Example ASD & LRFD Failure Checks
CE 331, Spring 2007
1/3
Example problem 1. A simply-supported beam, laterally braced full length.
Dead Load (D) = 2.0 klf
Beam is a steel wide flange, W12x53
wL =4.0 klf wD =2.0 klf
Live Load (L) = 4.0 klf
Yield Stress (Fy) = 50 ksi 3
S = 70.6 in
15 ft
3
2
D
D
M
=
w L
=
w L
8
Desig Desig n Metho Metho d
klf
=
2
L
L
M
8
2.0 (15 ) 2 ft
8 4.0 (15 )2 klf
=
ft
8
Z = 77.9 in
= 56.3k − ft = 112.5k − ft
Load Effect
ASD
Allowable Stress Design
D + L
M
f b
=
f b
=
f b
= 21.7ksi
I in
53.8
=
21.7
ksi
23.8
ksi
= 0.91 < 1, OK
4
unity check
φ Mn = (φ) Fy Z
L
k-ft
Mu = 1.2 (5.63
f b F b
F b = 0.66 Fy = 23.8 ksi
k ft k ft in (5.63 − + 14.06 − ) 4.94
Mu = 1.2 M + 1.6 M
Load and Resistance Factor Design
Failur Failur e Check
allowable bending stress
y
D
LRFD
Resis Resis tance
k-ft
) + 1.6 (14.06
)
k-ft
M u
φ Mn = (0.9) 36 12.6in / 12 ksi
3
in/ft
φ M n
=
k ft 29.3 −
34.0k − ft
= 0.86 < 1, OK
φ Mn = 34.0k-ft
Mu = 29.3
nominal moment capacity
factored moment
strength reduction factor
Example ASD & LRFD Failure Checks
CE 331, Spring 2007
2/3
Example problem 2. A simply-supported beam, laterally braced full length.
Dead Load (D) = 2.0 klf wL wD
Live Load (L) = 4.0 klf D
M
M L
= =
w D L2
8 w L L2
8
= =
2.0klf (15 ft ) 2 8 4.0klf (15 ft ) 2 8
Yield Stress (Fy) = 50 ksi k − ft
= 56.3
Allowable Stress Design
LRFD
3
S = 70.6 in
15 ft
section modulus =I / y
= 112.5k − ft
Desig n Metho d ASD
Beam is a steel wide flange, W12x53
3
Z = 77.9 in
Load Effect
Resis tance
D + L
f b
=
f b
=
M
S
(56.3 k − ft
+ 112.5 k − ft ) 12 in / ft 70.6 in
3
= 28.7 ksi
Failur e Check
Lc = 9.0ft, Lu = 15.9ft (p 132 FE Ref) unbraced length of compression flange ksi Lc < (L b = 15ft ) < Lu f b 28.7 ksi = = 0.96 < 1 ksi ∴ F = 0.60 F = 0.6 (50 ) b
y
F b = 30ksi L p = 8.8ft, Lr = 25.6ft (p 128 FE Ref) L p < (L b = 15ft) < Lr
F b
30
OK
Example ASD & LRFD Failure Checks
CE 331, Spring 2007
2/3
Example problem 2. A simply-supported beam, laterally braced full length.
