Beams - Bending Stress
BEAMS: BENDING STRESS
Axial Stress (tension and compression) and the Shear Stress (vertical and
horizontal) which develop in a loaded beam depend on the values of the
Bending Moments and the Shear Forces in the beam. Determining the axial
stress - which is often known as the Bending Stress in a beam; and determining
the shear stress - often called the Horizontal Shear Stress (for reasons we will
discuss) is important in two ways. First, it will enable us to determine if a
particular loaded beam is safe under the applied loading. Second, it will enable
us to select the best beam (from a table of beams) for a particular loading.
Both of these are very important processes for the safety and efficiency of a
beam.
The Bending Stress: We will first develop a relationship for the bending stress
which develops in a loaded beam. This relationship is known as the Flexure
Formula. In Diagram 1 we have shown a simply supported beam loaded at the
center. It deflects (or bends) under the load.
In Diagram 2, we have shown the left end section of the beam. As discussed
previously, when examining bending moments, horizontal forces act on the
cross sectional face of the beam section. We have shown only the horizontal
forces along the top and bottom in Diagram 2a, but the forces act across the
whole cross section as shown in the side view in Diagram 2b. The horizontal
forces decrease from maximum at the outer edges to zero at the neutral axis
(an axis running through the centroid of beam cross section).
We will now go through a relatively brief derivation to arrive at the flexure
formula. We first write an expression for the bending moment produced by the
horizontal forces with respect to the neutral axis (which is a line passing
through the centroid of the beam cross sectional area - shown in Diagram 2a).
The expression for the bending moment is simply the sum of the forces times
the perpendicular distance to the neutral axis, or:
We now note that we can express the force Fx as
; that is, the force
acting on any small horizontal strip of area (dA) is the product of axial stress at
that point and the amount of area (dA). (This simply comes from the definition
of axial stress = Force/Area). We can now rewrite the expression for the
bending moment as:
We now will rewrite this expression one more time by noting that the
horizontal forces and accompanying stresses increase linearly from zero at the
neutral axis to a maximum value at an outer edge. We can then write:
, where y is the distance from the neutral axis to area dA,
and ymax is the distance from the neutral axis to the outer edge of the beam
cross-section, as shown in Diagram 2a. We can then write the stress at an
arbitrary y as:
. We now substitute this expression into our
relationship for the bending moment and obtain
; then rewriting slightly and factoring out
the
term from the summation sign (which we may since it is a
constant), we find:
; and finally we recognize that the summation term
remaining is simply the Moment of Inertia (I) about the centroid of the beam
cross section.. We now rewrite one last time, arranging terms and isolating the
stress term by itself, we finally obtain:
M ymax / I
That is, the maximum "Bending Stress" at some location along the beam is
equal to the bending moment, M, at that location "times" the distance, y, from
the neutral axis to the outer edge of the beam "divided" by the moment of
inertia, I, of the beam cross sectional area. If this seems somewhat confusing,
it will become clearer as we work through several examples.
While the formula above was derived for the maximum stress, it actually holds
for the stress at any point in the beam cross section and is known as the
Flexure Formula.
Flexure Formula:
Example 1
My/I
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
bending (axial) stress which develops in the beam due to the loading.
Step 1: Out first step in solving this problem is, of course, 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 now apply the flexure formula: Bending Stress = M y / I
We wish to find the maximum bending stress, which occurs at the outer edge of
the beam so:
M = maximum bending moment = 50,000 ft-lb. = 600,000 in-lb. (from
bending moment diagram)
y = distance from the neutral axis of the cross section to outer edge of
beam = 2 inches
I = moment of inertia of cross section; for rectangle I = (1/12) bd3 = 1/12 (2"
* 4"3) = 10.67 in4.
Then, Maximum Bending Stress = M y / I = (600,000 in-lb)*(2 in)/(10.67 in4) =
112, 500 lb/in2
Example 2
A loaded, simply supported W 10 x 45 beam is shown in Diagram 1. For this
beam we will first determine the maximum bending moment (and where it
occurs in the beam). Then we will determine the maximum bending stress at
that location, and also the bending stress at that location along the beam and 8
inches from the bottom of the beam cross section.
STEP 1: Apply Static Equilibrium Principles and determine the external support
reactions:
1.) Draw Free Body Diagram of structure (See Diagram 2)
2.) Resolve all forces into x and y components
3.) Apply equilibrium conditions:
Sum Fx = 0 none
Sum Fy = By + Dy - 2,000 lbs/ft (4 ft) - 5,000 lbs = 0
Sum TB = 5,000 lbs (4 ft) - 2,000 lbs/ft (4 ft) (6 ft) + Dy(8 ft) = 0
Solving: By = 9,500 lbs; Dy = 3,500 lbs
STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the
Beam.
STEP 3: We will now Apply the Flexure Formula to determine the maximum
bending stress for the beam. We may use Flexure Formula:
M y / I, or a
special form of the Flexure Formula:
M / s, where s is what is known as
the section modulus. If we rewrite the standard flexure formula several times
as follows for the maximum stress:
M (ymax / I) = M / (I / ymax) = M / s , we then see that the section
modulus is defined as s = I / ymax. That is, this special form of the flexure
formula can only be used to find the maximum bending stress, and uses the
section modulus, where the section modulus is equal to the moment of inertia
of the beam cross section divided by the maximum distance from the neutral
axis of the beam to an outer edge of the beam.
