcrt
Mcr
1.13

Comment:
Including 0.6% reinforcment increased the cracking moment by 13%
B
a
h
z
a
d

section properties with 4‐No. 9 (No. 29) bars in Figure P4.1c based on a concrete tensile capacity of
7.5vf’c.
f'c 6000psi

28
8
8
c) f’c = 6000 psi
Es 29000000psi

b 28in
 h 36in
 bw 8in
 hf 8in

yb
b
hf
2
2
 bw h hf

 

h hf

 
2
hf









b hf
 bw h hf

 


13.0 in



Ec 57000 f'c psi
 4415 ksi



Ig
b hf
3

12
bw h hf

 3

12
 b hf
 yb
hf
2







2

 bw h hf

 

h hf

2
hf
 yb







2

 52117in
4



Mcr
7.5 f'c psi

 Ig

 
yb
2329 in kip



 Based on Gross Section
Calculations based on transformed section
n
Es
Ec
6.6

 As 4 As9
 4.00 in
2


 n As
 26.3 in
2

 ys 2.5in

yb
b
hf
2
2
 bw h hf

 

h hf

 
2
hf








 n As
 ys


b hf
 bw h hf

 

 n As


12.418in



Igc
b hf
3

12
bw h hf

 3

12
 b hf
 yb
hf
2







2

 bw h hf

 

h hf

2
hf
 yb







2

 52269in
4



Ig Igc n As
 yb ys

 2

 54853in
4



Ig
Igc
1.049
 Reinforcement adds 12% to Ig
Reinforcement adds about
10% to the cracking moment
calculation
Mcrt
7.5 f'c psi

 Ig

 
yb
2566 in kip




Mcrt
Mcr
1.102

B
a
h
z
a
d

section properties with 2‐No. 11 (No. 36) bars in Figure P4.1d based on a concrete tensile capacity
of 7.5vf’c.
3 6
3 6
8
4
d ) f’c = 3 00 0 p si
2 ‐½ ”
typ
Es 29000000psi

f'c 3000psi

Ec 57000 f'c psi
 3.122 10
6
 psi


b 36in
 h 36in
 bw 8in
 hf 4in

yb
bw
h hf

 2
2
 b hf
 h
hf
2









bw h hf

 
 b hf


22.48 in



Ig
bw h hf

 3

12
b hf
 3

12
 bw h hf

 
 yb
h hf

2







2

 b hf
 h
hf
2
 yb







2

 51897 in
4



Mcr
7.5 f'c psi

 Ig

yb
948 in kip




n
Es
Ec
9.3

 As 2 As11
 3.12 in
2


 n As
 29 in
2

 ys 2.5in

yb
bw
h hf

 2
2
 b hf
 h
hf
2








 n As
 ys


bw h hf

 
 b hf

 n As


21.13 in



Igc
bw h hf

 3

12
b hf
 3

12
 bw h hf

 
 yb
h hf

2







2

 b hf
 h
hf
2
 yb







2

 52626 in
4



Ig Igc n As
 yb ys

 2

 62685 in
4



Reinforcement increases the cracking
moment about 30 %.
Mcrt
7.5 f'c psi

 Ig

 
yb
1219 in kip




3.4 Transformed Section
Alternatively you can use the results from the gross section properties instead of working from scratch.
B
a
h
z
a
d

4.5 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 7.5vf’c.
3 6
3 6
8
4
d ) f’c = 3 00 0 p si
2 ‐½ ”
typ
Es 29000000
psi
 fc 300psi

f'c 3000psi

Ec 57000 f'c psi
 3.122 10
6
 psi


b 36in
 h 36in
 bw 8in
 hf 4in

yb
bw
h hf

 2
2
 b hf
 h
hf
2









bw h hf

 
 b hf


22.48in



Ig
bw h hf

 3

12
b hf
 3

12
 bw h hf

 
 yb
h hf

2







2

 b hf
 h
hf
2
 yb







2

 51897in
4



Mcr
7.5 f'c psi

 fc

  Ig

yb
1641 in kip




Mcr0
7.5 f'c psi

 Ig

yb
948 in kip




Mcr
Mcr0
1.73
 Adding 300 psi prestressing compression to the section increases the cracking
moment by nealy 75%
B
a
h
z
a
d

4.6 Determine the service level moment capacity of the sections in Figure P4.1 if the allowable stress
for concrete is 0.45 f’c and the allowable stress for the reinforcement is 30,000 psi. Use the areas of
reinforcement from Problem 4.2.
f'c 4000psi
 Es 29000000
psi

24“
36
2‐½”
typ
a) f’c = 4000 psi
Ec 57000 f'c psi
 3.605 10
6
 psi


b 36in
 h 24in
 Ig
b h
3

12
41472in
4



ys 2.5in

d h ys
 21.5 in



fc 0.45f'c 1800psi

 fs 30000psi

Calculation of cracked section
n
Es
Ec
 As 4 As11
 6.24 in
2


 n As
 50.2 in
2


ρ
As
b d

0.00806

 k ρ n

( )
2
2 ρ
 n

 ρ n

 0.301


j 1 k
 0.699


Msc
1
2
fc k
 d
 b
 d
k d

3







 4057 in kip




Mss As fs
 d
k d

3







 3621 in kip




Ms min Msc Mss

  3621 in kip




COMMENT:
The maximimn service load is the minimum value of either the concrete stress or the steel stress.
B
a
h
z
a
d

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’c and the allowable stress for the reinforcement is 30,000 psi. Use the
areas of reinforcement from Problem 4.2.
3 6
1 2 f'c 5000psi
 Ec 57000 f'c psi
 4.031 10
6
 psi


Es 29000000psi

b 12in
 h 36in

ys 2.5in

d h ys
 33.5 in



fc 0.45f'c 2250psi

 fs 30000psi

n
Es
Ec
7.2

 As 2 As10
 2.54 in
2


 n As
 18.3 in
2

 ys 2.5in

ρ
As
b d

0.00632

 k ρ n

( )
2
2 ρ
 n

 ρ n

 0.259


j 1 k
 0.741


Msc
1
2
fc k
 d
 b
 d
k d

3







 3591 in kip




Mss As fs
 d
k d

3







 2332 in kip




Ms min Msc Mss

  2332 in kip




COMMENT:
The maximimn service load is the minimum moment based on either the concrete stress or the
steel stress. In this example, the steel stress is the governing criteria.
B
a
h
z
a
d

4.8 Determine the service level moment capacity of the section in Figure P4.1c if the allowable
stress for concrete is 0.45 f’c and the allowable stress for the reinforcement is 30,000 psi. Use the
areas of reinforcement from Problem 4.3.
28
8
8
c) f’c = 6000 psi
f'c 6000psi

Ec 57000 f'c psi
 4415 ksi



Es 29000000
psi

b 28in
 h 36in
 bw 8in
 hf 8in

ys 2.5in

d h ys
 33.5 in



fc 0.45f'c 2700psi

 fs 30000psi

Calculate cracked section properties
n
Es
Ec
6.6

 As 4 As9
 4.00 in
2



ρ
As
bw d

0.01493

 k ρ n

( )
2
2 ρ
 n

 ρ n

 0.355


j 1 k
 0.645


Msc
1
2
fc k
 d
 bw
 d
k d

3







 3798 in kip




Mss As fs
 d
k d

3







 3544 in kip




Ms min Msc Mss

  3544 in kip




COMMENT:
The maximimn service load is the minimum moment from either the concrete stress or the steel
stress. In this example, the steel stress is critical.
B
a
h
z
a
d

4.9 Determine the nominal moment capacity of the sections in in Fig. 4.1a using the
reinforcement areas from problem 4.1. The reinforcement yield stress is 60,000 psi.
f'c 4000psi

