Solutions manual Reinforced Concrete Structures : Analysis and Design using ACI 318-25, 3rd Edition, by David A. Fanella

Solution Manual
Problem 2.1
Determine the modulus of elasticity, Ec , for a normalweight concrete mixture with a unit weight of
145 lb/ft3 (2,320 kg/m3) and f c′ equal to (a) 3,000 psi (21 MPa), (b) 4,000 psi (28 MPa), and (c) 5,000 psi
(35 MPa).

Solution
Modulus of elasticity is determined in accordance with ACI 19.2.2.1.
• Part (a)

Ec  wc1.5 33 f c
(145)1.5
33
3, 000  3, 155, 924 psi

In S.I.:

Ec  wc1.5 0.043 f c
(2, 320)1.5
0.043
21  22, 020 MPa

• Part (b)

Ec  (145)1.5
33
4, 000  3, 644, 147 psi

In S.I.:

Ec  (2, 320)1.5
0.043
28  25, 426 MPa

• Part (c)

Ec  (145)1.5
33
5, 000  4, 074, 281 psi

In S.I.:

Ec  (2, 320)1.5
0.043
35  28, 427 MPa


Problem 2.2
Repeat Problem 2.1 assuming a lightweight concrete mixture with an equilibrium density of 110 lb/ft3
(1,760 kg/m3).

Solution
• Part (a)

Ec  wc1.5 33 f c
(110)1.5
33
3, 000  2, 085, 276 psi

In S.I.:

Ec  wc1.5 0.043 f c
(1, 760)1.5
0.043
21  14, 550 MPa

• Part (b)

Ec  (110)1.5
33
4, 000  2, 407 , 870 psi




1



17_Fanella_Solutions_p001-164.indd 1 8/8/2025 5:00:30 PM

, 2 Solution Manual

In S.I.:

Ec  (1, 760)1.5
0.043
28  16, 800 MPa
• Part (c)

Ec  (110)1.5
33
5, 000  2, 692, 080 psi

In S.I.:

Ec  (1, 760)1.5
0.043
35  18, 783 MPa


Problem 2.3
Determine the modulus of rupture, f r , for a normalweight concrete mixture with a unit weight of
145 lb/ft3 (2,320 kg/m3) and f c′ equal to (a) 3,000 psi (21 MPa), (b) 4,000 psi (28 MPa), and (c) 5,000 psi
(35 MPa).

Solution
Modulus of rupture is determined in accordance with ACI 19.2.3.1, and λ is determined in accordance
with ACI Table 19.2.4.1(a).
• Part (a)

f r  7.5
f c
7.5
1.0
3, 000  411 psi

In S.I.:

f r  0.62
f c
0.62
1.0
21  2.84 MPa

• Part (b)
f r  7.5
1.0
4, 000  474 psi

In S.I.:

f r  0.62
1.0
28  3.28 MPa

• Part (c)

f r  7.5
1.0
5, 000  530 psi

In S.I.:

f r  0.62
1.0
35  3.67 MPa


Problem 2.4
Repeat Problem 2.3 for a sand-lightweight concrete mixture.

Solution
The factor λ is determined from ACI Table 19.2.4.1(b).
• Part (a)

f r  7.5
f c
7.5
0.85
3, 000  349 psi

In S.I.:
f r  0.62
f c
0.62
0.85
21  2.42 MPa

• Part (b)
f r  7.5
0.85
4, 000  403 psi




17_Fanella_Solutions_p001-164.indd 2 8/8/2025 5:00:38 PM

, Solution Manual 3
In S.I.:

f r  0.62
0.85
28  2.79 MPa

• Part (c)
f r  7.5
0.85
5, 000  451 psi

In S.I.:
f r  0.62
0.85
35  3.12 MPa


Problem 2.5
Identify the bar size, type of reinforcing steel, and the grade of reinforcing steel in both inch-pound
and metric units for a steel reinforcing bar with the following markings: C, 9, W, and 60.

