Steel Beam Design

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Nov 15, 2013 (3 years and 8 months ago)

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Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

1
of
8

Revised 4 May 2011

Steel Beam Design

Six Easy Steps

Steel beam design is about selecting the lightest steel beam that
will support the load without exceeding the bending strength or
shear strength of the material, and without exceeding the max
i-
mum allowable deflection for th
e beam. We want the lightest
beam because it is generally the cheapest. We can solve these
pro
b
lems with a 6
-
step process.

Step 1
Identify all loads and design constraints (yield strength,
maximum allowable deflection
Δ
max
, beam length L, etc.).

Step 2
Draw the load diagram and calculate all reactions.

Step 3
Draw the shear and moment diagrams, and calculate V
max

and M
max
. If the loading conditions are right, use the Formula
Method to find these values.

Step 4
Calculate
the plastic section modulus Z
x
required to su
p-
port the applied moment. Select the lightest steel beam from the
Appendix that su
p
ports M
max
and has enough stiffness to limit Δ
max

(if deflection is a constraint).

Step 5
Include the beam weight in new drawin
gs of the load,
shear, and moment diagrams. Check that the beam can support the
applied loads and its own weight, and that it still meets the max
i-
mum d
e
flection constraint.

Step 6
Calculate the shear strength of the selected beam, and
check that the beam w
ill support more shear load than is applied.


Example #1

Select the lightest W
-
beam that will support a uniformly distri
b-
uted load of 3 kip/ft. on a simply
-
supported span of 20 ft. The
beam is rolled high
-
strength, low
-
alloy steel (HSLA).

Step 1
We know t
he loading and length; the steel has a yield
strength
!
Y
S

50
ks
i
. The maximum beam deflection Δ
max
is not
specified.


Step 2
The total load on the beam is
3
ki
p
f
t
.
20
f
t
.

60
ki
ps
. Since
the loading is symmetrical,
R
A

R
B

30
ki
ps
.


Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

2
of
8

Revised 4 May 2011

Step 3
The shear diagram for a uniform distributed load is two
triangles. The moment diagram is a parabola, where M
max
is the
area of the shear diagram up to the midspan, or the area of the left
-
hand triangle. Since th
e area of a triangle is the base times the
height divided by two,
M
m
a
x

30
ki
ps
!
10
f
t
.
2

150
ki
p
f
t
.







Beam

Z
x
(in.
3
)



W18×40

78.4



W12×50

72.4



W10×54

66.6



W16×36

64.0



W12×40

57.5


Step 4
The moment strength of a steel beam is
M
R

0.6
!
Y
S
Z
x
.
We can rewrite the equation to find the value of Z
x
requir
ed to
support the applied moment.

Re
qui
r
e
d Z
x

M
0.6
!
Y
S

1.67
M
!
Y
S

1.67
"
150
ki
p
f
t
.
i
n
.
2
50
ki
ps
12
i
n
.
f
t
.

60.1
i
n
.
3

Appendix I lists W
-
beams in decreasing order of plastic section
modulus Z
x
. Look for a beam with a slightly larger Z
x
than the
r
e
quired value. In this case, the lightest beam is
W16×36, with a
w
eight of 36 lb./ft., or 0.036 kips/ft.





Step 5
We can add the beam weight to the applied uniform di
s-
tributed load, for a total of 3.036 kips/ft. Th
e total load on the
beam is
3.036
ki
p
f
t
.
20
f
t
.

60.72
ki
ps
. Since the loading is sy
m-
metrical,
R
A

R
B

30.36
ki
ps
. The maximum moment is
M
m
a
x

30.36
ki
ps
!
10
f
t
.
2

151.8
ki
p
f
t
.

Re
qui
r
e
d Z
x

1.67
!
151.8
ki
p
f
t
.
i
n
.
2
50
ki
ps
12
i
n
.
f
t
.

60.8
i
n
.
3
, which
is less than Z
x
of the selected beam. As long as we
have more than
we need, the beam will survive. If the new required Z
x
had been
66 in.
3
, then we would have to select a different beam.


Step 6
We know the beam will support the load without exceeding
its bending strength; now w
e need to check shear strength. For
wide
-
flange steel W
-
beams,
V
a
p
p
l
i
e
d
!
0.4
"
Y
S
dt
w
where d is the
beam depth and t
w
is the thickness of the web. Find these dime
n-
sions in Appendix A. A W
16×36 beam can support a shear load of
0.4
!
50
ki
ps
i
n
.
2
!
15.86
i
n
.
!
0.295
i
n
.

