Example 1-4
The interior column of a multi-storey building is subjected to the following axial loads:
Dead load = 900 kN
Reduced live load from floors = 1100 kN
Roof live load = 220 kN
Fluid pressure = 110 kN
Wind loads:
Compression = 560 kN
Tension = 460 kN
Earthquake load:
Compression = 260 kN
Tension = 310 kN
Determine the governing factored column load, Pu, using the LRFD combination. Assume that factor f₁ = 0.50.
Solution
D = 900kN (Dead load)
L = 1100kN (Reduced live load)
Lr = 220kN (Roof live load)
F = 110kN (Fluid pressure)
Wc = 560kN (Wind load, compression)
Wt = 460kN (Wind load, tension)
Ec = 260kN (Earthquake load, compression)
Et = 310kN Earthquake load, tension
f₁= 0.5
Eq. 1-1:
Pu1 = 1.4(D + F)
Pu1 = 1.4(900 + 110) = 1414kN
Eq. 1-2:
Pu2 = 1.2(D + F) + 1.6L + 0.5L
Pu2 = 1.2(900 + 110) + 1.6(1100) + 0.5(220) = 3082kN
Eq. 1-3:
Pu3a =1.2 D + 1.6 L r + f 1 L
Pu3a =1.2(900) + 1.6(220) + 0.5(1100) = 1982kN
Pu3b =1.2 D + 1.6 Lr + 0.5 Wc
Pu3b =1.2(900) +1.6(220) + 0.5(560) = 1712 kN
Eq. 1-4:
Pu4a =1.2 D + 1.0 Wc + f1L + 0.5 L r
Pu4a =1.2(900) + 1(560) + 0.5(1100) + 0.5(220) = 2300kN
Pu4b =1.2 D - 1.0 Wt + f1L + 0.5 Lr
Pu4b =1.2(900) - 1(460) + 0.5(1100) + 0.5(220) =1280 kN
Eq. 1-5:
Pu5a =1.2 D + 1.0 Ec + f1L
Pu5a =1.2(900) + 1.0(260) + 0.5(1100) = 1890 kN
Pu5b =1.2 D - 1.0 Et + f1L
Pu5b =1.2(900) - 1.0(310) + 0.5(1100) = 1320 kN
Eq. 1-6:
Pu6a = 0.9 D + 1.0 Wc
Pu6a = 0.9(900) + 1.0(560) = 1370 kN
Pu6b = 0.9 D - 1.0 Wt
Pu6b = 0.9(900) - 1.0(460) = 350 kN
Eq. 1-7:
Pu7a = 0.9 D + 1.0 Ec
Pu7a = 0.9(900) + 1.0(260) = 1070 kN
Pu7b = 0.9 D - Et
Pu7b = 0.9(900) - 1.0(310) = 500 kN
The governing factored load combination is Pu2 = 3082 kN answer
Example 1-5
The four-span continuous beam shown in Figure 1.5 is simply supported at A, B, C, D and E. The beam span is l = 8 m
and will be subjected to service dead load of 20 kN/m and service live load of 14 kN/m. Considering pattern loading
for live load as given in Figure 1.2, determine the following critical factored forces:
a) The maximum support reaction at E.
b) The maximum shear at D.
c) The maximum bending moment at C.
d) The maximum shear and bending moment in the continuous beam.
e) The maximum positive bending moment in the continuous beam
Solution
Span, l = 8m
Service dead load, wd = 20 kN/m
Service live load, wl = 14 kN/m
Factored loads:
For dead and live load combination, the critical combination is
Eq. 1-2, U = 1.2D + 1.6L
Factored dead load, wdu = 1.2 wd = 24 kN/m
Factored live load, wlu = 1.6 wl = 22.4 kN / m
a) Maximum reaction at E (or A)
RE = 0.393 Wdu • l + 0.446 wlu • l
RE = 0.393(24)(8) + 0.446(22.4) (8) = 155.379 kN answer
b) Maximum shear at D (or B)
VD = 0.607 wdu • l + 0.620 Wlu • l
VD = 0.607(24)(8) + 0.620 (22.4)(8) = 227.648 kN answer
c) Maximum bending moment at C.