Solution
The second identification mark is the bar size (in S.I.: 29).
The third identification mark is the type of steel. From Fig. 2.5, “W” corresponds to A706 steel.
The fourth identification mark is the grade mark. The “60” corresponds to Grade 60 bars (in S.I.: “4”
corresponds to Grade 420 bars).


Problem 3.1
For the floor system in Fig. 3.4, determine the factored bending moments and shear forces in the beam
along line A given the design data in Examples 3.1 through 3.3.

Solution
The dead loads supported by the beams are equal to the following:

84 25.0
Weight of the wide-module joist system
1.1 kips/ft
1, 000 2
22.0
20.5 150
Weight of the beam
0.5 kips/ft
144 1, 000
10 25.0
Superimposed dead load
0.2 kips/ft (conservatively rounded up from 0.13 kips/ft)
1, 000 2

Total dead load  1.1
0.5
0.2  1.8 kips/ft
50
15 25.0
Live load
0.8 kips/ft
1, 000 2

In S.I.:
7.62
Weight of the wide-module joist system  4.02
15.3 kN/m
2
550
515
Weight of the beam
2, 400
0.0098  6.7 kN/m
1, 000 2
7.62
Superimposed dead load  0.48
1.8 kN/m
2
Total dead load  15.3
6.7
1.8  23.8 kN/m
7.62
Live load  (2.39
0.72)
11.9 kN/m
2
Maximum total factored load, wu , is determined by ACI Equation (5.3.1b) in ACI Table 5.3.1:

wu  1.2wD
1.6wL  (1.2
1.8)
(1.6
0.8)  3.4 kips/ft




17_Fanella_Solutions_p001-164.indd 3 8/8/2025 5:00:45 PM

, 4 Solution Manual

In S.I.:
wu  1.2wD
1.6wL  (1.2
23.8)
(1.6
11.9)  47.6 kN/m

Check if the five conditions of ACI 6.5.1 are satisfied:
• The beams have a constant cross-section over the entire length, which makes them prismatic.
• The dead and live loads are uniformly distributed over the length of the beams.
• Live-to-dead load ratio  0.8/1.8  0.4
3 (in S.I.: 11.9/23.8  0.5
3 ).
• There are two spans.
• Ratio of adjacent span lengths = 1.0.
Because all five conditions are satisfied, the simplified method in ACI 6.5 is permitted to determine
the factored bending moments and shear forces.
The clear span length of the beams is equal to the following:

22
 n  20
18.2 ft
12
In S.I.:
6.10  (550/1, 000)
5.55 m
The negative bending moments at the interior faces of the exterior supports are equal to the
following (see ACI Table 6.5.2 and Fig. 3.3):
wu  2n 3.4
18.22
Mu


70.4 ft-kips
16 16
In S.I.:
47.6
5.552
Mu

91.6 kN-m
16
The positive bending moments are equal to the following:
wu  2n 3.4
18.22
Mu


80.4 ft-kips
14 14
In S.I.:
47.6
5.552
Mu

104.7 kN-m
14
Because there are only two spans, the negative moments at the exterior faces of the first interior
supports are equal to the following (see Note 1 in Fig. 3.3):
wu  2n 3.4
18.22
Mu


125.1 ft-kips
9 9
In S.I.:
47.6
5.552
Mu

162.9 kN-m
9
The shear forces are equal to the following:
1.15wu  n 1.15
3.4
18.2
At the face of the first interior support: Vu    35.6 kips
2 2
In S.I.:
1.15wu  n 1.15
47.6
5.55
Vu    151.9 kN
2 2
wu  n 3.4
18.2
At the faces of all the other supports: Vu    30.9 kips
2 2
In S.I.:
47.6
5.55
Vu   132.1 kN
2




17_Fanella_Solutions_p001-164.indd 4 8/8/2025 5:00:53 PM