24“
36
2‐½”
typ
a) f’c = 4000 psi
fy 60000psi

b 36in
 h 24in

ys 2.5in

d h ys
 21.5 in



Calculation of Nominal capacity
As 4 As11
 6.24 in
2



a
As fy

0.85f'c b

3.06 in



Mn As fy
 d
a
2







 7477 in kip




B
a
h
z
a
d

4.10 Determine the nominal moment capacity of the sections in in Fig. 4.1b using the
3 6
1 2 f'c 5000psi

fy 60000psi

b 12in
 h 36in

ys 2.5in

d h ys
 33.5 in



Calculate Nominal moment capacity
As 2 As10
 2.54 in
2



a
As fy

0.85f'c b

2.99 in



Mn As fy
 d
a
2







 4878 in kip




B
a
h
z
a
d

4.11 Determine the nominal moment capacity of the sections in in Fig. 4.1c using the
28
8
8
c) f’c = 6000 psi
f'c 6000psi

fy 60000psi

b 28in
 h 36in
 bw 8in
 hf 8in

ys 2.5in

d h ys
 33.5 in



β1 0.75

Calculation of Nominal moment capacity
As 4 As9
 4.00 in
2


 ρ
As
bw d

0.015


a
As fy

0.85f'c bw

5.88 in


 c
a
β1
7.843in



c
d
0.234

Mn As fy
 d
a
2







 7334 in kip




B
a
h
z
a
d

4.12 Determine the nominal moment capacity of the sections in in Fig. 4.1d using the
3 6
3 6
8
4
d ) f’c = 3 00 0 p si
2 ‐½ ”
typ
f'c 3000psi

fy 60000psi

b 36in
 h 36in
 bw 8in
 hf 4in

ys 2.5in

d h ys
 33.5 in



Calculate Nominal moment capacity
As 2 As11
 3.12 in
2



a
As fy

0.85f'c b

2.04 in


 a is less than hf, therefore this section can be treated as a rectangular
section.
Mn As fy
 d
a
2







 6080 in kip




B
a
h
z
a
d

4.13 Determine the required area of reinforcement and the corresponding reinforcement ratio for
the sections in Fig P4.1a if the ultimate moment is a) 10,000 in‐kip and b) 5,000 in‐kip. The
reinforcement yield strength is 60,000 psi.
24“
36
2‐½”
typ
a) f’c = 4000 psi
f'c 4000psi
 ϕ 0.90

fy 60000psi
 β1 0.85

b 36in
 h 24in

ys 2.5in

d h ys
 21.5 in



Mu1 10000in kip


Mu2 5000in kip


Try: a1 5in

Calculation of As for Mn1
As1
Mu1
ϕ fy
 d
a1
2








9.75 in
2


 n
As1
As10
7.674

 Try 8 No. 10 (No. 32)
bars
Check
As 8 As10
 10.16 in
2



a
As fy

0.85f'c b

4.98 in



c
a
β1
5.859 in



c
d
0.273
 .375 OK
ϕMn ϕ As
 fy
 d
a
2







 10430 in kip



 Mu1 10000 in kip

 Solution, use 8 No.
10 (No. 32) bars
Calculation of As for Mn2 Try: a1 3in

As2
Mu2
fy d
a1
2








4.17 in
2


 n
As2
As11
2.671

 Try 3 No. 11 (No. 36) bars
Check As 3 As11
 4.68 in
2



a
As fy

0.85f'c b

2.29 in



Solution, use 3 No. 11 (No.
36) bars
4-No. 10 also is OK but
less efficient
ϕMn ϕ As
 fy
 d
a
2







 5144 in kip



 Mu2 5000 in kip


B
a
h
z
a
d

4.14 . Determine the required area of reinforcement and the corresponding reinforcement
ratio for the section in Fig P4.1 b if the ultimate moment is (a)
7,000 in-kips (b) 3500 in-kips. f y = 60,000 psi.
3 6
1 2 f'c 5000psi
 ϕ 0.90

fy 60000psi

β1 0.80

b 12in
 h 36in

ys 2.5in

d h ys
 33.5 in



Mu1 7000in kip


Mu2 3500in kip



As1
Mu1
ϕ fy
 d
a1
2








4.32 in
2



n
As1
As11
2.77

 Try 3 No. 11 (No. 36) bars
Check
As 3 As11
 4.68 in
2



a
As fy

0.85f'c b

5.51 in



c
a
β1
0.175m


ϕMn ϕ As
 fy
 d
a
2







 7770 in kip




c
d
0.205
 0.375 OK
Mu1 7000in kip

 Solution, use 3 No. 11 (No. 36) bars

As2
Mu2
ϕ fy
 d
a1
2








2.03 in
2


 n
As2
As9
2.025

 Try 2 No. 9 (No. 29) bars
Check As 2 As9
 2 in
2



a
As fy

0.85f'c b

2.35 in



Solution, use 2 No. 9
(No.29) bars and solution
is within 1%
ϕMn ϕ As
 fy
 d
a
2







 3491 in kip



 = Mu2 3500in kip


B
a
h
z
a
d

4.15 Determine the required area of reinforcement and the corresponding reinforcement ratio for the
sections in Fig P4.1c if the ultimate moment is a) 10,000 in‐kip, b) 5,000 in‐kip. The reinforcement
yield strength is 60,000 psi.
f'c 6000psi
 β1 0.75

28
8
8
c) f’c = 6000 psi
fy 60000psi

ϕ 0.90

b 28in
 h 36in
 bw 8in
 hf 8in

ys 2.5in

β1 0.75

d h ys
 33.5 in



Mu1 10000in kip


Mu2 5000in kip



As1
Mu1
ϕ fy
 d
a1
2








6.28 in
2


 n
As1
As10
4.943

 Try 5 No. 10 (No. 32)
bars
Check
As 2 As10
 3As11
 7.22 in
2



a
As fy

0.85f'c bw

10.62in


 c
a
β1
14.2in


c
d
0.423
 0.375 must adjust ϕ
εt 0.003
d c

c
 0.0041

 0.004 OK
ϕ 0.9 0.25
0.005 εt

0.003







 0.825


ϕMn ϕ As
 fy
 d
a
2







 10074in kip



 Mu1 10000in kip

 Solution, use 2 No. 10 (No. 32) plus
3 No. 11 (No. 36) bars

As2
Mu2
ϕ fy
 d
a1
2








3.21 in
2


 n
As2
As11
2.056

 Try 2 No. 11 (No. 36) bars
Check As 2 As11
 3.12 in
2



a
As fy

0.85f'c b

1.31 in



ϕMn ϕ As
 fy
 d
a
2







 5072 in kip



 Mu2 5000in kip

 Solution, use 2 No. 11 (No. 36) bars
B
a
h
z
a
d

4.16. Determine the required area of reinforcement and the corresponding reinforcement ratio for the
section in Fig P4.1 d if the ultimate moment is (a) 10,000 in-kips and (b) 5000 in-kips. f y = 60,000
psi. Comment on your solutions.
3 6
3 6
8
4
d ) f’c = 3 00 0 p si
2 ‐½ ”
typ
f'c 3000psi
 β1 0.85

fy 60000psi
 ϕ 0.90

b 36in
 h 36in
 bw 8in
 hf 4in

ys 2.5in

d h ys
 33.5 in



SOLUTION
Mu1 10000in kip

 Mu2 5000in kip


Assume that the compressin block is in the flange. Let a hf 4 in



As1
Mu1
ϕ fy
 d
a
2








5.879in
2


 As2
Mu2
ϕ fy
 d
a
2








2.939in
2



Try No. 11 (No. 36) bars Try No. 11 (No. 36) bars
n1
As1
As11
3.769

 n2
As2
As11
1.884


Use As1 4As11 6.24in
2

 Use As2 2As11 3.12in
2


Check assumption Check assumption
a1
As1 fy

0.85f'c b

4.078in


 a2
As2 fy

0.85f'c b

2.039in



This is close enought to 4 to be acceptable This is less than 4 OK
c
a1
β1
4.798in

 c
d
0.143
 0.375 OK
ϕMn2 ϕ As2
 fy
 d
a2
2







 5472 in kip




ϕMn1 ϕ As1
 fy
 d
a1
2







 10601in kip




Mu2 5000in kip

 OK
Mu1 10000in kip


OK
B
a
h
z
a
d

1/2
Problem 4.17 A rectangular beam made using concrete with f’c = 6000 psi and steel with fy = 60,000 psi
had a width b = 20 in., and an effective depth of d = 17.5 in and an h =20 in. The Concrete modulus of
rupture fr = 530 psi. The elastic modulus of the steel and concrete are, respectively Ec = 4,030,000 psi
and Es = 29,000,000 psi. The area of steel is four No. 11(No. 36) bars.
(a) Find the maximum service load that can be resisted without stressing the concrete above 0.45 f’c
or the steel above 0.40 fy.
(b) Determine if the beam will show cracking before reaching the service load
(c) Compute the nominal moment capacity of the beam
(d) Compute the ratio of the nominal capacity of the beam to the maximum service level capacity
and compare your findings to the ACI load factors and strength reduction factor.
Reinforcement sizes
Given data
As 4 As11