Mc = - 0.0714 Wdu • l2 - 0.0714 Wlu • l2
Mc = - 0.0714(24)(8)2 - 0.0714(22.4)(8)2 = - 212.029kN-m answer
d) Maximum shear and bending moment
Maximum shear:
The maximum shear will occur at point B or D.
From the result of (b), Vmax = 227.648 kN
Maximum bending moment:
The maximum bending moment will occur at B or D.
Mmax = - 0.1071 Wdu • l2 - 0.1205 Wlu • l2
Mmax = - 0.1071(24)(8)2 - 0.1205(22.4)(8)2 = - 337.254 kN-m answer
e) Maximum positive bending moment will occur at end span AB (or DE).
Approximately;
Mmax = 0.0772 Wdu • l2 + 0.0996 Wlu • l2
Mmax = 0.0772(24)(8)2 +0.0996(22.4)(8)2 = 261.366 kN-m
Exact solution:
Wu = Wdu + Wlu = 24 + 22.4 = 46.4 kN m
RA = 0.393 Wdu • l + 0.446 Wlu • l
RA = 0.393(24)(8) + 0.446(22.4) (8) = 155.379kN
VA = RA = 155.379 kN
x = RA / Wu = 155.379/46.4 = 3.349 m
A1 = (1/2) VA • x = 1/2 (155.379)(3.349) = 260.158kN-m
Example 2-12
Compute the LRFD design tensile strength and ASD allowable tensile strength of the plate in Example 2-3. All steel are
A36 with Fu = 400 MPa and Fy = 250 MPa.
From Example 2-3:
Gross area, Ag = 280 • 12 = 3360 mm2
Path ABCF:
Net Area = 3360 - 2(29 mm) • 12 + (100)2/4(80) • 12 = 3039 mm2
Shear lag factor, U = 1 (Table 2-2, Case 1)
Effective net area, Ae = U • An = 3039 mm2
Fy = 250 MPa
Fu = 400 MPa
Gross section yielding:
Pn = Fy • Ag = 250(3360) = 840000N = 840kN
LRFD: ø = 0.90
Pu_max1 = ø • Pn = 0.90 (840) = 756 kN
ASD: Ω = 1.67
Pa_max1 = Pn /Ω = 840/1.67 = 503 kN
Net area rapture:
Pn = Fu • Ae = 400 (3039) = 1215600 N = 1215.6 kN
LRFD: ø = 0.75
Pu_max2 = ø • Pn = 0.75 (1215.6) = 911.7 kN
ASD: Ω = 2.00
Pa_max2 = Pn / Ω = 1215.6/2= = 607.8 kN
Finally:
Pu_max = min (Pu_max1, Pu_max2) = 756 kN
Pa_max = min ((Pa_max1, Pa_max2) = 503 kN
Example 2-13
Compute the LRFD design tensile strength and ASD allowable tensile strength of the angle section in Example 2-10. All
steel are A36 with Fu = 420 MPa and Fy = 250 MPa.
Solution:
Properties of L203 x 152 x 25.4
Area, A = 8390 mm2
Net area, An = Ag = 8390 mm2
Effective net area, Ae = U • An
U = 1 - x/l
x (From the back of long leg) = 41.9 mm
U = 1 - 41.9/150 = 0.721
Ae = 0.721 • 8390 = 6049.2 mm2
Thickness, t = 25.4mm
Fu = 420 MPa
Fy = 250 MPa
Gross section yielding:
Pn = Fy • Ag = 250(8390) = 2097500N = 2097.5kN
LRFD: ø = 0.9
Pu_max1 = ø • Pn = 0.90 (2097.5) = 1887.75kN
ASD: Ω = 1.67
Pa_max1 = Pn/Ω = 2097.5/1.67 = 1255.99kN
Net area rapture:
Pn = Fu • Ae = 420 (6049.2) = 2540664 N = 2540.66 kN
LRFD: ø = 0.75
Pu_max2 = ø • Pn = 0.75 (2540.66) = 1905.5 kN
ASD: Ω = 2.00
Pa_max2 = Pn / Ω = 2540.66/2= = 1270.33 kN
Finally:
Pu_max = min (Pu_max1, Pu_max2) = 1887.75
Pa_max = min ((Pa_max1, Pa_max2) = 1255.99kN
Example 2-22
Determine the LRFD design and ASD allowable block shear strengths of the PL 12 mm x 280 mm shown in Figure 2.30.
