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Design code: AISC Allowable Stress Design 9^{th} edition, 1989.
Maximum bending stress, f_{b} must not exceed allowable stress, F_{b}.
Deflection should not exceed allowable limit.
Maximum shear stress, fv shall not exceed allowable shear stress.
Design procedure:
Calculate design load.
Calculate design moment, M and bending stress, f_{b}.
Select a trial beam size and calculate allowable bending stress, F_{b} (see below)
Calculate deflection and check with allowable deflection ratio.
Calculate design shear and shear stress, f_{v}.
Calculate allowable shear stress, F_{v}.
Bending stress shall be
determined as
f_{b}= M/S where M is design moment, S is section modulus.
W, I shape and channel hot-roll section bending about its major axis or shear center
Compact section: allowable bending stress,
F_{b} = 0.66 F_{y} if L_{b} £ L_{c}
where
F_{y} is yield strength of steel members.
L_{b} is laterally unsupported length of the compression flanges,
L_{c} is the smaller of 76 b_{f}/ÖF_{y} or L_{c} = 20,000/[(d/A_{f})F_{y}]
b_{f} is the width of the flange, A_{f} is area of the flange.
Non-compact section: allowable bending stress
F_{b} = 0.60 F_{y} if L_{b} £ 76 b_{f}/F_{y}
Compact or non-compact section, F_{b} is the larger of the following
When
[ASD F1-6]
When
[ASD F1-7]
For any value of l/r_{T}.
[ASD F1-8]
Note: for channels bent, allowable stress is determined from [F1-8].
Where
r_{T} (inch) is radius of gyration of a section comprising the compression flange plus 1/3 of the compression web area, taken about an axis in the plan of web. (note: r_{T} is available in AISC steel table for most W and I section)
A_{f} is area of flange,
C_{b} = 1.75+1.05(M_{1}/M_{2})+0.3(M_{1}/M_{2}) but not more than 2.3.
M_{1} and M_{2} are the smaller and the larger applied moments
M_{1}/M_{2} is positive if M_{1} and M_{2} have the same sign.
W and I shape hot-roll section bending about its minor axis,
Compact section: allowable bending stress,
F_{b} = 0.75 F_{y}
Non-compact section: allowable bending stress
F_{b} = 0.60 F_{y}
Maximum width-thickness for compression flange for W, I and Channel section
Compact section: b_{f}/t £ 65/ÖF_{y}.
Non-compact secton: b_{f}/t £ 95/ÖF_{y}.
Simply supported beams
For simply supported beam, the top flange is in compression. If the beam is directly attached to roof deck or floor slab, the compression flange is fully supported. The unsupported length L_{b} is 0.
When the beam supporting joists or other beams, and its flange is directly attached to the supported joists or other beams, the unsupported length is the spacing of the joists or other beams.
Cantilever beams:
For cantilever beam, the compression flange is at the bottom of the beam. If the bottom flange is unbraced, the unsupported length is the length of the cantilever beam.
If bracing is provided at the bottom flange, the unsupported length is the spacing between bracings.
Continuous beams:
For the positive moment portion of the beam, the compression flange is at the top of the beam. The unsupported length is determined as a simply supported beam.
For the negative moment portion of the beam, the compression flange is at the bottom of the beam. The unsupported length is determined as a cantilever beam.
Shear stress, f_{v} =V/(t_{w}d) £ Allowable
shear stress, F_{v} =
0.4 F_{y}
Where V is applied shear force, d is the depth of beam, t_{w} is thickness of web.
Deflection limits listed in International Building Code 2003 Table 1604.3are
Construction |
L |
S or W |
D+L |
Roof members Supporting plaster ceiling Supporting non-plaster ceiling Not supporting ceiling |
L/360 L/240 L/180 |
L/360 L/240 L/180 |
L/240 L/180 L/120 |
Floor members |
L/360 |
- |
L/240 |
Exterior walls and interior partitions With brittle finishes With flexible finishes |
- - |
L/240 L/120 |
- - |
For cantilever beam, L is 2 time the length of cantilever.
Example 1:
Situation:
A structural steel beam is supporting a roof as shown in the figure. The
beam is simply supported at each end.
Design Code: AISC ASD 9th edition_{}
Roof live load: W_{L} =
12 psf
Roof dead load: W_{D} =
20 psf
Length of beam: L
= 35 ft
Unsupported
length (Joist spacing): L_{b} =
5 ft
Tributary width:
TriB = 35 ft
Material: ASTM
A36, yield strength, Fy = 36 ksi
Requirements: Select
a W24 beam
Solution:
Total load on
beam: W = (W_{L}+W_{D}) TriB = 1120 lb/ft.