Dead Load (D) = 2.0 klf wL wD
Live Load (L) = 4.0 klf D
M
M L
= =
w D L2
8 w L L2
8
= =
2.0klf (15 ft ) 2 8 4.0klf (15 ft ) 2
Yield Stress (Fy) = 50 ksi k − ft
= 56.3
Desig n Metho d
Allowable Stress Design
section modulus =I / y 3
Z = 77.9 in
Load Effect
Resis tance
D + L
ASD
3
S = 70.6 in
15 ft
= 112.5k − ft
8
Beam is a steel wide flange, W12x53
f b
=
f b
=
M
S
(56.3 k − ft
+ 112.5 k − ft ) 12 in / ft 70.6 in
3
= 28.7 ksi
Failur e Check
Lc = 9.0ft, Lu = 15.9ft (p 132 FE Ref) unbraced length of compression flange ksi Lc < (L b = 15ft ) < Lu f b 28.7 ksi = = 0.96 < 1 ksi ∴ F = 0.60 F = 0.6 (50 ) b
y
30
F b
F b = 30ksi
OK
L p = 8.8ft, Lr = 25.6ft (p 128 FE Ref)
LRFD
L p < (L b = 15ft) < Lr
Load and Resistance Factor Design
D
∴ φ Mn = C b [φ M p – BF(L b – L p)]
L
Mu = 1.2 M + 1.6 M k-ft
Mu = 1.2 (56.3
k-ft
) + 1.6 (112.5
)
k-ft
Mu = 248
CE 331, Spring 2007
<φ
M p C b = 1.0 (always for this class)
φ M p = 292k-ft (p 128 FE Ref) BF = 4.78k ( p 128 FE Ref)
M u
φ Mn = (1) [292kpft – (4.78k (15ft – 8.8ft)] φ Mn = 262k-ft (< 292k-ft = φ M p)
φ M n
=
248 k − ft 262
k − ft
= 0.95 < 1 OK
Example ASD & LRFD Failure Checks
3/3
Example problem 3. A single column, laterally braced at mid-height for buckling in the weak direction, no lateral support for buckling in the strong direction.
Dead Load k (D) = 25 Live Load k (L) = 75
k
P L = 75
k
P D = 25
Beam is a steel wide flange, W6x20 Yield Stress (Fy) = 50 ksi 4
Ixx = 41.4 in
4
Iyy = 13.3 in 9 ft
2
A = 5.87 in
Lu_x = 18 ft (unbraced length for buckling in the strong direction) 9 ft
Lu_y = 9 ft (unbraced length for buckling in the weak direction)
Design
Load Effect
Resis tance
Failu re Check
Example ASD & LRFD Failure Checks
CE 331, Spring 2007
3/3
Example problem 3. A single column, laterally braced at mid-height for buckling in the weak direction, no lateral support for buckling in the strong direction.
Dead Load k (D) = 25
Beam is a steel wide flange, W6x20
k
P L = 75
Yield Stress (Fy) = 50 ksi
k
P D = 25
4
Ixx = 41.4 in
4
Live Load k (L) = 75
Iyy = 13.3 in
2
A = 5.87 in
9 ft
Lu_x = 18 ft (unbraced length for buckling in the strong direction) 9 ft
Lu_y = 9 ft (unbraced length for buckling in the weak direction)
Design Method ASD
Load Effect
D + L
P
f a
=
f a
= 17.0ksi
A
=
Resis tance
25k + 75k 5.87 in 2
r x
=
r y
=
I x A I y
r x k y Lu _ y r y
allowable axial stress
LRFD
Pu = 1.2 PD + 1.6 PL Pu = 1.2 (25k ) + 1.6 (75 k ) Pu = 150k
=
A
k x Lu _ x
slenderness ratio
=
=
41.1in 4
= 2.65
13.3in 4 2
1
in ft 12
1 ft
1.50in
18.7ksi
= 0.91 < 1, OK
= 81.5
controls
unity check
)
= 72.0
Fa = 18.7ksi [ from Table on pg 134 FE Ref. for Fy = 50ksi and kL/r = 81.5 ]
⎛ k L ⎞ = 81.5 (from above) ⎜ ⎟ ⎝ r ⎠ max φ c F cr = 26.16ksi (Table p 131 FE Ref.) φcFcr = 26.16ksi
factored axial force
=
)
2.65in
=
F a
17.0ksi
= 1.50in
in ft 12 (1.0)(18 ft
(1.0)(9
f a
in
5.87 in 2
5.87 in
Failu re Check
[from Table on pg 131 FE Ref. for Fy = 50ksi and kL/r = 81.5 ]
φ Pn = φc Fcr A = 26.2ksi 5.87in2 φ Pn =153.8kt nominal axial capacity
Pu
φ Pn
=
150 k 153.8 k
= 0.98 < 1, OK
