Beams -Horizontal Shear Stress
In addition to the bending (axial) stress which develops in a loaded
beam, there is also a shear stress which develops, including both a
Vertical Shear Stress, and a Horizontal (longitudinal) Shear
Stress. It can be shown that at any given point in the beam, the
values of vertical shear stress and the horizontal shear stress must be
equal, at that point, for static equilibrium. As a result it is usual to
discuss and calculate the horizontal shear stress in a beam (and simply
remember that the vertical shearing stress is equal in value to the
horizontal shear stress at any given point).We will take a moment to
derive the formula for the Horizontal Shear Stress. In Diagram 1, we
have shown a simply supported loaded beam.
In Diagram 2a, we have cut a section dx long out of the left end of the
beam, and have shown the internal horizontal forces acting on the
section.
In Diagram 2b, we have shown a side view of section dx. Notice that
the bending moment is larger on the right hand face of the section by
an amount dM. (This is clear if we make the bending moment diagram
for the beam, in which we see the bending moment increases from a
value of zero at the left end to a maximum at the center of the beam.)
In Diagram 2c, we have shown a top slice of section dx. Since the
forces are different between the top of the section and the bottom of
the section (less at the bottom) there is a differential (shearing) force
which tries to shear the section, shown in Diagram 2c, horizontally.
This means there is a shear stress on the section, and in terms of the
shear stress, the differential shearing force, F, can be written as F
=
times the longitudinal area of the section (b dx). A second way of
expressing the shear force is by expressing the forces in terms of the
bending stress, that is F1 = (My/I) dA, and F2 = (M+dM)y/I dA,
then the differential force is (dM y/I)dA. If we now combine the two
F = expressions, we have:
F=
* b dx = (dM y/I)dA, and then rewriting to solve for the
shear stress:
= [(dM/dx)/Ib] y dA, however dM/dx is equal to the shear force
V (as discussed in the previous topic), and y dA is the first moment
of the area of the section, and may be written as A y', where A is the
area of the section and y' is the distance from the centroid of the area
A to the neutral axis of the beam cross section. Rewriting in a final
form we have:
Horizontal Shear Stress:
= VAy'/Ib, where
V = Shear force at location along the beam where we wish to
find from the horizontal shear stress
A = cross sectional area, from point where we wish to find the
shear stress at, to an outer edge of the beam cross section (top
or bottom)
y' = distance from neutral axis to the centroid of the area A.
I = moment of inertia for the beam cross section.
b = width of the beam at the point we wish to determine the
shear stress.
(In some texts, the product Ay' is given the symbol Q and used in the
shear stress equation)
If we consider our shear relationship a little, we observe that the
Horizontal Shear Stress is zero at the outer edge of the beam since the area A is zero there. The Horizontal Shear Stress is
(normally) a maximum at the neutral axis of the beam. This is
the opposite of the behavior of the Bending Stress which is maximum
at the other edge of the beam, and zero at the neutral axis.
To help clarify the Horizontal Shear Stress equation we will now look at
at several example of calculating the Horizontal Shear Stress.
Example 1
In Diagram 1, we have shown a simply supported 20 ft. beam with a
load of 10,000 lb. acting downward at the center of the beam. The
beam used is a rectangular 2" by 4" steel beam. We would like to
determine the maximum Horizontal Shear Stress which develops in the
beam due to the loading. We will also determine the Horizontal Shear
Stress 3 inches above the bottom of the beam at the position in the
beam where the shear force is a maximum.
Step 1: Out first step in solving this problem is to apply static
equilibrium conditions to determine the external support reactions. In
this particular example, because of the symmetry of the problem, we
will not go through the statics in detail, but point out that the two
support forces will support the load at the center equally with forces of
5000 lb. each as shown in Diagram 2.
Step 2: The second step is to draw the shear force and bending
moment diagrams for the beam.
Step 3. We will now apply the Horizontal Shear Stress formula:
Shear Stress = Vay'/Ib
We wish to find the maximum shear stress, which occurs at the
neutral axis of the beam:
V = maximum shear force = 5,000 ft-lb. (from the shear force
diagram)
I = moment of inertia of cross section; for rectangle
I = (1/12) bd3 = 1/12 (2" * 4"3) = 10.67 in4.
b = width of beam section where we wish to find shear stress
at; b= 2 in.
a = area from point we wish to find shear stress at (neutral
axis) to an outer edge of beam
a= (2" x 2")= 4 in2.
y' = distance from neutral axis to the centroid of the area "a"
which we used; y'= 1 in.
