Lecture 8 – Bending & Shear Stresses on Beams - WordPress.com

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Nov 15, 2013 (3 years and 10 months ago)

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Lecture 8 - Page 1 of 9
Lecture 8 – Bending & Shear Stresses on Beams

Beams are almost always designed on the basis of bending stress and, to a
lesser degree, shear stress. Each of these stresses will be discussed in detail as
follows.

A) Bending Stresses


A bending stress is NOT considered to be a simple stress
. In other
words, it is not load divided by area. The formula for bending stress
, σ
b
,
is as follows:

I
My
b



where: M = moment acting on beam from moment diagram (kip-in or lb-in)
y = distance from neutral axis to extreme edge of member (in)
I = moment of inertia about the axis (in
4
)

Recalling that
y
I
S
=
, the bending stress formula could be re-written as:

S
M
b



where: S = section modulus about the axis (in
3
)

Bending stress is distributed through a beam as seen in the diagram
below:


So, in reality, bending stresses are tensile or compressive stresses in the
beam! A simply-supported beam always has tensile stresses at the
bottom of the beam and compressive stresses at the top of the beam.
Lecture 8 - Page 2 of 9
Example 1
GIVEN
: A nominal 2x10 (actual dims. 1½” x 9¼”) is used as a simply-supported
beam with loading as shown. The allowable bending stress is 1200 psi.
REQUIRED
:
a) Determine the maximum moment on the beam.
b) Determine the maximum actual bending stress on the beam
c) Determine if the beam is acceptable based upon allowable bending
stress.









The maximum bending moment, M
max
, on a simply-supported, uniformly
loaded beam is:

8
2
max
wL
M =


8
)'11PLF)( 140(
2
max
=M


M
max
= 2117.5 lb-ft


The actual bending stress is:

S
M
b



3
39.21
/ft)ft(12"-lb 2117.5
in
b




σ
b
= 1187.9 PSI


Since the actual bending stress of 1187.9 PSI is less than the allowable
bending stress of 1200 PSI, THE BEAM IS ACCEPTABLE
.
11’-0”
w = 140 PLF (includes beam weight)
See textbook appendix
Lecture 8 - Page 3 of 9
Example 2
GIVEN
: A W14x30 steel beam experiences a moment of 95 kip-feet. The
allowable bending stress for the steel beam is 24 KSI.
REQUIRED
:
a) Determine the maximum ALLOWABLE moment based on the allowable
bending stress (leave answer in units of kip-ft).
b) Determine if the beam is acceptable or not based upon allowable bending
moment.

Using the bending stress formula above, re-write it to solve for moment:

S
M
b



SM
b
σ
=


Substituting 24 KSI for σ
b
and using
S
= 42.0 in
3
(from textbook appendix)
gives:

M
all
= 24 KSI(42.0 in
3
)
= 1008 kip-in

Dividing by 12 to get into units of kip-ft:

M
all
= 84 kip-ft


Since the actual bending moment of 95 kip-ft is more than the allowable
bending moment of 84 kip-ft, THE BEAM IS UNACCEPTABLE
.

Lecture 8 - Page 4 of 9

Example 3
GIVEN
: The beam problem in Example 2.
REQUIRED
: Determine the lightest weight W12 beam that can carry a moment
of 95-kip-feet based upon allowable bending stress of 24 KSI.

Using the bending stress formula above, re-write it to solve for section
modulus:

S
M
b
=
σ

b
M
S
σ
=


Substituting 24 KSI for
σ
b
and 95 kip-ft for
M
gives:

KSI24
/ft)ft(12"-kip 95
=S



S
= 47.5 in
3


Looking at the textbook appendix, the lightest weight W12 that has a section
modulus of at least 47.5 in
3
is a W12x40 (S
x
= 51.9 in
3
> 47.5 in
3
)
.


