000_[Jonathan_Ochshorn]_Structural_Elements_for_Architecture_400

160 CHAPTER 8 Beams

Deflection

Deflection is based on the same criteria discussed earlier for wood beams and
involves a comparison of an allowable deflection, typically set at span/240 for total
loads and span/360 for live loads on floor beams, to the actual computed deflection.
Allowable deflection guidelines can be found in Table A-8.1; actual deflections can
be computed based on the coefficients in Table A-8.2.

Example 8.6 Design steel beam
Problem definition
Using A992 steel, design the typical beam and girder for the library stack area shown in Figure
8.27. Use the generic dead load for steel floor systems. Assume that the beams are continu-
ously braced by the floor deck and that the girders are braced only by the beams framing
into them.

Solution overview
Find loads; compute maximum bending moment and shear force; use appropriate tables to
select beams for bending; then check for shear and deflection.

Problem solution
1. Find loads:

From Table A-2.1, the dead load, D ϭ 47 psf.
From Table A-2.2, the live load, L ϭ 150 psf

Beam design
1. Create load, shear, and moment diagrams as shown in Figure 8.28 to determine critical

(i.e., maximum) shear force and bending moment. The total distributed load, w ϭ (dead ϩ
live)(tributary area for 1 linear foot) ϭ (47 ϩ 150)(6) ϭ 1182 lb/ft ϭ 1.18 kips/ft. Live load
reduction would not apply even if the “influence” area was not less than 400 ft2, because

FIGURE 8.27
Framing plan for Example 8.6

Steel 161

of the library stack occupancy (i.e., the probability of full loading makes live load reduction
a dangerous assumption).
2. Find allowable bending stress: since the beam is laterally braced by the floor
deck and the cross section is assumed to be compact, use Equation 8.12 to find
Zreq ϭ ΩMmax /Fy ϭ 1.67Mmax/Fy. From Table A-3.12, Fy ϭ 50 ksi for A992 steel, so
Zreq ϭ 1.67(399)/50 ϭ 13.33 in3.
3. From Table A-8.4, select a W12 ϫ 14 with actual Zx ϭ 17.4 in3 Ն Zreq; this section is, by
definition, OK for bending.
4. Check section for shear: from Table A-4.3, the actual web area, Aw ϭ d ϫ tw ϭ 11.9 ϫ 0.2
0 ϭ 2.38 in2.
5. From Equation 8.14, the required Aw ϭ V/Fv ϭ 8.86/(0.36 ϫ 50) ϭ 0.49 in2 where,
from Table A-3.13, the allowable shear stress, Fv ϭ 0.36Fy (and not the usual value of
Fv ϭ 0.40Fy) because the beam web is unusually slender. Beams requiring such reduced
allowable shear stresses are noted in Table A-4.3. Since the actual web area is greater than
the required web area, the beam is OK for shear.
6. From Table A-8.1, find the allowable total-load deflection for a floor beam: Δallow ϭ span/
240 ϭ 12(15)/240 ϭ 0.75 in.; and the allowable live-load deflection for a floor joist:
Δallow ϭ span/360 ϭ 12(15)/360 ϭ 0.5 in.
7. From Table A-8.2, the actual total-load deflection, Δactual ϭ CPL3/(EI), where:

C ϭ 22.46.
L ϭ 15 ft.
P ϭ wL ϭ (150 ϩ 47)(6)(15) ϭ 17,730 lb ϭ 17.73 kips.
E ϭ 29,000 ksi (Table A-3.12, Note 1).
I ϭ 88.6 in4 (Table A-4.3).
Δactual ϭ 22.46(17.73)(153)/(29,000 ϫ 88.6) ϭ 0.523 in.
Since Δactual ϭ 0.523 in. Յ Δallow ϭ 0.75 in., beam is OK for total-load deflection.
8. From Table A-8.2, the actual live-load deflection, Δactual ϭ CPL3/(EI), where:
C ϭ 22.46.
L ϭ 15 ft.
P ϭ wL ϭ (150 ϫ 6)15 ϭ 13,500 lb ϭ 13.5 kips. (Use live load only!)
E ϭ 29,000 ksi (Table A-3.12, Note 1).

FIGURE 8.28
Load, shear, and moment diagrams for beam in Example 8.6

162 CHAPTER 8 Beams

FIGURE 8.29
Load, shear, and moment diagrams for girder in Example 8.6

I ϭ 88.6 in4 (Table A-4.3).
Δactual ϭ 22.46(13.5)(153)/(29,000 ϫ 88.6) ϭ 0.398 in.
Since Δactual ϭ 0.398 in. Յ Δallow ϭ 0.5 in., beam is OK for live-load deflection.
9. Conclusion: The W12 ϫ 14 section is OK for bending, shear, and deflection. Therefore, it is
acceptable.