93.6
ki
ps
. Since the actual
shear load of 30.36 kips is less than 93.6 kips, the beam will not
fail in shear.


Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

3
of
8

Revised 4 May 2011

Example #2

Select the lightest W
-
beam that will support a uniformly distri
b-
uted load of 3 kip/ft. on a simply
-
supported span of 20 ft.
and
d
e-
flect no
more than 0.6 inches. The beam is rolled high
-
strength,
low
-
alloy steel (HSLA).

Steps 1
-
4
The first few steps are identical to Example #1 because
the beam loading and length are the same. However, we have an
additional constraint of
!
m
a
x

0.6
i
n
.
From Appendix H, Case #1,
the maximum deflection for a simply
-
supported beam with a un
i-
form distributed load is
!
m
a
x

5
w
L
4
384
E
I
. We can rewrite the equ
a-
tion to find the moment of inertia required to limit the maximum
deflection.








Beam

Z
x
(in.
3
)

I
x
(in.
4
)



W18×40

78.4

612



W12×50

72.4

394



W10×54

66.6

303



W16×36

64.0

448



W12×40

57.5

310


Re
qui
r
e
d I

5
w
L
4
384
E
!
m
a
x

5
384
3
ki
p
f
t
.
20
f
t
.


4
i
n
.
2
30
"
10
3
ki
p
0.6
i
n
.
12
i
n
.


3
f
t
.
3

600
i
n
.
4

The moment of inertia of a
W16×36 is 448 in.
4
, which is not
enough. Instead, we need to select a beam with a moment of ine
r-
tia greater than 600 in.
4
, such as W18×40, which has a weight of
40 lb./ft. or 0.040 k
ip/ft.






Step 5
Add the beam weight to the applied uniform di
s
tributed
load, for a total of 3.040 kips/ft. The total
load on the beam is
3.040
ki
p
f
t
.
20
f
t
.

60.8
ki
ps
. Since the loading is sy
m
metrical,
R
A

R
B

30.4
ki
ps
. The maximum moment is
M
m
a
x

30.4
ki
ps
!
10
f
t
.
2

152
ki
p
f
t
.

Re
qui
r
e
d Z
x

1.67
!
152
ki
p
f
t
.
i
n
.
2
50
ki
ps
12
i
n
.
f
t
.

60.9
i
n
.
3
, which
is less than Z
x
of the selected beam, so the beam is stron
g enough
in bending.

Checking for deflection,

Re
qui
r
e
d I

5
384
3.04
ki
p
f
t
.
20
f
t
.


4
i
n
.
2
30
!
10
3
ki
p
0.6
i
n
.
12
i
n
.


3
f
t
.
3

608
i
n
.
4

Since the beam’s moment of inertia is more than the required
value, the beam meets the deflection criterion.


Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

4
of
8

Revised 4 May 2011

Step 6
Use
V
a
p
p
l
i
e
d
!
0.4
"
Y
S
dt
w
to f
ind the shear strength. A
W
18×40 beam can support a shear load of
0.4
!
50
ki
ps
i
n
.
2
!
17.9
i
n
.
!
0.315
i
n
.

113
ki
ps
. Since the actual
shear load of 30.4 kips is less than 113 kips, the beam will not fail
in shear.


Examples #1 and #2 are the easiest types to solve because the
weight of the beam is a uniform distributed load, therefore the
load, shear, and moment diagrams have the same shape after the
beam weight is included.


Example #3

Select the lightest W
-
beam that will support a point load of 40 kips
at the midspan of a si
mply
-
supported 30 foot span.

Step 1

P

40
ki
ps
,
L

30
f
t
.
,
!
Y
S

50
ks
i
, Δ
max
is not spec
i-
fied.

Step 2
The total load on the beam is 40 kips. Since the loading is
symmetrical,
R
A

R
B

40
ki
ps
2

20
ki
ps
.

S
tep 3
The shear diagram for a point load at the midspan is two
rectangles. The moment diagram is a triangle, where M
max
is the
area of the shear diagram up to the midspan, or the area of the left
-
hand rectangle:
M
m
a
x

20
ki
ps
!
15
f
t
.

300
ki
p
f
t
.