 As 6.24 in
2

 Es 29000000psi

b 20in
 d 17.5in
 h 20in

f'c 6000psi
 fy 60000psi
 Ec 57000 f'c psi

 Ec 4415 ksi


fr 7.5 f'c psi

 fr 581 psi
 n
Es
Ec
 n 6.6

(a) Find the maximum service load that can be resisted without stressing the concrete
above 0.45 f'c or the steel above 0.40 fy.
fc 0.45f'c
 fc 2700 psi

fs 0.60fy
 fs 36000 psi

ρ
As
b d

 ρ 0.018

k ρ n

( )
2
2ρ n

 ρ n


 k 0.381

j 1
k
3

 j 0.873

Moment due to concrete limits
Msc
1
2
fc
 b
 k
 d
 d
k d

3









Msc 229 ft kip



Moment due to steel limit
Mss As fs
 j
 d


Mss 286 ft kip



The maximum service moment is the minimum of the two values.
B
a
h
z
a
d

2/2
Ms min Mss Msc

 
 Ms 229 ft kip



(b) Determine if the beam will show cracking before reaching the service load
Ig
b h
3

12
 Ig 13333 in
4


Mcr1
fr Ig

h
2
 Mcr1 64.5 ft kip



This is less than the service load so the section cracks. To demonstrate that the transformed
section does not affect this conclusion, the following checks the cracked transformed section.
Δy
n As
 d
h
2








n As
 b h



Δy 0.697 in


Iut Ig b d
 Δy
2

 n As
 d
h
2
 Δy







2



Iut 15400 in
4


Mcr2
fr Iut

h
2
Δy


Mcr2 80.1 ft kip


 Mcr2
Mcr1
1.242

(c) Determine the nominal moment capacity of the section.
a
As fy

0.85f'c b

 a 3.67 in


Mn As fy
 d
a
2









Mn 489 ft kip



(d) Compute the ratio of the nominal capacity of the beam to the maximum service level capacity and
compare your findings to the ACI load factors and strength reduction factor.
Ratio
Mn
Ms
 Ratio 2.13

First, the extra calculation of the uncracked transformed area gives only a 18% increase in the
cracking moment. Comparing the cracking moment to the service moment shows that the service
moment is almost 3 time the cracking moment. Therefore, unless the service moment is very
close to the service moment, you can be assured that the section will crack based on the gross
section calculation.
Second, the margin of safety between the service moment and the nominal capacity is 2.11. This
is greater than the ultimate load factors and phi factors from ASCE-7 and ACI (1.6/0.9 =1.78 if the
entire load is classified as live load) indicating that this service level design is more conservative
than LRFD design.
B
a
h
z
a
d

4.18. A rectangular, tension-reinforced beam is to be designed for dead load of 500 lb/ft plus
self-weight and service live load of 1200 lb/ft, with a 22 ft simple span. Material strengths will be fy =
60 ksi and fc・= 3 ksi for steel and concrete, respectively. The total beam depth must not exceed 16
in. Calculate the required beam width and tensile steel requirement, using a reinforcement
ratio of 0.60 ρ0.005 . Use ACI load factors and strength reduction factors. The
effective depth may be assumed to be 2.5 in. less than the total depth.
16
b
2‐½ ”
typ
wd 500plf
 f'c 3000psi
 β1 0.85

εu 0.003

wl 1200plf
 fy 60000psi

L 22ft

SOLUTION
Assume girder dead load and check
assumption at the end of the problem.
wo 500plf
 h 16in

wu 1.2 wd wo

 
 1.6 wl

 3120plf


Calculate the maximum reinforcement ratio
and then 60% of that ratio.
Mu
wu L
2

8
188.8ft kip



ρ005 0.85β1
f'c
fy

εu
εu 0.005







 0.014


ρ 0.60ρ005 0.0081


Calculate the resistance factor needed for the reinforcement ratio.
R ρ fy
 1 0.588ρ

fy
f'c








 441psi

 Because ρ is 0.6ρ0.005, it is reasonable to assume ϕ 0.90

bd2
Mu
ϕ R

5.706 10
3
 in
3


d h 2.5in
 13.5in

 Solve for b
b
bd2
d
2
31.309in


Use b = 32 in. and then check assumptions. wo
150 pcf

144
in
2
ft
2
b
 d
 440.3plf


This is less than assume, OK
As ρ b
 d
 3.44in
2


Use 3 As10
 3.81in
2
 b=32 in and d= 16 in.
B
a
h
z
a
d

1/2
Problem 4.19 A rectangular reinforced concrete section has dimension b=14 in., d=25 in, and h =
28 in., and is reinforced with 3 No. 10 (No. 32) bars. The material strengths are f'c = 5000 psi, fy =
60,000 psi.
(a) Find the moment that will produce first cracking at the bottom surface of the section
basing your calculations on Ig, the moment of inertial of the gross section.
(b) Repeat the calculation using Iut, the uncracked transformed moment of inertia.
(c) Determine the maximum moment that can be carried without the concrete stress
exceeding 0.45 f'c or the steel stress exceeding 0.60 fy.
(d) Determine the nominal moment capacity of the section.
(e) Compute the ratio of nominal moment capacity from part (d) to the service level moment
from part (c)
(f) Comment on your results with particular attention to comparing parts (a) and (b) and
comparing part (e) to established load factors.
Reinforcement sizes
Given data
As 3 As10

 As 3.81 in
2

 Es 29000000psi

b 14in
 d 25in
 h 28in

f'c 5000psi
 fy 60000psi
 Ec 57000 f'c psi

 Ec 4031 ksi


fr 7.5 f'c psi

 fr 530 psi
 n
Es
Ec
 n 7.2

(a) Find the moment that will produce first cracking at the bottom surface of the section
basing your calculations on Ig, the moment of inertial of the gross section.
Ig
b h
3

12
 Ig 25611 in
4

 Mcr1
fr Ig

h
2



(b) Repeat the calculation using Iut, the uncracked transformed moment of inertia.
Δy
n As
 d
h
2








n As
 b h


 Δy 0.719 in


Iut Ig b d
 Δy
2

 n As
 d
h
2
 Δy







2


 Iut 28689 in
4


Mcr2
fr Iut

h
2
Δy



 Mcr2
Mcr1
1.181

(c) Determine the maximum moment that can be carried without the concrete stress exceeding
0.45 f'c or the steel stress exceeding 0.60 fy.
fc 0.45f'c
 fc 2250 psi

B
a
h
z
a
d

2/2
fs 0.60fy
 fs 36000 psi

ρ
As
b d

 ρ 0.011
 k ρ n

( )
2
2ρ n

 ρ n


 k 0.325
 j 1
k
3

 j 0.892

Moment due to concrete limits Moment due to steel limit
Msc
1
2
fc
 b
 k
 d
 d
k d