Hole diameter is 24 mm. All steel are A36 with Fu = 400 MPa and Fy = 250 MPa.
Solution
Fu = 400 MPa
Fy = 250 MPa
Plate thickness, t = 12mm
Effective hole diameter, de = dh + 2 = 24mm + 2mm = 26mm
Path ABCDE:
Gross shear area, Agv = 160 mm • (12 mm) 2 = 3840 mm2
Net shear area, Anv = Agv – 1.5 det • 2 = 3840 - 1.5 (26 mm)(12 mm)2 = 2904 mm2
Net tension area: Ant = ((160 mm -2(26 mm) + ((60 mm)2 / 4(80 mm)) 2))12 mm
Ant = 1566 mm2
Ubs =1.0
Rn1 = 0.6 Fu • Anv + Ubs • Fu • Ant
Rn1 = 0.6 (400)(2904) + 1.0 (400)(1566)= 1323360 N = 1323.36 kN
Rn2 = 0.6 Fy • Agv + Ubs • Fu • Ant
Rn2 = 0.6 (250)(3840) + 1.0 (400)(1566) = 1202400 N = 1202.4 kN
Rn_ABCDE = min (Rn1, Rn2) = 1202.4 kN
Path ABCDF
Gross shear area, Agv =160 mm • (12 mm) = 1920 mm2
Net shear area; Anv = Agv – 1.5 det = 1920 -1.5(26 mm)(12 mm) = 1452 mm2
Net tension area: Ant = ((220 mm - 2.5(26 mm) + ((60 mm)2 / 4(80 mm)) 2))12 mm = 2130 mm2
Ubs =1.0
Rn1 = 0.6 Fu • Anv + Ubs • Fu • Ant
Rn1 = 0.6(400)(1452) + 1(400)(2130) = 1200480 N = 1200.48 kN
Rn2 = 0.6 Fy • Agv + Ubs • Fu • Ant
Rn2 = 0.6(250)(1920) + 1(400)(2130) = 1140000 N = 1140 kN
Rn_ABCDF = min (Rn1, Rn2) = 1140 kN
Path ABCG:
Agv =160 mm • (12 mm) = 1920 mm2
Net shear area, Anv = Agv – 1.5 det = 1920 - 1.5 (26mm)(12 mm) = 1452 mm2
Net tension area: Ant = ((220 mm - 1.5(26 mm) + ((60 mm)2 / 4(80 mm)))12 mm = 2307 mm2
Ubs =1.0
Rn1 = 0.6 Fu • Anv + Ubs • Fu • Ant
Rn1 = 0.6(400)(1452) + 1(400)(2307) = 1271280 N = 1271.28 kN
Rn2 = 0.6 Fy • Agv + Ubs • Fu • Ant
Rn2 = 0.6(250)(1920) + 1(400)(2307) = 1210800 N = 1210.8 kN
Rn_ABCG = min (Rn1, Rn2) = 1210.8 kN
Rn_ABCG = min (Rn_ABCDE, Rn_ABCDF, Rn_ABCG) = 1140 kN
Governed by Path ABCDF
LRFD: ø = 0.75
Ru_max = ø • Rn = 0.75
Ru_max = 0.75 (1140)
Ru_max = 855 kN answer
ASD: Ω = 2
Ra_max = Rn / Ω
Ra_max = 1140 / 2
Ra_max = 570 kN answer
Example 2-23
Determine the LRFD design and ASD allowable block shear strengths of the coped W410 x 60 Section (t w = 7.75mm)
shown in Figure 2.32. Hole diameter is 24 mm. Steel is A572 Grade 50 steel with F y = 345 MPa and Fu = 450 MPa.