(See Diagram 4)
Placing the values into the equation, we find:
Maximum Horizontal Shear Stress = Vay'/Ib = (5000 lb)*(4
in2)*(1 in)/ (10.67 in4)(2 in)= 937 lb/in2
Part II We now would also like to determine the Horizontal Shear
Stress 3 inches above the bottom of the beam at the position in the
beam where the shear force is a maximum (which is actually through
out the beam, since the value of shear force is either +5000 lb., or 5000 lb. through out the beam.)
We again apply the Horizontal Shear Stress formula: Horizontal
Shear Stress = Vay'/Ib
We wish to find the shear stress 3 inches above the bottom of the
beam cross section. (See Diagram 5)
V = shear force = 5,000 ft-lb. (from the shear force diagram)
I = moment of inertia of cross section; for rectangle I = (1/12)
bd3 = 1/12 (2" * 4"3) = 10.67 in4.
b = width of beam section where we wish to find shear stress
at; b = 2 in.
a = area from point we wish to find shear stress at (3" above
bottom of the beam) to an outer edge of beam. We will go to
the top edge of the beam, then a = (2" x 1")= 2 in2.
y' = distance from neutral axis to the centroid of the area "a"
which we used; y'= 1.5 in. (See Diagram 5)
Then the horizontal shear stress 3 inches above the bottom of the
beam is:
Horizontal Shear Stress = Vay'/Ib = (5000 lb)*(2 in2)*(1.5
in)/ (10.67 in4)(2 in)= 703 lb/in2
Notice, as we expect, the horizontal shear stress value becomes
smaller as we move toward the outer edge of the beam cross section.
Horizontal Shear Stress - Example 2
A loaded, simply supported beam is shown in Diagram 1. For two
different beam cross sections (a WT 8 x 25 T-beam, and a W 10 x 45
beam) we will determine the maximum Horizontal Shear Stress which
would develop in the beam due to the loading. We will also determine
the Horizontal Shear Stress 3 inches above the bottom of the beam at
the position in the beam where the shear force is a maximum.
STEP 1: Apply Static Equilibrium Principles and determine the external
support reactions:
1.) FBD of structure (See Diagram 2)
2.) Resolve all forces into x/y components
3.) Apply equilibrium conditions:
Sum Fx = 0 none
Sum Fy = By + Cy - 1,000 lbs/ft (4 ft) - 1,500 lbs/ft (4 ft) = 0
Sum TB = 1,000 lbs/ft (4 ft) (2 ft) - 1,500 lbs/ft (4 ft) (8 ft) +
Cy(6 ft) = 0
Solving: By = 3,330 lb.; Cy = 6,670 lb.
Step 2: The second step is to draw the shear force and bending
moment diagrams for the beam
Beam Table for WT 8 x 25
Designation Area
of T Width thick
thick
-
x-x
axis
x-x
axis
x-x
axis
x-x
axis
-
A
d
bf
tf
tw
d/tw
I
S
r
y
-
in2
in
in
in
in
-
in4
in3
in
in
WT 8x25
7.36 8.13 7.073 0.628 0.380 21.40 42.20 6.770 2.400 1.890
Step 3. For the WT 8 x 25 T-beam (table above) we will now apply
the Horizontal Shear Stress formula:
Shear Stress = Vay'/Ib, to find the maximum shear stress, which
occurs at the neutral axis of the beam:
V = maximum shear force = 6,000 lb. (from the shear force
diagram)
I = moment of inertia of cross section, from beam table; I =
42.20 in4.
b = width of beam where we wish to find shear stress (neutral
axis for maximum) from table; b = .38 in.
a = area from point we wish to find shear stress at (neutral
axis) to an outer edge of beam. In this case we will go to
bottom of beam. Then a = (.38" * 6.24" )= 2.37 in2.
y' = distance from neutral axis to the centroid of the area "a" ;
y' = 3.12 in. (See Diagram 5) Then placing values into our
expression we find:
Maximum Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(2.37
in2)*(3.12 in)/ (42.20 in4)(.38 in) = 2770 lb/in2
We now would also like to determine the Horizontal Shear Stress 3
inches above the bottom of the beam at the position in the beam
where the shear force is a maximum
We again apply the Horizontal Shear Stress formula: Shear Stress =
Vay'/Ib
We wish to find the shear stress 3 inches above the bottom of the
beam cross section, where the shear force is a maximum. (See
Diagram 6)\
V = maximum shear force = 6,000 lb. (from the shear force
diagram)
I = moment of inertia of cross section, from beam table; I =
42.20 in4.
b = width of beam where we wish to find shear stress (3"
above bottom of beam) from table; b = .38 in.
a = area from point we wish to find shear stress at (neutral
axis) to an outer edge of beam. In this case we will go to
bottom of beam. Then a = (.38" * 3" )= 1.14 in2. (See Diagram
6)
y' = distance from neutral axis to the centroid of the area "a" ;
y' = 4.74 in. (See Diagram 6)
Then the horizontal shear stress 3 inches above the bottom of the
beam is:
Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(1.14 in2)*(4.74
in)/ (42.2 in4)(.38 in)= 2020 lb/in2
Notice, as we expect, the horizontal shear stress value becomes
smaller as we move toward an outer edge of the beam cross section.