3








 Msc 238 ft kip


 Mss As fs
 j
 d

 Mss 255 ft kip



The maximum service moment is the minimum of the two values.
Ms min Mss Msc

 
 Ms 238 ft kip



(d) Determine the nominal moment capacity of the section.
a
As fy

0.85f'c b

 a 3.84 in

 Mn As fy
 d
a
2








 Mn 440 ft kip



(e) Compute the ratio of nominal moment capacity from part (d) to the service level moment from
part (c)
Ratio
Mn
Ms
 Ratio 1.85
 0.9 Ratio
 1.66

Ratio1
Ms
Mcr1
 Ratio1 2.942

(f) Comment on your results with particular attention to comparing parts (a) and (b) and comparing
part (e) to established load factors.
First, the extra calculation of the uncracked transformed area gives an 18% increase in the
cracking moment. Comparing the cracking moment to the service moment, Ratio1, shows that
the service moment is almost 3 time the cracking moment. Therefore, unless the service moment
is very close to the service moment, you can be assured that the section will crack.
Second, the margin of safety between the service moment and the nominal capacity is 1.8, 1.6 if
a  factor is included. This is greater than the ultimate load factors from ASCE-7 indicating that
this service level design is more conservative than LRFD design.
B
a
h
z
a
d

1/2
4.20. A singly reinforced rectangular beam is to be designed, with effective depth approximately
1.5 times the width, to carry a service live load of 2000 lb/ft in addition to its own weight, on a 24
ft simple span. The ACI Code load factors are to be applied as usual. With f y = 60,000 psi and fc
= 4000 psi, determine the required concrete dimensions b, d, and h, and steel reinforcing bars ( a )
for ρ = 0.60 ρ 0.005 and ( b ) for ρ = ρ 0.005 . Include a sketch of each cross section drawn to
scale. Allow for No. 4 (No. 13) stirrups. Comment on your results.
Given properties
fy 60000psi
 f'c 4000psi
 l 24 ft

 wl 2000
lbf
ft

 wc 150
lbf
ft
3


Estimate beam dimensions for self weight
determination b
d
=
1.5
b
b 12 in

 d 18 in

 h 21 in


wo b h
 wc


wu 1.2 wo
 1.6 wl


 wu 3.515
kip
ft


Mu wu
l
2
8

 Mu 253.1 kip ft



β1 0.85
0.05 f'c 4000 psi


 
1000 psi

 0.85


ρmax 0.85 β1

f'c
fy

0.003
0.003 0.004

 0.0206


a) For  = 0.6 max, find the section properties to carry
Mu.
ρ 0.6 ρmax

 ρ 0.0124
 From table R ρ fy
 1 0.588
ρ fy

f'c










R 662 psi

 Since 0.005, then ϕ 0.90

b
3
Mu
2.25 ϕ
 R

13.134 in


 d 1.5 b

 d 19.7 in


As ρ b
 d

 As 3.2 in
2


A solution is to use 4-#8 , b= 14, d= 20.5, and h=22.
As 4 0.79
 in
2
 b 14 in

 d 20.5 in

 giving a design capacity of
a
As fy

0.85 f'c
 b

3.98 in



ϕMn ϕ As
 fy
 d
a
2








 ϕMn 263 ft kip


 Mu Mu 253.1 ft kip


 OK
B
a
h
z
a
d

2/2
b)
ρ005 0.85 β1

f'c
fy

0.003
0.003 0.005



ρ ρ005
 ρ 0.0181

R ρ fy
 1 0.588 ρ

fy
f'c









 R 911 psi


b
3
Mu
1.5
2
ϕ
 R

11.8 in


 d 1.5 b

 d 17.7 in


As ρ b
 d

 As 3.78 in
2


This is satisfied by 4-#9 As = 4.0 in2. Beam dimensions would be b= 12, d= 18 and h=21
(or more when two layers of steel are used). The increase in depth accounts for the smaller area of
steel.
Check:
b 12 in

 d 18 in

 As 4 1.0
 in
2
4.0 in
2



a
As fy

0.85 f'c
 b

 a 5.88 in


ϕMn ϕ As
 fy
 d
a
2








 ϕMn 271.1 ft kip


 Mu OK
Comment: While there is a small savings in concrete using solution b, the final selection
of reinforcement can often result in a reinforcement ratio greater than the code allows. In
which case, the solution in a) is preferable.
B
a
h
z
a
d

1/2
4.21   A four span continuous beam of constant rectangular cross‐section is supported
at A, B, C, D, and E.  The factored moments resulting from analysis are
At Supports, ft‐kip At midspan ft‐kip
Ma = 138
Mb = 220
Mc = 200
Md = 220
Me = 138
Mab = 158
Mbc = 138
Mcd = 138
Mde = 158
Determine the required final concrete dimensions for this beam using d=1.75 b
and determine the reinforcement requirements at each critical moment section.
Your final reinforcement ratio should not exceed = 0.6 005, fy = 60,000 psi and
f’c = 6000 psi
Reinforcement Details
Given Properties
f'c 6000psi
 fy 60000psi
 εu 0.003

β1 0.85 0.05
f'c 4000psi

1000psi

 0.75

 ϕ 0.90

Solution Approach:
Solve the first section using the Resistance Factors. Then, having selected a section, determine
the reinforcement requirements for the remaining sections. Choose reinforcement bars to meet
your criteria.
Mnreqd
220ft kip

ϕ
 Mnreqd 244 ft kip



ρ005
0.85 β1
 f'c
fy
εu
εu 0.005







 0.024

 ρ 0.6ρ005 0.014


R ρ fy
 1 .588
ρ fy

f'c








 R 788 psi

b
3
Mnreqd
1.75
2
R

10.7 in



d 1.75 b
 18.7 in



As b d
 ρ
 2.86 in
2



n
As
As8
3.6


Try the following dimensions and then check the solution
b 10in
 h 22in
 d h 4in

 Additional cover is required because the bars have to be in
two layers, See Table A.7 in text.
As 4 As8

 As 3.16 in
2

 Note: 3#10 (No. 36) also works and allows the steel to be in one
layer but require more steel.
B
a
h
z
a
d

2/2
a
As fy

0.85 f'c
 b

3.72 in



Mn1 As fy
 d
a
2








 Mn1 255 ft kip



Mn1
Mnreqd
1.04

OK
The next largest moment is 200 ft-kip. This is 10 percent less than the first interior joint so no
adjustment of reinforcement is needed.
The largest positive moment is 158 ft-kip. This is 72 percent of the maximum moment so the steel
requirements will be approximately:
As2 0.72As 2.28 in
2



n2
As2
As8
 n2 2.88
 use: As 3As8

From Table A.7 use 3-#8 bars in one level
d2 h 2.5in


a
As fy

0.85f'c b

2.79 in


 Mnab As fy
 d
a
2







 197 ft kip




Check total Capacity ϕ Mnab

158ft kip

1.12
 OK
Design end negative moment at A and E
As2
138
158
As
 2.07 in
2


 Use 3#8 As 3As8 2.37 in
2



a
As fy

0.85f'c b

2.79 in



Mn2 As fy
 d
a
2








 Mn2 197 ft kip



Check total Capacity ϕ Mn2

158ft kip

1.12
 OK
4#8
In two layers
3-#8 3-#8
3-#8
10
22
B
a
h
z
a
d

1/3
Problem 4.22 A two span continuous beam is supported on three concrete walls spaced 30 ft.
on centers. A service live load of 1.5 kip/ft is to be carried in addition to the self weight of the
beam and is to be applied in a pattern loading. The dimensions of the beam should be
approximately d=2 b, and the reinforcement is to be varied according to the demand. Determine
concrete dimensions at all critical sections but select a constant section for the beam. Allow for
No. 4 (No. 13) stirrups. Use a span to depth ratio of 15 for your first estimate of the depth.
Adjust the depth if the reinforcement ratio is too high. Include sketches drawn to scale. Use f'c
= 6000 psi and fy = 60,000 psi.
Reinforcement Details
Given Properties
f'c 6000psi
 fy 60000psi
 ϕ 0.90
 β1 0.85 0.05
f'c 4000psi

