CE 331, Spring 2007 Example ASD & LRFD Failure Checks 1 / 3

Example problem 1. A simply-supported beam, laterally braced full length.

Dead Load (D) = 2.0 klf

Live Load (L) = 4.0 klf

ftk

ftklfLL

ftkftklfD

D

LwM

LwM

−

−

===

===

5.1128

)15(0.48

3.568

)15(0.28

22

22

Beam is a steel wide flange, W12x53

Yield Stress (Fy) = 50 ksi

S = 70.6 in3

Z = 77.9 in3

Design Method Load Effect Resistance Failure Check

ASD

Allowable Stress Design

ksib

in

inftkftk

b

LD

b

f

f

IyMf

7.218.53

94.4)06.1463.5(4

=

+=

=

−−

+

Fb = 0.66 Fy = 23.8 ksi ksi

ksi

b

b

Ff

8.237.21

= = 0.91 < 1, OK

LRFD

Load and Resistance Factor Design

Mu = 1.2 MD + 1.6 ML

Mu = 1.2 (5.63k-ft) + 1.6 (14.06k-ft)

Mu = 29.3k-ft

φ Mn = (φ) Fy Z

φ Mn = (0.9) 36ksi 12.6in3 / 12 in/ft

φ Mn = 34.0k-ft

ftk

ftk

n

u

MM

−

−

=0.343.29

φ= 0.86 < 1, OK

wD = 2.0 klf wL = 4.0 klf

15 ft

strength reduction factor

allowable bending stress

unity check

factored moment nominal moment capacity

CE 331, Spring 2007 Example ASD & LRFD Failure Checks 2 / 3

Example problem 2. A simply-supported beam, laterally braced full length.

Dead Load (D) = 2.0 klf Live Load (L) = 4.0 klf

ftkftklfL

L

ftkftklfD

D

LwM

LwM

−

−

===

===

5.1128

)15(0.48

3.568

)15(0.28

22

22

Beam is a steel wide flange, W12x53

Yield Stress (Fy) = 50 ksi

S = 70.6 in3

Z = 77.9 in3

Design Method Load Effect Resistance Failure Check

ASD

Allowable Stress Design ksi

ftinftkftk

b

LD

b

inf

SMf

7.286.70

12)5.1123.56(3

/

=+

=

=

−−

+Lc = 9.0ft, Lu = 15.9ft (p 132 FE Ref)

Lc < (Lb = 15ft ) < Lu ∴Fb = 0.60 Fy = 0.6 (50ksi) Fb = 30ksi

ksi

ksi

b

b

Ff

307.28

= = 0.96 < 1

OK

LRFD

Load and Resistance Factor Design

Mu = 1.2 MD + 1.6 ML

Mu = 1.2 (56.3k-ft) + 1.6 (112.5k-ft)

Mu = 248k-ft

Lp = 8.8ft, Lr = 25.6ft (p 128 FE Ref) Lp < (Lb = 15ft) < Lr ∴φ Mn = Cb [φ Mp – BF(Lb – Lp)] < φ Mp Cb = 1.0 (always for this class) φ Mp = 292k-ft (p 128 FE Ref) BF = 4.78k (p 128 FE Ref) φ Mn = (1) [292kpft – (4.78k(15ft – 8.8ft)] φ Mn = 262k-ft (< 292k-ft = φ Mp)

ftk

ftk

n

u

MM

−

−

=262248

φ= 0.95 < 1

OK

wD wL

15 ft

unbraced length of compression flange

section modulus = I / y

CE 331, Spring 2007 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 (D) = 25k

Live Load (L) = 75k

Beam is a steel wide flange, W6x20 Yield Stress (Fy) = 50 ksi Ixx = 41.4 in4

Iyy = 13.3 in4 A = 5.87 in2

Lu_x = 18 ft (unbraced length for buckling in the strong direction)

Lu_y = 9 ft (unbraced length for buckling in the weak direction)

Design Method

Load Effect Resistance Failure Check

ASD

ksia

kkLD

a

f

inAPf

0.17

87.57525

2

=

+==

+

0.7250.1

)1

129)(0.1(

5.8165.2

)1

1218)(0.1(

50.187.53.13

65.287.5

1.41

_

_

2

4

2

4

==

==

===

===

in

ft

inft

y

yuy

in

ft

inft

x

xux

inyy

inxx

rLk

rLk

inin

AI

r

inin

AIr

Fa = 18.7ksi [ from Table on pg 134 FE Ref. for Fy = 50ksi and kL/r = 81.5 ]

ksi

ksi

a

a

Ff

7.180.17

= = 0.91 < 1,

OK

LRFD

Pu = 1.2 PD + 1.6 PL

Pu = 1.2 (25k) + 1.6 (75k)

Pu = 150k Ref.) FE131 p (Table 16.26

above) from(5.81max

ksicrcF

rLk

=

=⎟⎠⎞

⎜⎝⎛

φ

φcFcr = 26.16ksi [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

k

k

n

u

PP

8.153150

=φ

= 0.98 < 1,

OK

PD = 25k

9 ft

PL = 75k

9 ft

unity check

allowable axial stress

slenderness ratio controls

factored axial force

nominal axial capacity