4-9 UNSYMMETRICAL BEAM SECTIONS
3. Compute
Note that the moment calculation is based on the lever arm measured vertically(parallel to the plane of loading).
4. Check if .Again, for 3000 psi concrete, is less than 200 psi.
So, the value of from Eq. (4-11) is
■ As,min = 200 psi
fy bwd = 200 psi
60,000 psi112 in.2121.5 in.2 = 0.86 in.2 6 As1o.k.2 As,min
32fcœ As » As,min
= 2700 k-in. = 225 k-ft
fMn = 0.9s2.58 in.2 * 60 ksia21.5 in. -6.46 in.
3 bt fMn = fsAsfyad - g
3bt FMnFto Eq. (4-43):
di⫽ 22.4 in.
18.8 in.
Fig. 4-51 Beam section for Example 4-7.
4-1 Figure P4-1 shows a simply supported beam and the cross section at midspan. The beam supports a uniform service (unfactored) dead load consisting of its own weight plus 1.4 kips/ft and a uniform ser- vice (unfactored) live load of 1.5 kips/ft. The con- crete strength is 3500 psi, and the yield strength of the reinforcement is 60,000 psi. The concrete is normal-weight concrete. Use load and strength- reduction factors from ACI Code Sections 9.2 and 9.3. For the midspan section shown in Fig. P4-1b, compute fMnand show that it exceeds Mu.
4-2 A cantilever beam shown in Fig. P4-2 supports a uniform service (unfactored) dead load of 1 kip/ft plus its own dead load and a concentrated service (unfactored) live load of 12 kips, as shown. The concrete is normal-weight concrete with
psi and the steel is Grade 60. Use load and strength-reduction factors from ACI Code Sections 9.2 and 9.3. For the end section shown in Fig. P4-2b, compute and show that it ex- ceeds
4-3 (a) Compare for singly reinforced rectangu- lar beams having the following properties. Use strength reduction factors from ACI Code Sections 9.2 and 9.3.
fMn Mu.
fMn
= 4000
fcœ
PROBLEMS
Fig. P4-1
Beam b d
No. (in.) (in.) Bars (psi) (psi)
1 12 22 3 No. 7 4000 60,000
2 12 22 2 No. 9 plus 1 No. 8 4000 60,000
3 12 22 3 No. 7 4000 80,000
4 12 22 3 No. 7 6000 60,000
5 12 33 3 No. 7 4000 60,000
(b) Taking beam 1 as the reference point, discuss the effects of changing and d on (Note that each beam has the same properties as beam 1 except for the italicized quantity.)
(c) What is the most effective way of increasing What is the least effective way?
fMn? fMn.
As,fy,fcœ, fy fc¿
(a)
(b) Fig. P4-2
4-4 A 12-ft-long cantilever supports its own dead load plus an additional uniform service (unfactored) dead load of 0.5 kip/ft. The beam is made from normal-weight 4000-psi concrete and has in., in., and in. It is reinforced with four No. 7 Grade-60 bars. Compute the max- imum service (unfactored) concentrated live load that can be applied at 1 ft from the free end of the cantilever. Use load and strength-reduction factors from ACI Code Sections 9.2 and 9.3. Also check As,min.
h = 18 d = 15.5
b = 16
Video Solution
4-5 and 4-6 Compute and check for the beams shown in Figs. P4-5 and P4-6, respec- tively. Use psi for Problem 4-5 and 4000 psi for Problem 4-6. Use psi for both problems.
fy = 60,000 fcœ = 4500
As,min fMn
48 in.
6 in.
19 in.22 in.
12 in.
Fig. P4-5
4-7 Compute the negative-moment capacity, and check for the beam shown in Fig. P4-7. Use
psi and fy = 60,000 psi.
fcœ = 4000 As,min
fMn,
Fig. P4-6
48 in.
6 in.
16 in. 19.5 in.
12 in.
Fig. P4-7
4-8 For the beam shown in Fig. P4-8, psi and psi.
(a) Compute the effective flange width at midspan.
(b) Compute for the positive- and negative- moment regions and check for both sections. At the supports, the bottom bars are in one layer; at midspan, the No. 8 bars are in the bottom layer, the No. 7 bars in a second layer.
4-9 Compute and check for the beam
shown in Fig. P4-9. Use psi and
psi.
(a) The reinforcement is six No. 8 bars.
(b) The reinforcement is nine No. 8 bars.
4-10 Compute and check for the beam
shown in Fig. P4-10. Use psi and
psi.
fy = 60,000
fcœ = 5000 As,min fMn
fy = 60,000
fcœ = 4000 As,min fMn
As,min fMn
fy = 60,000
fcœ = 3500
2 No. 8 bars at ends Support (negative bending)
Midspan (positive bending)
22 ft
3 No. 8 plus 2 No. 7 bars at midspan
Fig. P4-8
Video Solution
30 in.
5 in.
5 in.
5 in.
5 in. 35 in.
32.5 in.
Fig. P4-9
42 in.
6 in.
6 in.
6 in.
23.5 in. 20 in.
Fig. P4-10
2.5 in. 12 in.
3.5 in.
Fig. P4-11
4-11 (a) Compute for the three beams shown in
Fig. P4-11. In each case, psi,
ksi, in., in., and
in.
(b) From the results of part (a), comment on whether adding compression reinforcement is a h = 36
d = 32.5 b = 12
fy = 60
fcœ = 5000
fMn cost-effective way of increasing the strength,
of a beam.
4-12 Compute for the beam shown in Fig. P4-12.
Use psi and psi. Does the
compression steel yield in this beam at nominal strength?
fy = 60,000 fcœ = 4500
fMn fMn,
10 in.
5 in.
5 in.
2.5 in.
2.5 in.
20 in.
5 in.
Fig. P4-12
Video Solution
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