1000psi

 0.75


wl 1.5
kip
ft
 L 30ft
 γc 150pcf

Solution: Begin by estimating a beam depth and width, then compute the girder load.. Compute the
maximum negative moment due to both spans being loaded then the maximum positive moment
with only one span loaded with live load. Try:
h
L
15
24 in


 d h 2.5in
 21.5 in


 b
h
2
12.0 in



wg γc b
 h
 0.30
kip
ft


 wu 1.2wg 1.6wl
 2.76
kip
ft



Mneg
wu L
2

8
310 ft kip




By trial estimate a, then compute As for the negative moment over the support, select final As and
check Mn
a 4.0in
 As
Mneg
ϕ fy
 d
a
2








3.54 in
2


 Try 3-#10 As 3 As10
 3.81 in
2



a
As fy

0.85 f'c
 b

3.74 in


 Mn As fy
 d
a
2







 374 ft kip



 ϕ Mn
 337 ft kip


 OK
c
a
β1
4.98 in



c
d
0.232
 3/8 therefore phi = 0.9 is OK
By trial, compute the required As for the positive moment condition.
The maximum positive moment occurs with dead load on both spans and live load on one span
only. The maximum moment for the DL occurs at 3/8L in from the end and the maximum moment
for live load occurs 7/16L in from the end. Conservatively, take the sum of the two conditions and
apply them at the same location.
B
a
h
z
a
d

2/3
Mpos
49 1.6
 wl L
2

512
9 1.2
 wg
 L
2

128
 229 ft kip




a 3.0in
 As
Mpos
ϕ fy
 d
a
2








2.55 in
2


 Try 3-#9 As 3 As9
 3.00 in
2



a
As fy

0.85 f'c
 b

2.94 in


 Mn As fy
 d
a
2







 300 ft kip



 ϕ Mn
 270 ft kip


 OK
c
a
β1
3.92 in



c
d
0.182
 3/8 therefore phi = 0.9 is OK
12
24
21.5
21.5
3-#9 (#29)
3-#10 (#32)
Section at midspan Section at interior support
#3 (# 10)
stirrup, typ
B
a
h
z
a
d

3/3
Comments on Problem 4.22
Load
Shear
Moment
3/8l
3/8 wl
l
7/16 wl
7/16l
M+=½ *3/8 * 3/8 *wl2
=9wl
2
/128
M+=½ *7/16 * 7/16 *wl2
=49wl
2
/512
The combined effect of the two loadings with wD continuous and wl on one span
The effective location of the maximum moment is then:
(3/8 wDl +7/16 wll)l
x’ =‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐
wd+wl
Both wd and wl are factor loads so the maximum positive moment is:
M+ = ½(7/16 wl + 3/8 wd)lx’
This will be greater than placing the full dead and live load on both spans. While placing the live load on
both spans is a quick solution, this problem is intended to point out the effects of checkerboard loading
in addition to concrete design.

B
a
h
z
a
d

1/2
4.23. A rectangular concrete beam of width b = 24 in. is limited by architectural considerations to a
maximum total depth h = 16 in. It must carry a total factored load moment M u = 400 ft-kips. Design
the flexural reinforcement for this member, using compression steel if necessary. Allow 3 in. to the
center of the bars from the compression or tension face of the beam. Material strengths are fy =
60,000 psi and fc''・= 4000 psi. Select reinforcement to provide the needed areas, and show a
sketch of your final design, including provision for No. 4 (No. 13) stirrups.
16
24 2.5 typ. top and
bottom
f'c 4000psi
 fy 60000psi

Mu 400kip ft


d 16in 2.5in
 13.5 in



d' 3in

b 24in


εu 0.003

Es 29000000psi
 β1 0.85

SOLUTION
First try a singly reinforced beam with the maximum reinforcement ratio
ρmax 0.0181
 for ρ0.005 to assure ϕ 0.90

As ρmax b
 d
 5.86 in
2



a
As fy

0.85f'c b

4.312 in



c
a
β1
5.07 in



Mn As fy
 d
a
2







 332.6 kip ft




The capacity is less than the applied load, therefore compression reinforcement must be provided.
From table 4.2 the compression reinforcement will be less than the yield stress.
f's Es εu

c d'

d
 13359 psi


M'
Mu
ϕ
Mn
 111.8 ft kip




A's
M'
f's d d'

( )

9.57 in
2


 Try No. 11 (No. 36) bars n
A's
As11
6.132


Use 7 No. 11 (No. 36) bars
A's 7 As11
 10.92 in
2



B
a
h
z
a
d

2/2
Positive Moment reinforcement
Asp As A's
f's
fy

 8.296 in
2



Try No. 10 (No. 32) bars n
Asp
As10
6.532


Use 7 No. 10 (No.32) bars.
Because the positive moment steel provided is slightly more than that needed for rho max, check
the net tensile strain and strength reduction factor.
Try
As 7 As10
 8.89 in
2


 Assume f's 16ksi

a
As A's
f's
fy








fy

0.85f'c b

4.396 in


 c
a
β1
5.171 in



Check f's
f's Es εu

c d'

d






 13993 psi

 OK
εt εu
d c

c






 4.832 10
3



 0.004 still OK
ϕ 0.9 0.25
0.005 εt

0.001







 0.858


ϕMn ϕ fy
 As A's
f's
fy








 d
a
2







 ϕ A's
 f's
 d d'

( )

 422.2 ft kip




This is a classic problem of compression reinforced shallow beams. Finding a combination of positive
and negative reinforcement that still meets the net tensile strain requirement and the nominal moment
capacity by trial is time consuming. The inefficiency of the compression reinforcement is also seen
since more compression reinforcement than tension reinforcement is required.
7‐No. 11 (No. 36)
7‐No. 10 (No. 32)
B
a
h
z
a
d

1/2
Problem 4.24 For the beam with a triangular cross section shown in Figure P4.24,
determine a) the balanced reinforcement ratio and b) the maximum reinforcement ratio if
t =0.005. The dimensions of the triangle are such that the width of the triangle equals
the distance from the apex. The width at the effective width b equals the effective depth
d. Draw the strain distribution, stress distribution, and define your notation.
b
c
d
As strain stress
SOLUTION
a) Find the balanced reinforcement ratio
From Equilibrium
C=T
0.85f’c (1/2 a ba) = Asfy
Substitution As =  b d, where b = d and ba = a gives
0.85fc’ a2
/2 =  d2
fy
Solve for 
2
)
(
'
2
85
.
0
d
a
f
f
y
c


The relationship between a and c is a = 1c. From the relationship of plane
sections remain plane, the correlation of c/d is
t
u
u
d
c





0.85fc’
eu = 0.003
a
d c
As
t fs
C
T
B
a
h
z
a
d

2/2
Substitution for a and then c/d gives
2
2
1 )
(
'
2
85
.
0
t
u
u
y
c
f
f







For the balanced condition, t = y and b is
2
2
1 )
(
'
2
85
.
0
y
u
u
y
c
b
f
f







Using grade 60 steel, so y is 0.002 and u = 0.003, the balanced ratio becomes.
y
c
b
f
f '
153
.
0 2
1

  =Solution for balanced ratio
b) Find the reinforcement ratio for t = 0.005.
For t = 0.005 and u = 0.003 the reinforcement ratio is
y
c
f
f '
060
.
0 2
1
005
.
0 
  =Solution for reinforcement ratio when t = 0.005
B
a
h
z
a
d

4.25. A precast T beam is to be used as a bridge over a small roadway. Concrete dimensions are b =
48 in., b w = 16 in., h f = 5 in., and h = 25 in. The effective depth d = 20 in. Concrete and steel
strengths are 6000 psi and 60,000 psi, respectively. Using approximately one-half the maximum
tensile reinforcement permitted by the ACI Code (select the actual size of bar and number to be used),
determine the design moment capacity of the girder. If the beam is used on a 30 ft simple span, and if
in addition to its own weight it must support railings, curbs, and suspended loads totaling 0.475 kip/ft,
what uniform service live load limit should be posted?
b=48
25
20
5
16
f'c 6000psi
 β1 0.75

fy 60000psi

L 30ft

wD 475plf

hf 5in
 bw 16in
 b 48in

d 20in

SOLUTION
Determine the effective width of the flange using 4 bw for an isolated beam
beff min
L
4
4bw
 b