Lecture 8 - Page 5 of 9
B) Shear Stresses


It is easy to imagine vertical shear on a beam that was made up of
concrete blocks:







This type of shear is called “transverse” shear, and occurs if there is no
bending stresses present
. The transverse shear stress =
A
V


However, almost all real beams have bending stresses present. In this
case, beams are more like a deck of cards and bending produces sliding
along the horizontal planes at the interfaces of the cards as shown below:









This type of shear is called “longitudinal” or horizontal shear. The
formula used for determining the maximum longitudinal shear stress, f
v
, is
as follows:

Ib
VQ
f
v
=


where: V = vertical shear, usually from shear diagram (lb. or kip)
Q = first moment of area
= Ay
A = area of shape above or below the neutral axis (in
2
)
y = distance from neutral axis to centroid of area “A” (in)

I
= moment of inertia of shape (in
4
)
b = width of area “A” (in)


Load
Load
Lecture 8 - Page 6 of 9
Example 4
GIVEN
: The nominal 2x10 wood beam from Example 1. The allowable
horizontal shear stress is 95 PSI.
REQUIRED
:
a) Determine the maximum horizontal shear stress on the beam.
b) Determine if the beam is acceptable based upon allowable horizontal
shear stress.




































-770 lbs.
y = 2.31”
1½”
N.A
4.625”
4.625”
9 ¼”
0
770 lbs.
11’-0”
w = 140 PLF (includes beam weight)
Shear diagram

End reaction = ½(140 PLF)(11’)
= 770 lbs.
1½”
N.A
4.625”
4.625”
Area “A” shown
shaded
Beam X-Section

0
Lecture 8 - Page 7 of 9
Using the horizontal shear stress formula:

Ib
VQ
f
v
=


and substituting in the values:

V = 770 lbs. (from shear diagram)
Q = Ay

where: A = (4.625”)(1.5”)
= 6.94 in
2


y = ½(4.625”)
= 2.31”

Q = (6.94 in
2
)(2.31”)
= 16.03 in
3


I
= 98.93 in
4
(from textbook appendix, or calculate
12
3
bh
I =
)

b = 1½”


)"5.1)(93.98(
)03.16)(770(
4
3
in
inlbs
f
v
=


f
v
= 83.2 PSI


Since the actual horizontal shear stress of 83.2 PSI is less than the
allowable horizontal shear stress of 95 PSI, THE BEAM IS

ACCEPTABLE.


Lecture 8 - Page 8 of 9
Example 5
GIVEN
: A wood built-up beam is constructed using a nominal 2X8 web and 2X6
flanges as shown below. The moment of inertia about the strong axis,
I
, = 366.5
in
4
. The maximum vertical shear is shown below.
REQUIRED
: Determine the spacing of 12d nails connecting the top & bottom
flanges to the web. Assume the allowable shear strength of each nail is 140 lbs.






















-1100 lbs.
0
1100 lbs.
11’-0”
w = 200 PLF (includes beam weight)
Shear diagram

End reaction = ½(200 PLF)(11’)
= 1100 lbs.
0
Beam Cross-Section

Lecture 8 - Page 9 of 9
The shear flow, “q” can be used to determine the horizontal shear along
the length of the beam in terms of force per unit length. In particular,

I
VQ
q =


Where: V = Maximum vertical shear (from shear diagram)
= 1100 lbs.

Q = A
p
y

A
p
= Area of piece to be fastened, in
2

= (1½” x 5½”)
= 8.25 in
2


y =
Distance from beam N.A. to centroid of piece, in.
= 4.375”

Q = (8.25in
2
)(4.375”)
= 36.1 in
3


Substituting:
I
VQ
q =


4
3
5.366
)1.36.)(1100(
in
inlbs
q =


=
inch
lbs
.
3.108


The allowable shear capacity per nail = 140 lbs., So then the nail
spacing can be determined as follows:

q
NailperShearAllowable
Spacing
__.
=


=
inch
lbs
nailperlbs
.
3.108
__.140


Spacing = 1.29 inches → use 1¼”

Space the 12d nails at 1¼” apart along entire length of beam.