Lecture 5 - Page 1 of 9

Lecture 5 – Beams – Design for Shear & Deflection

Steel beams are usually designed solely on the basis of moment. This means

that bending stresses are the critical design factor. However, under certain

circumstances, shear and deflection must also be checked.

1. Design for Shear

Shear in steel beams generally does not control the design EXCEPT in

the following two situations:

• Reduced beam cross-sectional area, as with “coped” beams

• Very heavy loads on short-span beam

Short span

Reduced Shear plane

Normal beam

Shear plane

Coped beam

Very heavy loads

Lecture 5 - Page 2 of 9

Shear in steel beams is assumed to be carried entirely by the area of the

web, A

w

:

Design for shear is dictated in AISC Spec. G p. 16.1-64 as follows:

LRFD Factored Design shear strength = φ

v

V

n

ASD Service Allowable shear strength =

v

n

V

Ω

where: φ

v

= 1.00 (LRFD)

Ω

v

= 1.50 (ASD)

V

n

= nominal shear strength

= 0.6F

y

A

w

C

v

A

w

= area of web (see sketch above)

= t

w

d

C

v

= Web shear coefficient

=

1.0 for webs of rolled “I” – shaped shapes (Conservative)

= see AISC Eq. G2-3, G2-4 and G2-5 p. 16.1-65

for other conditions

d

t

w

Beam X-Sect

A

w

= Shear area

in normal beam

(shaded)

A

w

= Shear area

in coped beam

(shaded)

Lecture 5 - Page 3 of 9

Example 1 (LRFD)

GIVEN

: A W14x26 A992 steel beam.

REQUIRED

: Determine the FACTORED design shear strength, φ

v

V

n

for

the beam.

Step 1 – Determine area of web, A

w

:

A

w

= t

w

d

= 0.255 in(13.9 in)

= 3.54 in

2

Step 2 – Determine FACTORED design shear strength

:

φ

v

V

n

= 1.00(0.6F

y

A

w

)C

v

= 1.00(0.6(50 KSI)(3.54 in

2

)1.0)

φ

v

V

n

= 106.2 KIPS

(NOTE

: The values of φ

v

V

n

can be found directly in the “Maximum

Total Factored Uniform Load” Table 3-6 AISC p. 3-67)

d

t

w

Lecture 5 - Page 4 of 9

Example 2 (ASD)

GIVEN

: The W12x30 A992 steel beam has a 4” cope and has the

SERVICE load as shown below. Disregard beam weight.

REQUIRED

:

1) Determine if the beam is acceptable on the basis of shear at the coped

end.

2) Determine if the beam is acceptable on the basis of shear at the

location of the point load.

W12x30

R1

R2

Lecture 5 - Page 5 of 9

Step 1 – Determine the reaction at the left support R1

:

R1 =

'10

)'7(120KIPS

= 84 KIPS

Step 2 – Determine ALLOWABLE shear strength in coped web at R1

:

v

n

V

Ω

=

50.1

6.0

vwy

CAF

A

w

= Area of web

= t

w

h

From properties, t

w

= web thickness

= 0.260 in.

d = 12.3 in.

h = d – cope

= 12.3” – 4”

= 8.3”

A

w

= t

w

h

= 0.260”(8.3”)

= 2.16 in

2

v

n

V

Ω

=

50.1

)0.1)(16.2(50(6.0

2

inKSI

v

n

V

Ω

= 43.2 KIPS < 84 KIPS → UNACCEPTABLE

h = d - cope

d

t

w

Lecture 5 - Page 6 of 9

Step 3 – Determine ALLOWABLE shear strength at point of load

:

Since the beam is not coped at this location, the design shear

strength

v

n

V

Ω

can be found from AISC Table 3-6 p. 3-71.

v

n

V

Ω

= 64.2 KIPS < 120 KIPS → UNACCEPTABLE

(Note

: This example looks ONLY at shear

. Very high loads also require a

detailed look at connections, which will be investigated later)

Possible Fixes for High Shear

:

By modifying any of the variables in the design shear equation, the

capacity may be increased. These include:

• Using a higher grade of steel (increase F

y

)

• Use a bigger beam (increase A

w

)

• Weld additional plates to the web (increase A

w

)

New plate welded

to web of beam

Lecture 5 - Page 7 of 9

2. Design for Deflection

Building codes (such as the IBC) require that deflections in beams be held

to a minimum for occupancy comfort as well as to reduce likelihood of

cracking ceiling finishes such as plaster. It is considered to be a

“serviceability” check. Allowable deflection limits are dictated by the

codes, such as L/360 of the span.

Actual deflection is calculated using SERVICE LOADS in the formulas

given in the AISC p. 3-208 thru 3-226. These actual deflections are then

compared against the allowable deflection.

The following allowable deflection limits for steel construction are used by

the IBC:

Construction

Live Load

Snow or Wind

Roof member supporting plaster ceiling L/360 L/360

Roof member supporting nonplaster ceiling L/240 L/240

Roof member supporting no ceiling L/180 L/180

Floor members L/360 -

It should be noted that the above-noted allowable deflections are

minimums. Architects and engineers often reduce the deflection limit to

L/480 or even L/540 to ensure that floors are not “bouncy.”

Lecture 5 - Page 8 of 9

Example 3

GIVEN

: The floor framing plan below. The total superimposed service

dead load = 86 PSF (not including beam weight) and the service live load

= 125 PSF.

REQUIRED

: Determine the maximum actual mid-span deflection and

compare with a Live load limit = L/480 and a Dead load + Live load limit =

L/240 on the W18x35 steel beam.

Step 1 – Determine the actual max. deflection considering Live load only

:

From AISC p. 3-211, the maximum deflection formula is:

∆

max

=

EI

wL

384

5

4

w = uniform live load on beam

= 8’(125 PSF)

= 1000 PLF

∆

max

=

)510)(29000000(384

)/"12'25(

12

1000

5

4

4

inPSI

ftx

PLF

⎟

⎠

⎞

⎜

⎝

⎛

= 0.59 in.

do

do

W18x35 beam

W24x62 girder

25’-0”

4@8’-0” = 32’-0”

Lecture 5 - Page 9 of 9

Step 2 – Determine allowable deflection of L/480 for live load only

:

∆

allow

=

480

L

=

480

)/"12("0'25 ft

−

= 0.625 in.

Since

∆

allow

= 0.625” > 0.59”

→

beam is acceptable

Step 3 – Determine max. defl. considering Deal load + Live load

:

∆

max

=

EI

wL

384

5

4

w = uniform dead load + live load on beam

= 8’(125 PSF + 86 PSF) + 35 PLF

= 1723 PLF

∆

max

=

)510)(29000000(384

)/"12'25(

12

1723

5

4

4

inPSI

ftx

PLF

⎟

⎠

⎞

⎜

⎝

⎛

= 1.02 in.

Step 4 – Determine allowable deflection of L/240 for LL + DL

:

∆

allow

=

240

L

=

240

)/"12("0'25 ft

−

= 1.25 in.

Since

∆

allow

= 1.25” > 1.02”

→

beam is acceptable

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