48in


Calculate girder self weight
wo
150pcf
144
in
2
ft
2
b hf
 bw 20
 in

 
 583plf


ρmax 0.0276
 From Table A.4
ρ
ρmax
2
0.014

 Assume that the compression block remains in the flange, then
As ρ b
 d
 13.248in
2


Try 8-No. 11 (No. 36) As 8As11 12.48in
2


a
As fy

0.85f'c b

3.059in

 less than 5 in hf, so proceed. c
a
β1
4.078in


Mn As fy
 d
a
2







 13831in kip


 c
d
0.204
 0.375, therefore
ϕ 0.90

B
a
h
z
a
d

Determine capacity and load rating
ϕ Mn
 1037 ft kip


Total Capacity
Dead Load
MD
1.2wD L
2

8
64 ft kip



Girder load
Mo
1.2wo L
2

8
79 ft kip



Moment available for live load ML ϕ Mn
 MD
 Mo
 894 ft kip



Live load wL
8ML
1.6 L
2

4969 plf

 Post 2.5 tons/ft
w
wL
b
1242 psf


Posted load
B
a
h
z
a
d

4.26 Using Eq. (4.27) and assuming that d = 0.9 h, show that A s is approximately equal to Mu /4h for
Grade 60 reinforcement and where Mu is in kip‐ft.
SOLUTION
MU = Mn = Asfy(d‐a/2)
Solve for As
As=Mu/(fy(d‐a/2))
Assume a is approximately 0.2d, multiply the moment by 12000 to use the moment in ft‐kip and the
other units in inches and pounds, and letting d = 0.9h
As = 12000 Mu/(0.9 * 60000*0.9h*(1‐0.2/2)) = Mu/3.6h

Ignoring the phi factor the approximation is As = Mu/4h
While approximate, the approach is a simple way to check design work.

B
a
h
z
a
d

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B
a
h
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d

PROBLEM 4‐28
f'c = 4000 psi
fy = 60000 psi
1= 0.85
u= 0.003
y= 0.0021
rbal 0.0289
r0.005= 0.0181
r0.004= 0.0206
et   Mn/bd2
Mn/bd2
0.050 0.90 0.0027 160 144
0.045 0.90 0.0030 176 158
0.040 0.90 0.0034 196 176
0.035 0.90 0.0038 221 198
0.030 0.90 0.0044 253 227
0.025 0.90 0.0052 296 266
0.020 0.90 0.0063 356 320
0.015 0.90 0.0080 448 403
0.010 0.90 0.0111 602 541
0.009 0.90 0.0120 646 581
0.008 0.90 0.0131 697 627
0.007 0.90 0.0145 757 681
0.0060 0.90 0.0161 827 744
0.0059 0.90 0.0162 835 751
0.0058 0.90 0.0164 843 758
0.0057 0.90 0.0166 851 766
0.0056 0.90 0.0168 859 773
0.0055 0.90 0.0170 867 780
0.0054 0.90 0.0172 876 788
0.0053 0.90 0.0174 884 796
0.0052 0.90 0.0176 893 804
0.0051 0.90 0.0178 902 812
0.0050 0.90 0.0181 911 820
0.0049 0.89 0.0183 920 821
0.0048 0.88 0.0185 930 821
0.0047 0.87 0.0188 940 822
0.0046 0.87 0.0190 949 822
0.0045 0.86 0.0193 960 823
0.0044 0.85 0.0195 970 823
0.0043 0.84 0.0198 980 824
0.0042 0.83 0.0201 991 824
0.0041 0.82 0.0204 1002 825
B
a
h
z
a
d

0.0040 0.81 0.0206 1013 825
0.0039 0.81 0.0209 1024 826 0.65%
0
200
400
600
800
1000
1200
0.000 0.005 0.010 0.015 0.020 0.025
M
n
/bd
2
(psi)
Reinforcement ratio, 
B
a
h
z
a
d

5.1 A rectangular beam secƟon is 10 in. wide and has a structural depth of 13.5 in. Given that the
concrete strength is 4000 psi and there is no shear reinforcement in the secƟon, determine the
nominal shear capacity allowed by the ACI Code.
≔
b 10 in ≔
d 13.5 in ≔
f'c 4000 psi ≔
λ 1.0
The maximum allowable shear stress for a section without shear reinforcement is 1/2 the regular shear
stress. Thus:
≔
vc =
λ ‾‾‾‾‾‾
⋅
f'c psi 63.2 psi
≔
Vn =
⋅
⋅
vc b d 8.54 kip
secƟon has No. 3 (No. 10) sƟrrups at 9 in., a specified concrete strength of 3000 psi, and fyt of 60,000
psi, calculate the factored shear capacity of the secƟon.
≔
b 14 in ≔
d 20.5 in ≔
f'c 3000 psi ≔
fyt 60000 psi
≔
Av =
⋅
2 0.11 in
2
0.22 in
2
≔
s 9 in ≔
ϕ 0.75 ≔
λ 1.0
≔
vc =
⋅
2 λ ‾‾‾‾‾‾
⋅
f'c psi 109.5 psi
≔
Vn =
+
⋅
⋅
vc b d ⋅
⋅
Av fyt ―
d
s
61.5 kip
≔
Vu =
⋅
ϕ Vn 46.1 kip
B
a
h
z
a
d

5.3 A rectangular beam secƟon is 16 in. wide and has a structural depth of 26 in. Given that the
secƟon has No. 3 (No. 10) sƟrrups at 12 in., a specified concrete strength of 4000 psi, and fyt of 60,000
psi, calculate the factored shear capacity of the secƟon.
≔
b 16 in ≔
d 26 in ≔
λ 1.0 ≔
f'c 4000 psi ≔
fyt 60000 psi
≔
Av =
⋅
2 0.11 in
2
0.22 in
2
≔
s 13 in ≔
ϕ 0.75
≔
vc =
⋅
2 λ ‾‾‾‾‾‾
⋅
f'c psi 126.5 psi
≔
Vn =
+
⋅
⋅
vc b d ⋅
⋅
Av fyt ―
d
s
79 kip
≔
Vu =
⋅
ϕ Vn 59.3 kip
5.4 A rectangular beam secƟon is 16 in. wide and has a structural depth of 26 in. Given that the secƟon
has No. 4 (No. 13) sƟrrups at 13 in., a specified concrete strength of 4000 psi, and fyt of 60,000 psi,
calculate the factored shear capacity of the secƟon.
≔
b 16 in ≔
d 26 in ≔
f'c 4000 psi ≔
fyt 60000 psi
≔
Av =
⋅
2 0.20 in
2
0.40 in
2
≔
s 13 in ≔
ϕ 0.75
≔
vc =
⋅
2 λ ‾‾‾‾‾‾
⋅
f'c psi 126.5 psi
≔
Vn =
+
⋅
⋅
vc b d ⋅
⋅
Av fyt ―
d
s
100.6 kip
≔
Vu =
⋅
ϕ Vn 75.5 kip
Note: Both problem 5.3 and 5.4 have the maximum allowable spacing of the stirrups equal to
d/2. Going from a No. 3 (No. 10) to a No. 4 (No. 13) stirrup increases the sectional capacity 27
percent but increases the steel area by over 80 percent. It is appropirate to ask the students to
compare the solutions to these two problems.
B
a
h
z
a
d