Maximum moment: M
= W L^{2}/8 = 171.5 kip-ft
Maximum
unsupported length, L_{b} =
5 ft
Try W24x55, From
AISC steel Table, d = 23.57 in, b_{f} =
7 in, t_{f} =0.5
in, A_{f} = b_{f} t_{f}=3.5
in^{2}, t_{w} =
5/16 in
Section modulus,
S = 114 in^{3}.
Moment of
inertia, I = 1350 in^{4}.
Calculate compact
length L_{c }= 76
bf/ÖFy
= 76 (7) / Ö36
= 7.4 ft or
Lc
=20000/[(d/Af)Fy] = 20000 /[23.75/ 3.5) 36 ] = 6.9 ft >
L_{b} = 5 ft
Allowable stress:
F_{b} = 0.66 F_{y} =
24 ksi
Bending stress: f_{b} =
M/S = 18.1 ksi <
F_{b} = 24 ksi O.K.
Check deflection:
Elastic modulus, E = 29000 ksi
Total deflection, D =
5 W L^{4}/(388 E I ) = 0.97 in
Deflection ratio,
L/D =
1/435 < 1/
240 O.K
Live load
deflection, D_{L} = D (12/32)
= 0.36 in,
Live load
deflection ratio, D_{L} /L
= 1/1160 < 1/360 O.K.
Check shear
stress,
Shear force, V =
WL/2 = 19.6 kips
Shear stress, fv
= V/(t_{w}d) = 2.7 ksi
Allowable shear
stress, F_{v} =
0.4 F_{y} = 14.4
ksi
Example 2:
Situation:
A structural steel beam with is supporting a roof as shown in the
figure. The
beam is simply supported at each end.
Design Code: AISC ASD 9th edition_{}
Roof live load: W_{L} =
12 psf
Roof dead load: W_{D} =
20 psf
Length of beam: L
= 36 ft
Length of
cantilever: a = 12 ft
Tributary width:
TriB = 24 ft
Material: ASTM
A992, yield strength, Fy = 50 ksi
Requirements: Select
a W18 beam
Solution:
Total load on
beam: W = (W_{L}+W_{D}) TriB = 768 lb/ft.
Maximum negative
moment at cantilever end: M_{neg} =
W a^{2}/2 = 55.3 kip-ft
Maximum
unsupported length at cantilever, L_{b} =
12 ft
Try W18x35, From
AISC Table, d = 17.7 in, b_{f} =
6 in, t_{f} =7/16
in, A_{f} = b_{f} t_{f}=2.62
in^{2},
t_{w} =
0.3 in, r_{T} =
1.49 in
Section modulus,
S = 57.6 in^{3}.
Moment of
inertia, I = 510 in^{4}.
Check cantilever
end:
Calculate compact
length L_{c }= 76
bf/ÖFy
= 76 (6) / Ö50
= 5.3 ft or
Lc
=20000/[(d/Af)Fy] = 20000 /[17.7/ 2.62) 50 ] = 4.9 ft <
L_{b} = 12 ft
Calculate a/r_{T} =
96
Calculate C_{b},
M_{1} = 0, M_{2} =
69.1 kip-ft, C_{b} =
1.75
Allowable stress
from Eq. F1-6:
Allowable stress
from Eq. F1-8:
Maximum bending
stress:
Bending stress: f_{b} =
M/S = 11.5 ksi <
F_{b} = 24.6 ksi O.K.
Maximum
deflection at cantilever end, E = 29000 ksi
From structural
analysis,
D =
(Wa/24EI)(4a^{2}L-L^{3}+3a^{3}) = -0.93 in D/2a
= 1/248 O.K.
Live load
deflection, DL=D(12/32)
= 0.35 in D/2a
= 1/825 O.K.
Check interior
span:
Maximum negative
moment is the same as cantilever end.
Unsupported
length is less. O.K.
by inspection.
Maximum positive
moment:
M_{pos} =
(W/8L^{2})(L+a)^{2}(L-a)^{2} =
122.8 ft-kip
Bending stress: f_{b} =
M_{pos}/S = 25.6 ksi
The span is fully
supported. Allowable stress: F_{b}=0.66F_{y}=33 ksi O.K.
Check deflection:
Maximum
deflection at x = (L/2)[1-(a/L)^{2}] = 16 ft
Deflection D =
(Wx/24E I L)(L^{4}-2L^{2}x^{2}+Lx^{3}-2a^{2}L^{2}+2a^{2}x^{2})=1.8
in
D/L
= 1/310 O.K.
Live load
deflection D_{L} =
(12/32)D =
0.54 in DL/L=
1/802 O.K.
Check shear
stress:
Shear force at
cantilever end, V_{1}=W a = 9.3 kip
Shear force at
simply supported end, V_{2} =
W (L+a)^{2}/2L =24.6 kips
Shear stress fv =
V_{2}/t_{w}d = 4.6 ksi
Allowable shear
stress, Fv = 0.4 Fy = 20 ksi O.K.