Girder design
1. Create load, shear, and moment diagrams as shown in Figure 8.29 to determine the critical

(i.e., maximum) shear force and bending moment. Each concentrated load is twice the typ-
ical beam reaction, or 17.73 kips. Alternatively, compute using tributary areas; that is, P ϭ
(47 ϩ 150)(15 ϫ 6) ϭ 17,730 lb ϭ 17.73 kips. Live load reduction does not apply even
though the “influence” area is greater than 400 ft2, because of the library stack occupancy
(i.e., the probability of full loading makes live load reduction a dangerous assumption).
2. Find allowable bending stress: the girder is not continuously braced by the floor deck; rather, it
is braced every 6ft by the beams framing into it, so the unbraced length, Lb ϭ 6ft. Use Table
A-8.5 to directly find the lightest cross section for bending, based on Mmax ϭ 212.76ft-kips,
Lb ϭ 6ft, and assuming (conservatively) that the “lateral-torsional buckling modifier,” Cb ϭ 1.0.
Find the intersection of moment and unbraced length (follow the dotted lines shown in Figure
8.30) and then move up or to the right to the first solid line representing the available moment
capacity of wide-flange beams. Select a W21 ϫ 44.
3. Check section for shear: from Table A-4.3, the actual web area, Aw ϭ d ϫ tw ϭ 20.7 ϫ 0.3
5 ϭ 7.25 in2.
4. From Equation 8.14, the required Aw ϭ V/Fv ϭ 26.595/(0.40 ϫ 50) ϭ 1.33 in2, where,
from Table A-3.13, the allowable shear stress, Fv ϭ 0.40Fy (the usual value of Fv ϭ 0.40Fy
can be used in this case). Since the actual web area is greater than the required web area,
the beam is OK for shear.
5. From Table A-8.1, find the allowable total-load deflection for a floor beam: Δallow ϭ span/
240 ϭ 12(24)/240 ϭ 1.2 in.; and the allowable live-load deflection for a floor joist:
Δallow ϭ span/360 ϭ 12(24)/360 ϭ 0.8 in.
6. From Table A-8.2, the actual total-load deflection, Δactual ϭ CPL3/(EI), where:

C ϭ 85.54.
span, L ϭ 24 ft.

Steel 163

FIGURE 8.30
Selection of W21 ϫ 44 beam based on available moment graphs (Table A-8.5) for Example 8.6

P ϭ (47 ϩ 150)(15 ϫ 6) ϭ 17,730 lb ϭ 17.73 kips.
E ϭ 29,000 ksi (Table A-3.12, Note 1).
I ϭ 843 in4 (Table A-4.3).
Δactual ϭ 85.54(17.73)(243)/(29,000 ϫ 843) ϭ 0.86 in.
Since Δactual ϭ 0.86 in. Յ Δallow ϭ 1.2 in., beam is OK for total-load deflection.
7. From Table A-8.2, the actual live-load deflection, Δactual ϭ CPL3/(EI), where:
C ϭ 85.54.
span, L ϭ 24 ft.
P ϭ 150(15 ϫ 6) ϭ 13,500 lb ϭ 13.5 kips. (Use live load only!)

E ϭ 29,000 ksi (Table A-3.12, Note 1).
I ϭ 843 in4 (Table A-4.3).
Δactual ϭ 85.54(13.5)(243)/(29,000 ϫ 843) ϭ 0.65 in.
Since Δactual ϭ 0.65 in. Յ Δallow ϭ 0.80 in., beam is OK for live-load deflection.
8. Conclusion: The W21 ϫ 44 section is OK for bending, shear, and deflection. Therefore, it is
acceptable.

Example 8.7 Analyze rectangular HSS (hollow structural section)
Problem definition
Determine whether a HSS12 ϫ 4ϫ1/4 can be used as a typical beam for the library stack area
shown in Example 8.6.
Solution overview
Find loads; compute maximum bending moment and shear force; check beam for bending,
shear, and deflection.

164 CHAPTER 8 Beams

Problem solution
1. Find loads and moment (same as Example 8.6):

The dead load, D ϭ 47 psf.
The live load, L ϭ 150 psf
Maximum moment, Mmax ϭ 399 in-kips
2. Find allowable bending stress: since the beam is laterally braced by the floor deck and
the cross section is assumed to be compact, use Equation 8.12 to find Zreq ϭ ΩMmax/
Fy ϭ 1.67Mmax/Fy. From Table A-3.12, Fy ϭ 46 ksi for HSS rectangular shapes (A500 grade
B), so Zreq ϭ 1.67(399)/46 ϭ 14.49 in3.
3. From Table A-4.6, the actual plastic section modulus for a HSS12 ϫ 4ϫ1/4, Zx ϭ 25.6 in3.
Since the actual Zx is greater than Zreq, this HSS section is OK for bending.
4. Check section for shear: from Table A-3.13 (Note 3), the web area, Aw, is taken as 2ht (where t
is the wall thickness of the web and h can be assumed to equal the nominal depth minus 3t).
From A-4.4, this web area, Aw ϭ 2ht ϭ 2(12 Ϫ 3 ϫ 0.233)(0.233) ϭ 5.27in2.
5. From Equation 8.14, the required Aw ϭ V/Fv ϭ 8.86/(0.36 ϫ 50) ϭ 0.49 in2, where, from
Table A-3.13, the allowable shear stress, Fv ϭ 0.36Fy (and not the value of Fv ϭ 0.40Fy
used for most wide-flange beams). Since the actual web area is greater than the required
web area, the HSS beam is OK for shear.
6. From Table A-8.1, find the allowable total-load deflection for a floor beam: Δallow ϭ span/
240 ϭ 12(15)/240 ϭ 0.75 in.; and the allowable live-load deflection for a floor joist:
Δallow ϭ span/360 ϭ 12(15)/360 ϭ 0.5 in.
7. From Table A-8.2, the actual total-load deflection, Δactual ϭ CPL3/(EI), where:

C ϭ 22.46.
span, L ϭ 15 ft.