5.5 A Tsec on shown in Figure P5.5 has a structural depth d = 22 in. a web width bw = 6 in. and a
flange width of bf = 36 in. If the sec on has a specified concrete strength of 5000 psi and No. 3 (No.
10) s rrups at 10 in, determine the factored shear capacity of the sec on.
≔
bw 6 in ≔
d 26 in ≔
λ 1.0 ≔
f'c 4000 psi ≔
fyt 60000 psi
≔
Av =
⋅
2 0.11 in
2
0.22 in
2
≔
s 10 in ≔
ϕ 0.75
≔
vc =
⋅
2 λ ‾‾‾‾‾‾
⋅
f'c psi 126.5 psi
≔
Vn =
+
⋅
⋅
vc bw d ⋅
⋅
Av fyt ―
d
s
54.1 kip
≔
Vu =
⋅
ϕ Vn 40.5 kip
B
a
h
z
a
d

factored shear on the secƟon is 90 kips, and a specified concrete strength of 4000 psi, determine the
required spacing of No. 4 (No. 13) sƟrrups.
≔
b 16 in ≔
d 26.5 in ≔
λ 1.0 ≔
f'c 4000 psi ≔
fyt 60000 psi
≔
Av =
⋅
2 0.20 in
2
0.40 in
2
≔
ϕ 0.75
≔
vc =
⋅
2 λ ‾‾‾‾‾‾
⋅
f'c psi 126.5 psi ≔
Vc =
⋅
⋅
vc b d 53.6 kip
≔
Vu 90 kip ≔
Vn =
―
Vu
ϕ
120 kip
≔
s =
――――
⋅
⋅
⋅
ϕ Av fyt d
−
Vu ⋅
ϕ Vc
9.58 in
Use s = 9.0 in. s= 9.5 in. is technically correct; however, field installation is typically to the
nearest inch.
B
a
h
z
a
d

5.7 A Tsec on shown in Figure P5.5 has a structural depth d = 24 in. a web width bw = 8 in. and a
flange width of bf = 36 in. If the sec on has a specified concrete strength of 5000 psi and a factored
shear of 50 kips, determine the spacing of No. 3 (No. 10) s rrups.
≔
bw 8 in ≔
d 24 in ≔
λ 1.0 ≔
f'c 5000 psi ≔
fyt 60000 psi
≔
Av =
⋅
2 0.11 in
2
0.22 in
2
≔
ϕ 0.75
≔
vc =
⋅
2 λ ‾‾‾‾‾‾
⋅
f'c psi 141.4 psi ≔
Vc =
⋅
⋅
vc bw d 27.2 kip
≔
Vu 50 kip ≔
Vn =
―
Vu
ϕ
66.7 kip
≔
s =
――――
⋅
⋅
⋅
ϕ Av fyt d
−
Vu ⋅
ϕ Vc
8.017 in
Use s = 8.0 in.
B
a
h
z
a
d

5.8 A simple span rectangular beam has and eﬀec ve length of 18 , a width of 14 in. and a structural
depth of 24 in. It is reinforced with 3 No. 9 (No. 29) bars longitudinally and No. 3 (No. 10) s rrups at
12 in. on center over the en re length. Determine the maximum factored load the beam can carry in
plf. The specified concrete strength is 5000 psi, fy = 60,000 psi, and and fyt = 40,000psi.
≔
b 14 in ≔
d 24 in ≔
l 18 ft ≔
f'c 5000 psi ≔
λ 1.0
≔
As 3.0 in
2
≔
Av ⋅
2 0.11 in
2
≔
fy 60000 psi ≔
fyt 40000 psi
Flexural capacity ≔
ϕ 0.90
≔
a =
――――
⋅
As fy
⋅
0.85 f'c b
3.025 in
≔
Mn =
⋅
⋅
As fy
⎛
⎜
⎝
−
d ―
a
2
⎞
⎟
⎠
⎛
⎝ ⋅
4.048 10
3 ⎞
⎠ ⋅
in kip
≔
wu =
―――
⋅
⋅
8 ϕ Mn
l
2
7.496 ――
kip
ft
Shear Capacity ≔
ϕ 0.75
≔
s 12 in
≔
vc =
⋅
2 λ ‾‾‾‾‾‾
⋅
f'c psi 141.4 psi
≔
Vn =
+
⋅
⋅
vc b d ⋅
⋅
Av fyt ―
d
s
65.1 kip
≔
Vu =
⋅
ϕ Vn 48.8 kip
≔
wu =
―――
2 Vu
−
l 2 d
6.98 ――
kip
ft
The factored load for the beam is 6.98 kip/ft.
Note that the maximum shear can be computed a distance d in from the end of the beam;
hence, the divisior is l - 2d for the shear capacity.
Revising fyt to 60,000 psi results in a flexure critical factored load.
B
a
h
z
a
d

5.9. A beam is to be designed for loads causing a maximum factored shear of
60.0 kips, using concrete with f c = 5000 psi. Proceeding on the basis that the
concrete dimensions will be determined by diagonal tension, select the appropriate
width and effective depth ( a ) for a beam in which no web reinforcement
is to be used, ( b ) for a beam in which only the minimum web reinforcement is
provided, as given by Eq. (5.13), and ( c ) for a beam in which web reinforcement
provides shear strength V s = 2 V c . Follow the ACI Code requirements,
and let d = 2 b in each case. Calculations may be based on the more approximate
value of V c given by Eq. (5.12 d ).
Vu 60kip

f'c 5000psi
 fy 60000psi
 fyt 60000psi
 ϕ 0.75
 λ 1.0

vc 2λ f'c psi
 141psi

 vu ϕ vc
 106psi


SOLUTION
a) No web reinforcement
Vu=1/2(vu)bwd
bwd
Vu
ϕ f'c psi


1131 in
2


 bw = 23.8 in; d=47.6 in
b) minimum web reinforcement
Vu=(vu)bwd
bwd
Vu
vu
566 in
2


 bw = 16.8 in; d = 33.6 in
Note, this area could be further reduced if the contribution of the minimum transverse
reinforcement is included.
c) Vs=2Vc
Vu=(vu +2 vu)bwd
bwd
Vu
3vu
189 in
2


 bw = 9.7 in; d = 19.4 in
B
a
h
z
a
d

5.10. A rectangular beam having b = 10 in. and d = 17.5 in. spans 15 ft face to face of
simple supports. It is reinforced for flexure with three No. 9 (No. 29) bars that
continue uninterrupted to the ends of the span. It is to carry service dead load
D = 1.27 kips/ft (including self-weight) and service live load L = 3.70 kips/ft,
both uniformly distributed along the span. Design the shear reinforcement,
using No. 3 (No. 10) vertical U stirrups. The more approximate Eq. (5.12 d )
for V c may be used. Material strengths are f c = 4000 psi and f y = 60,000 psi.
l 15ft
 d 17.5in
 b 10in

L 3.70
kip
ft
 D 1.27
kip
ft

λ 1.0
 ϕ 0.75

f'c 4000psi
 fy 60000psi
 fyt 60000psi

SOLUTION
wu 1.2 D
 1.6 L

 7.44
kip
ft


 Av 2As3 0.22 in
2



Vu
wu l

2
55.83kip



Vud Vu wu d

 44.97kip


 Shear at d in from end of beam
Vc 2λ f'c psi
 b
 d
 22.14kip



spacing required at d from end
sd
ϕ Av
 fyt
 d

Vud ϕ Vc


6.106in



s min 24in sd

d
2

Av fyt

50 psi
 b








6.106in


 The critical section is at d in from the end,
Use #3 (No. 10) at 6 in.
d
2
8.75 in


Av fyt

50 psi
 b

26.4 in


7.5'
Vu 55.8 kip
Vud 45 kip
Vc 16.6 kip
Vc/2 8.3 kip
d
5.27'
6.38'
Stirrups are no longer theoretically
needed at 5.27 in from the end and
may be terminated at vc/2 or 6.38 ft
in from the end.
As a practical design issue,
minimum spacing of 8 in. (based on
d/2) would run through the center
portion of the beam.
B
a
h
z
a
d

6.1 through 6.6 (based on the beam in Fig. 4.15, Example 4.13, fy = 60 ksi)
For the No. 6 bars:
( )
2.5 in.
1.5 0.5 0.75 2 2.375 in.
10 2 1.5 0.5 0.75 2 2 2.625 in.
Use: 2.375 in.
b
b
b
c
c
c
≤
≤ + + =
≤ − + + =
 