Beam

Z
x
(in.
3
)




W21×62

144




W14×74

126




W18×60

123




W21×50

110



Step 4
Calculate the needed plastic section modulus:

Re
qui
r
e
d Z
x

1.67
M
!
Y
S

1.67
"
300
ki
p
f
t
.
i
n
.
2
50
ki
ps
12
i
n
.
f
t
.

120.2
i
n
.
3

Select
W18×60, with a weight of 60 lb./ft., or 0.06 kips/ft.






Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

5
of
8

Revised 4 May 2011

Step
5
Redraw the load, shear, and moment diagrams to include
the weight of the beam. Add the beam weight to the point load for
a total load of
40
ki
ps

0.06
ki
p
f
t
.
20
f
t
.

41.8
ki
ps
. Since the
loading is sy
m
metrical,
R
A

R
B

20.9
ki
ps
.

V
1

R
A

R
B

20.9
ki
ps
V
2

V
1
!
0.06
ki
p
f
t
.
15
f
t
.

20
ki
ps
V
3

V
2
!
40
ki
ps

!
20
ki
ps
V
4

V
3

20.9
ki
ps

0
ki
ps

The maximum moment is the area of the left
-
hand trapezoid,
which is the average height times the base:

M
m
a
x

20.9
ki
ps

20
ki
ps
2
15
f
t
.

306.75
ki
p
f
t
.

Re
qui
r
e
d Z
x

1.67
!
306.75
ki
p
f
t
.
i
n
.
2
50
ki
ps
12
i
n
.
f
t
.

122.9
i
n
.
3
,
which is slightly less than Z
x
of the selected beam, so the beam is
strong enough in bending
.


Step 6
Use
V
a
p
p
l
i
e
d
!
0.4
"
Y
S
dt
w
to find the shear strength. A
W
18×60 beam can support a shear load of
0.4
!
50
ki
ps
i
n
.
2
!
18.24
i
n
.
!
0.415
i
n
.

151
ki
ps
. Since the actual
shear load of 20.9 kips is less than 151 kips, the beam will not fail
in shear.


Another way to solve this problem is to use the Formula Method
and Superposition. For
Step 2
and
Step 3
, Case #5 in Appendix H
gives us

R
A

R
B

V
m
a
x

P
2

40
ki
ps
2

20
ki
ps
M
m
a
x

P
L
4

40
ki
ps
!
30
f
t
.
4

300
ki
p
f
t
.


Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

6
of
8

Revised 4 May 2011

Solve
Step 4
as before. For
Step 5
, we can add the
reaction forces
for the two cases:

R
A

R
B

P
2

w
L
2

40
ki
ps
2

0.06
ki
ps
f
t
.
30
f
t
.
2

20.9
ki
ps

The maximum shear load occurs at the same place in both shear
diagrams (the ends of the beams) and is equal to the reactions, so
V
m
a
x

R
A

R
B

20.9
ki
ps
.

The maximum moment occurs at the sa
me place in both moment
diagrams (the midspan), so we can add the maximum moments for
both cases:

M
m
a
x

P
L
4

w
L
2
8

40
ki
ps
!
30
f
t
.
4

0.06
ki
ps
f
t
.
30
f
t
.


2
8

306.75
ki
p
f
t
.


This use of the Formula Method and Superposition
only
works for
shear loads when the maximum values o
f the two cases occur at
the same location; likewise for maximum moment. For example, if
the initial loading is a point load which is not at the midspan as in
Case #6, then the maximum shear load
V
m
a
x

R
A
, but the max
i-
mum moment is not a
t the mi
d
span. You can add the maximum
shear loads of Cases #1 and #6 because they coincide, but you
cannot add the maximum moments because they do not coi
n
cide.



Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

7
of
8

Revised 4 May 2011


Example #4

Select the lightest W
-
beam that will support a poin
t load of 5 kips
3 feet from the end of a 10
-
foot cantilever beam. The maximum
deflection is 0.50 inches.


Step 1

P

5
ki
ps
,
L

10
f
t
.
, location of the point load
a

3
f
t
.
,
!
Y
S

50
ks
i
,
!

0.5
i
n
.
.

Step 2
The total load on the beam is 5 kips, so the force reaction
R
B

P

5
ki
ps
. The point load is 7 feet from the wall, so the
moment reaction
M
B

5
ki
ps
!
7
f
t
.

35
ki
p
f
t
.