P ϭ wL ϭ (47 ϩ 150)(6)(15) ϭ 17,730 lb ϭ 17.73 kips.
E ϭ 29,000 ksi (Table A-3.12, Note 1).
I ϭ 88.6 in4 (Table A-4.6).
Δactual ϭ 22.46(17.73)(153)/(29,000 ϫ 119) ϭ 0.389 in.
Since Δactual ϭ 0.389 in. Յ Δallow ϭ 0.75 in., the HSS beam is OK for total-load deflection.
8. From Table A-8.2, the actual live-load deflection, Δactual ϭ CPL3/(EI), where:
C ϭ 22.46.
L ϭ 15 ft.
P ϭ wL ϭ (150 ϫ 6)15 ϭ 13,500 lb ϭ 13.5 kips. (Use live load only!)
E ϭ 29,000 ksi (Table A-3.12, Note 1).
I ϭ 88.6 in4 (Table A-4.6).
Δactual ϭ 22.46(13.5)(153)/(29,000 ϫ 119) ϭ 0.300 in.
Since Δactual ϭ 0.300 in. Յ Δallow ϭ 0.5 in., the HSS beam is OK for live-load deflection.
9. Conclusion: The HSS12 ϫ 4ϫ1/4 section is OK for bending, shear, and deflection.
Therefore, it is acceptable.

REINFORCED CONCRETE

Concrete beams are reinforced with steel rods (reinforcing bars) in order to resist
internal tension forces within the cross section. Unlike wood and steel, which can

Reinforced concrete 165

withstand substantial tension stress, concrete may be safely stressed only in compres-
sion.The pattern of steel reinforcement thus corresponds to the pattern of positive
and negative bending moments within the beam: in regions of positive bending, steel
is placed at the bottom of the cross section; in regions of negative bending, steel is
placed at the top (Figure 8.31). Like concrete columns, 2½ in. to 3 in. of cover, mea-
sured from the outside face of the beam to the centerline of the reinforcing steel, is
used to protect the steel from corrosion and provide adequate bond between the
steel and concrete (see Figure 7.9).

The strength, or capacity, of a reinforced concrete beam can be determined by
considering the equilibrium of tensile and compressive forces at any cross section.
Failure of the beam occurs either with crushing of the concrete within the compres-
sion region; or yielding of the tension steel, followed by compressive crushing of
the concrete. Since tension yielding is the preferred mode of failure—compressive
crushing of the concrete would be sudden and catastrophic, whereas yielding of
the steel provides warning signs of collapse—concrete beams are often deliberately
under-reinforced to guarantee that, in the case of failure, the steel reinforcing bars
begin to yield before the concrete in the compressive zone crushes.

At the point of failure, the stresses in a reinforced concrete cross section are as
shown in Figure 8.32. The curved distribution of stresses within the compressive

FIGURE 8.31
Relationship of bending moment and position of tension steel reinforcement

FIGURE 8.32
Strain and stress diagrams for tension-reinforced concrete beam at point of failure

166 CHAPTER 8 Beams

zone (above the neutral axis for “positive” bending) corresponds to the nonlinear

stress-strain curves characteristic of plain concrete, with a value of 0.85 f cЈ taken for
the strength of concrete corresponding to its behavior in an actual structure (Figure

8.33, curve b). Testing of many reinforced concrete beams has shown that the aver-

age stress within the compressive zone is 0.85β1 f cЈ, and the resultant location is
β1c/2 from the face of the concrete beam, as shown in Figure 8.34a. The coefficient
β1 ranges from 0.85 for f cЈ Յ 4000 psi, to 0.65 for f cЈ Ն 8000 psi (Figure 8.35). Thus,
for a cross section of width, b, the total compressive force, C, is

C ϭ 0.85β1 fcЈ bc (8.15)

Since the steel yields before the concrete crushes (assuming that the beam has

been designed to be under-reinforced), the steel stress is fy and the total tensile
force, T, is:

T ϭ As f y (8.16)

where As is the steel area. (As the steel is now used in the context of concrete
design, the designation for its yield stress changes from Fy to fy.)

Alternatively, a different, but equivalent, rectangular stress distribution can be

used in place of the actual nonlinear distribution, as shown in Figure 8.34b. In this

FIGURE 8.33

Stress-strain diagrams for plain concrete showing (a) fast loading characteristic of test cylinders
and (b) slow loading characteristic of actual structures (same as Figure 7.11)