 
For the No. 9 bars:
( )
2.75 in.
2.5 2 1.25 in.
1.5 0.5 1.128 2 2.564 in.
10 2 1.5 0.5 1.128 2 2 2.436 in.
Use: 1.25 in.
b
b
b
b
c
c
c
c
≤
≤ =
≤ + + =
≤ − + + =
 
 
No. 6 bars No. 9 bars
t
ψ 1.3 1.0
e
ψ 1.0 1.0
s
ψ 0.80 1.0
B
a
h
z
a
d

6.1 For the beam cross section shown in Fig. 4.15, what are the development lengths of the top No. 6
bars and bottom No. 9 bars for No. 4 (No. 13) stirrups with 1½ in. clear side cover spaced at 6 in.
using Eq. (6.4) and (6.5). Normalweight concrete, c
f ′ = 4000 psi. Comment.
No. 6 bars: Plane of splitting through side cover, s = 6 in., n = 1, Atr = 0.2 in
2
. t
ψ = 1.3, e
ψ = 1.0, s
ψ = 0.80,
λ = 1.0, b
c = 2.375 in.
( ) ( )
40 40 0.20 6 1 1.333
tr tr
K A sn
= = × × =
2.375 1.333
4.94 2.5
0.75
b tr
b
c K
d
 
+ +
 
= = ≤
   
 
 
Eq. (6.4):
3 3 60,000 1.3 1.0 0.8
0.75 22.2 in. 12.0 in.
40 40 2.5
λ 1 4000
y t e s
d b
b tr
c
b
f
d
c K
f
d
yyy
 
 
× ×
 
 
= = = ≥
 
 
 
′ + ×
 
 
 
 
 
 

Eq. (6.5):
60,000 1.3 1.0
0.75 37.0 in. 12.0 in.
25λ 25 1 4000
y t e
d b
c
f
d
f
yy
  × ×
 
= = = ≥
   
 
′ × ×
 
 

No. 9 bars: Plane of splitting between bars separated vertically with 2½-in. center-to center spacing, s = 6
in., n = 2, and Atr = 0.2 in
2
. t
ψ = 1.0, e
ψ = 1.0, s
ψ = 1.0, λ = 1.0, b
c = 1.25 in.
( ) ( )
40 40 0.20 6 2 0.667
tr tr
K A sn
= = × × =
1.25 0.667
1.7 2.5
1.128
b tr
b
c K
d
 
+ +
 
= = ≤
   
 
 
Eq. (6.4):
3 3 60,000 1.0 1.0 1.0
1.128 47.2 in. 12.0 in.
40 40 1.7
λ 1 4000
y t e s
d b
b tr
c
b
f
d
c K
f
d
yyy
 
 
× ×
 
 
= = = ≥
 
 
 
′ + ×
 
 
 
 
 
 

Eq. (6.5):
60,000 1.0 1.0
1.128 53.5 in. 12.0 in.
20λ 20 1 4000
y t e
d b
c
f
d
f
yy
  × ×
 
= = = ≥
   
 
′ × ×
 
 

Equation (6.4) takes advantage of the actual cover and the presence of the stirrups, resulting in
development lengths that are less than calculated by Eq. (6.5). The values of development length
obtained using the two equations are closer for the No. 9 bars because of the lower values of cb and Ktr
compared to those for the No. 6 bars.
B
a
h
z
a
d

using Eq. (6.4) and (6.5). Lightweight concrete, c
2
. t
ψ = 1.3, e
ψ = 1.0, s
ψ = 0.80,
λ = 0.75, b
c = 2.375 in.
( ) ( )
40 40 0.20 4 1 2.0
tr tr
K A sn
= = × × =
2.375 2.0
5.83 2.5
0.75
b tr
b
c K
d
 
+ +
 
= = ≤
   
 
 
Eq. (6.4):
3 3 60,000 1.3 1.0 0.8
0.75 29.6 in. 12.0 in.
40 40 2.5
λ 0.75 4000
y t e s
d b
b tr
c
b
f
d
c K
f
d
yyy
 
 
× ×
 
 
= = = ≥
 
 
 
′ + ×
 
 
 
 
 
 

Eq. (6.5):
60,000 1.3 1.0
0.75 49.3 in. 12.0 in.
25λ 25 0.75 4000
y t e
d b
c
f
d
f
yy
  × ×
 
= = = ≥
   
 
′ × ×
 
 

in., n = 2, and Atr = 0.2 in
2
. t
ψ = 1.0, e
ψ = 1.0, s
ψ = 1.0, λ = 0.75, b
c = 1.25 in.
( ) ( )
40 40 0.20 4 2 1.0
tr tr
K A sn
= = × × =
1.25 1.0
1.99 2.5
1.128
b tr
b
c K
d
 
+ +
 
= = ≤
   
 
 
Eq. (6.4):
3 3 60,000 1.0 1.0 1.0
1.128 53.8 in. 12.0 in.
40 40 1.99
λ 0.75 4000
y t e s
d b
b tr
c
b
f
d
c K
f
d
yyy
 
 
× ×
 
 
= = = ≥
 
 
 
′ + ×
 
 
 
 
 
 

Eq. (6.5):
60,000 1.0 1.0
1.128 71.3 in. 12.0 in.
20λ 20 0.75 4000
y t e
d b
c
f
d
f
yy
  × ×
 
= = = ≥
   
 
′ × ×
 
 

compared to those for the No. 6 bars. The relatively higher transverse reinforcement compared to
Problem 6.1 results in a greater difference between the two equations for the No. 9 bars, but not for the
No. 6 bars because the maximum value of Ktr governs for the smaller bars.
B
a
h
z
a
d

using Eq. (6.4) and (6.5). Normalweight concrete, c
2
. t
ψ = 1.3, e
ψ = 1.0, s
ψ = 0.80,
λ = 1.0, b
c = 2.375 in.
( ) ( )
40 40 0.20 6 1 1.333
tr tr
K A sn
= = × × =
2.375 1.333
4.94 2.5
0.75
b tr
b
c K
d
 
+ +
 
= = ≤
   
 
 
Eq. (6.4):
3 3 60,000 1.3 1.0 0.8
0.75 15.7 in. 12.0 in.
40 40 2.5
λ 1 8000
y t e s
d b
b tr
c
b
f
d
c K
f
d
yyy
 
 
× ×
 
 
= = = ≥
 
 
 
′ + ×
 
 
 
 
 
 

Eq. (6.5):
60,000 1.3 1.0
0.75 26.1 in. 12.0 in.
25λ 25 1 8000
y t e
d b
c
f
d
f
yy
  × ×
 
= = = ≥
   
 
′ × ×
 
 

in., n = 2, and Atr = 0.2 in
2
. t
ψ = 1.0, e
ψ = 1.0, s
ψ = 1.0, λ = 1.0, b
c = 1.25 in.
( ) ( )
40 40 0.20 6 2 0.667
tr tr
K A sn
= = × × =
1.25 0.667
1.7 2.5
1.128
b tr
b
c K
d
 
+ +
 
= = ≤
   
 
 
Eq. (6.4):
3 3 60,000 1.0 1.0 1.0
1.128 33.4 in. 12.0 in.
40 40 1.7
λ 1 8000
y t e s
d b
b tr
c
b
f
d
c K
f
d
yyy
 
 
× ×
 
 
= = = ≥
 
 
 
′ + ×
 
 
 
 
 
 

Eq. (6.5):
60,000 1.0 1.0
1.128 37.8 in. 12.0 in.
20λ 20 1 8000
y t e
d b
c
f
d
f
yy
  × ×
 
= = = ≥
   
 
′ × ×
 
 

compared to those for the No. 6 bars.
B
a
h
z
a
d

Instructor's Solutions Design of Concrete Structures – Arthur H. Nilson – 15th Edition

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Instructor's Solutions Design of Concrete Structures – Arthur H. Nilson – 15th Edition