Part 2: W 10 x 45 beam: We would like to again determine the
maximum horizontal shear stress, and the shear stress 3 inches above
the bottom of the beam (at the point where the shear force is a
maximum), but now find these values for a W 10 x 45 I-beam.
Beam Table for W 10 x 45 I-Beam
-
-
-
Flange Flange Web
Designation Area Depth Width
Cross Section Info. Cross Section Info.
thick
thick
x-x
axis
x-x
axis
x-x
axis
y-y
axis
y-y
axis
y-y
axis
-
A
d
bf
tf
tw
I
S
r
I
S
r
-
in2
in
in
in
in
in4
in3
in
in4
in3
in
13.30
2.00
W 10 x 45
13.20 10.12 8.022
0.618 0.350 249.0
49.1
4.33 53.20
We have already done the statics, and the shear force and bending
moment diagrams are shown in the first part of this example above, so
we continue at the point where we apply the horizontal shear stress
formula to find the values we desire.
For the WT 8 x 25 T-beam we apply the Horizontal Shear Stress
formula: Shear Stress = Vay'/Ib, however since we are looking for
the maximum shear stress in the I-Beam, we can use the
approximate formula for I-beam, Maximum Horizontal Shear
Stress = Vmax/Aweb. This says the approximate maximum shear
stress in an I - Beam is equal to the maximum shear force divided by
the area of the web of the I-Beam. Applying this we have:
Vmax = maximum shear force = 6,000 lb. (from the shear force
diagram)
Amax = area of web: A = (.35" * 8.88" )= 3.11 in2. (See Diagram 7)
Then Maximum Horizontal Shear Stress = (6000 lb)/(3.11 in2)
= 1930 lb/in2
As long as this approximate value is reasonably below the allowable
shear stress for the beam material there is no need to use the exact
formula for the maximum shear stress. Please remember, however,
the approximate formula is only for the maximum horizontal shear
stress (which occurs are the neutral axis) in an I-Beam. If we need to
know the shear stress at any other location, we must use the standard
formula – as we will do in the next part.
We now wish to find the shear stress 3 inches above the bottom of the
beam cross section, where the shear force is a maximum. (See
Diagram 8). To do so, we apply the standard horizontal shear stress
formula: Shear Stress = Vay'/Ib
V = maximum shear force = 6,000 lb. (from the shear force diagram)
I = moment of inertia of cross section, from beam table; I = 249.0
in4.
b = width of beam where we wish to find shear stress (3" above
bottom of beam) from table; b = .35 in.
a = area from point we wish to find shear stress at (3" from the
bottom) to an outer edge of beam. In this case we will go to bottom of
beam. Notice that the area is composed of the area of the flange (A1)
and part of the area of the web (A2). (See Diagrams 8 and 9.) Then a
= (A1 + A2)= (.618" x 8.022") + (2.383 in2 x .35in2) = 4.96 in2
+ .834 in2= 5.794 in2 (See Diagram 9)
y' = distance from neutral axis to the centroid of the area "a" Notice
that in this case, for an I-Beam, this is not a entirely simple matter.
The area we wish to find the centroid of is not a simple rectangle, but
rather two rectangles. To find the centroid of this compound area we
use: y' = (A1 y1 + A2 y2)/(A1 +A2); where A1 and A2 are the two
areas, and y1 and y2 are the distances from the neutral axis of the
beam to the centroid of each of the respective areas. (Which is simply
the distance from the neutral axis to the center of each of the
respective areas, since for a rectangle the centroid is at the center.)
Using the values shown in Diagram 9, we have:
y' = (A1 y1 + A2 y2)/(A1 +A2) = (4.96 in2 x 4.75 in + .834 in2 x 2.94 in)/(4.96 in2 +
.834 in2) = 4.49 in.
Then the horizontal shear stress 3 inches above the bottom of the beam is:
Horizontal Shear Stress = Vay'/Ib = (6000 lb)*(5.794 in2)*(4.49 in)/ (249.0 in4)(.35
in)= 1790 lb/in2
Notice, as we expect, the horizontal shear stress value becomes smaller as we
move toward an outer edge of the beam cross section.