Step 3
The shear diagram is a rectangl
e. The moment diagram is a
triangle, where M
max
is the area of the shear diagram:
M
m
a
x

!
5
ki
ps
"
7
f
t
.

!
35
ki
p
f
t
.








Beam

Z
x
(in.
3
)

I
x
(in.
4
)



W8×24

23.2

82.8



W12×16

20.1

103



W6×25

18.9

53.8



W10×12

12.6

53.8


Step 4
Calculate the needed plastic section modulus:

Re
qui
r
e
d Z
x

1.67
M
!
Y
S

1.67
"
35
ki
p
f
t
.
i
n
.
2
50
ki
ps
12
i
n
.
f
t
.

14.0
i
n
.
3

W6×25 has a Z
x
of 18.9 in.
3
, which m
eets the requirements, but
W12×16 is 36% lighter, with a weight of 16 lb./ft., or 0.016
kips/ft.






We also need to calculate the requir
ed moment of inertia so that
the beam does not deflect more than 0.50 inches. From Case #13
in Appendix H,
!
m
a
x

P
b
2
6
E
I
3
L
"
b


. Recalculating, we get

Re
qui
r
e
d I

P
b
2
6
E
!
m
a
x

5
ki
ps
7
f
t
.


2
6
i
n
.
2
30
"
10
3
ki
p
0.5
i
n
.
12
i
n
.


2
f
t
.
2

0.39
i
n
.
4
The selected beam easily passes this requirement.


Notes for Strength of Materials, ET 200


Beam Design

© 2011 Barry Dupen

8
of
8

Revised 4 May 2011

Step 5
Redraw the load, shear, and moment diagrams to include
the weight of the beam. Draw an Equivalent Load Diagram to find
R
B
and M
B
. Add the beam weight to the point load for a total load
of
R
B

5
ki
ps

0.016
ki
p
f
t
.
10
f
t
.

5.16
ki
ps
. The moment rea
c-
tion
M
B

5
ki
ps
!
7
f
t
.

0.16
ki
ps
!
5
f
t
.

35.8
ki
p
f
t
.

The shear diagram consists of a triangle and a trapezoid.

V
1

!
0.016
ki
p
f
t
.
3
f
t
.

!
0.048
ki
ps
V
2

V
1
!
5
ki
ps

!
5.048
ki
ps
V
3

V
2
!
0.016
ki
p
f
t
.
7
f
t
.

!
5.16
ki
ps
V
4

V
3

5.16
ki
ps

0
ki
ps

The moment diagram consists of two parabolas.

M
1

!
0.048
ki
p
2
3
f
t
.

!
0.072
ki
p
f
t
.
M
2

M
1
!
5.048
ki
ps

5.16
ki
ps
2
7
f
t
.

!
35.8
ki
p
f
t
.
M
3

M
2

35.8
ki
p
f
t
.

0
ki
p
f
t
..

The maximum moment is
M
B

35.8
ki
p
f
t
.
.

Re
qui
r
e
d Z
x

1.67
!
35.8
ki
p
f
t
.
i
n
.
2
50
ki
ps
12
i
n
.
f
t
.

14.3
i
n
.
3
, which
is less than Z
x
of the selected beam, so the beam is strong enough
in bending.


Step 6
Use
V
a
p
p
l
i
e
d
!
0.4
"
Y
S
dt
w
to find the shear strength. A
W
12×16 beam can support a shear load of
0.4
!
50
ki
ps
i
n
.
2
!
11.99
i
n
.
!
0.220
i
n
.

53
ki
ps
. Since the actual
shear load of 5.16 kips is less than 53 kips, the beam will not fail
in shear.



Symbols, Terminology, & Typical Units

Δ

Beam deflection

in.

mm

σ
YS

Yield strength

psi, ksi

MPa

a

Distance along a beam from the left end to the point l
oad

ft., in.

m, mm

b

Distance along a beam from the point load to the right end

ft., in.

m, mm

d

Beam depth

in.

mm

E

Young’s modulus (modulus of elasticity)

psi, ksi

GPa

I

Moment of inertia

ft.lb., ft.kips

kNm

L

Length

lb., kips

N, kN

M

Moment

lb./ft
., kip/ft.

N/m, kN/m

P

Point load

lb., kips

N, kN

S

Section modulus

in.
3

mm
3

t
w

Web thickness (of a beam)

in.

mm

Z

Plastic section modulus

in.
3

mm
3