Solution
Fy = 345 MPa
Fu= 450MPa
Web thickness, tw = 7.75mm
Hole diameter, dh= 24mm
Effective hole diameter; de = dh + 2mm = 24 + 2 = 26mm
Gross shear area, Agv =280 mm • (7.75 mm) = 2170 mm2
Net shear area, Anv = Agv - 4.5(26 mm)(7.75 mm) = 1263.25 mm2
Net tension area: Ant =(70 mm - 0.5(26 mm)) • 7.75 mm = 441.75 mm2
Ubs =1.0
Rn1 = 0.6 Fu • Anv + Ubs • Fu • Ant
Rn1 = 0.6(450)(1263.25) + 1(450)(441.75) = 539865 N = 539.865 kN
Rn2 = 0.6 Fy • Agv + Ubs • Fu • Ant
Rn2 = 0.6(345)(2170) + 1(450)(441.75) = 647977.5 N = 647.978 kN
Rn = min (Rn1, Rn2) = 539.865 kN
LRFD: ø = 0.75
Ru_max = ø • Rn = 0.75
Ru_max = 0.75 (539.865)
Ru_max = 404.899 kN answer
ASD: Ω = 2
Ra_max = Rn / Ω
Ra_max = 539.865 / 2
Ra_max = 269.933 answer
Example 4-3
Determine the plastic moment of the propped beam shown in Figure 4-27 using the (a) collapse method and (b)
equilibrium method
(a) Collapse method.
Let the rotation at A be 9, then the displacement at C is or θ • 3 = 3θ
The rotation at B is Δ/5 or 3θ / 5 and the total rotation
at C is θA + θB = 8θ / 5
By virtual work method:
Internal work = external work
Internal work = MP x Σθ (at plastic hinges)
External work = 2 Force x displacement
MP (θ+8θ/5) = 50(3θ) = 57.692 kN–m
(b) Equilibrium method
Mc = M P MC right = RB (5) = Mp
RB = M P / 5
MA = - MP M A = RB(8) - 50(3) = - MP
(MP/5)(8) + MP = 150
MP = 57.692 kN–m
RB = M P / 5 RB = 57.692 / 5 = 11.5384 kN
ΣFv = 0 RA = 50 - 11.538 = 38.462kN
Example 4-9
A W400 x 60 of A992 (Fy = 345MPa) steel beam is simply supported on a span of 10 m. The beam supports a reinforced
concrete floor slab that provides continuous lateral support of the compression flange. The loads on the beam are:
Superimposed dead load = 7kN / m
Live load = 8kN / m
Properties of W400 x 60
d = 407mm Mass = 60 kg/m
bf = 178mm Sx = 1060 x 103 mm3
tf = 12.8mm Zx = 1190 x 103 mm3
tw = 7.7mm
Determine the adequacy of the beam based on moment strength.
Solution
Determine whether the section is compact, non-compact, or slender.
Flange:
bf/(2tf) = 178/(2(12.8)) = 6.956=3
0.38√(E/Fy) = 0.38√(200000/345) = 9.15 ∴ the flange is compact
Web:
h/tw = (407 - 2(12.8))/7.7 = 49.53
3.76√(E/Fy) = 3.76√ (200000/345) = 90.53 ∴ the web is compact
Therefore, the section is compact.
Since the beam is fully laterally supported;
Mn = MP = Fy • Zx = 345(1190 x 103) = 410550000 N-mm = 410.55 kN-m
Total dead load:
wd = 7 + (60 • 9.81) / 1000 = 7.589 kN/m
LRFD: ø = 0.9
Design strength, øbMn = 0.9(410.55) = 369.495 kN-m
wu= 1.2wd + 1.6wl = 1.2(7.589) + 1.6(8) = 21.907 kN/m
Mu = wuL2 / 8
Mu = 21.907 • (10)2 / 8
Mu = 273.838 kN-m
Mu < øbMn ∴ the section is adequate
Example 4-17
A roof system consists of trusses spaced 5 m apart. ASTM A * C(36,125) * 10 purlins are used with sag rods located at
the center of each purlin. The purlins used are and to support service dead load of 720 Pa and service live load of
580 Pa. Angle of theta = 25 deg, a ) Determine the adequacy of the purlins based on flexure and shear (LRFD only), if
they are spaced at 0.8 m apart, and b) determine the maximum safe spacing of the purlins based on LRFD flexure
only.
Properties of C125 x 10
d = 127mm Ix = 3.11 x 106 mm4
bf = 44.5mm Iy = 0.196 x 106 mm4
Sx = 49 x 10 mm
3 3 tf = 8.1mm
Sy = 6.1 x 103 mm3 tw = 4.8mm
Zx = 58.17 x 103 mm3 Zy = 12.41 x 103 mm3
J = 22.85 x 10 mm
3 4 ho = 119mm
ry = 12.42mm Cw = 0.596 x 109 mm6
rts =14.83 mm Mass = 10 kg/m