As an example we apply this form to determine the maximum bending stress in
our beam. First we determine the maximum bending moment from our bending
moment diagram - which we observe from Diagram 4 is: Mmax = 20,000 ft.-lb.,
and occurs at x = 4 ft. (We will drop the negative sign which simply tells us that
the beam is bent concave facing downward at this point. This means the top of
the beam is in tension and the bottom of the beam is in compression.) We also
then find the x-x axis section modulus of the beam as listed in the beam table
below, s = 49.1 in3
-
-
-
Flange Flange Web
Cross Section Info.
Cross Section Info.
Designation Area Depth Width
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
49.1
4.33
53.20
13.30
2.00
W 10x45
13.20 10.12 8.022 0.618 0.350 249.0
in3
lb./in2.
Now
M / s = (20,000 ft-lb.)(12 in./ft.)/ 49.1
= 4, 890
(Notice that we had to convert the bending moment in ft.-lb. to in.-lb. for the
units to be consistent) We have thus determined the maximum bending stress
(axial stress) in the beam. Since the beam is symmetric about the neutral axis,
the stress at the top of the beam and at the bottom of the beam are equal in
value (4,890 psi.) with the top in tension and the bottom in compression.
Finally, we will determine the bending stress at 4 ft (where the maximum
bending moment occurs), and 8 inches above the bottom of the beam. For this
we need to use the flexure formula in the form
M y / I, where M = 20,000
ft-lb. = 240,000 in-lb., I = moment of inertia of beam = 249 in4, and y =
distance from the neutral axis to point at which we wish to find the bending
stress. Since we wish to find the bending stress 8 inches above the bottom of
the beam, and since the neutral axis is 5.06 inches above the bottom of the
beam (at the beam center), then y = 8 - 5.06 = 2.94 inches. Then
240,000 in-lb. * 2.94 in. / 249 in4. = 2830 lb./in2. And since the location is
above the beam centroid (and the bending moment is positive), this is a tensile
stress.
Example 3
A simply supported WT 8 x 25 T-beam is loaded is shown in Diagram 1. For this
beam we will determine the maximum bending stress in the beam. We will also
determine the bending stress 4 ft from the left end of the beam and 2 inches
above the bottom of the beam.
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 lbs; Cy = 6,670 lbs
STEP 2: Draw both the Shear Force and Bending Moment Diagrams for the
Beam.
STEP 3: We will now Apply the Flexure Formula to determine the maximum
bending stress for the beam. We may use Flexure Formula:
M y / I, or a
special form of the Flexure Formula:
M / s, where s is the section
modulus. This special form of the flexure formula can only be used to find the
maximum bending stress, and uses the section modulus, where the section
modulus is equal to the moment of inertia of the beam cross section divided by
the maximum distance from the neutral axis of the beam to an outer edge of
the beam.
As an example we apply this form to determine the maximum bending stress in
our beam. First we determine the maximum bending moment from our bending
moment diagram - which we observe from Diagram 4 is: Mmax = 12,000 ft.-lb,
and occurs at x = 10 ft. (We will drop the negative sign which simply tells us
that the beam is bent concave facing downward at this point. This means the
top of the beam is in tension and the bottom of the beam is in compression.)
We also then find the x-x axis section modulus of the beam as listed in the
beam table, s = 6.77 in3
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
WT8x25
7.36
8.13
7.073
0.628
0.380
21.40
42.20
6.770
2.400 1.890
Now
M / s = (12,000 ft-lb)(12 in./ft.)/ 6.77 in3 = 21,270 lb/in2. (Notice
that we had to convert the bending moment in ft.-lb. to in.-lb. for the units to
be consistent) We have thus determined the maximum bending stress (axial
stress) in the beam. Please note that this maximum stress is compressive and
occurs at the bottom of the stem, since that is the outer edge of the beam
which is farthest from the neutral axis. (The stress at the top of the tee is less
than that at the bottom of the stem, since the top of the tee is closer to the
neutral axis than the bottom of the stem.)
Finally, we will determine the bending stress at 4 ft from the left end of the beam
and 2 inches above the bottom of the beam. For this we need to use the flexure
formula in the form
M y / I., where M = -8,000 ft-lb = -96,000 in-lb (which
we determine from the bending moment graph shown in Diagram 4), I = moment
of inertia of beam = 42.2 in4, and y = distance from the neutral axis to point
at which we wish to find the bending stress. The neutral axis is 1.89 inches
below the top of the beam (from the beam data table), then the neutral axis is
8.13"-1.89" = 6.24" from the bottom of the beam. Since we wish to find bending
stress 2 inches above the bottom of the beam, then y = 6.24" - 2" = 4.24" from
the neutral axis to where we wish to determine the bending stress. Thus
96,000 in-lb * 4.24 in. / 42.2 in4. = 9,450 lb/in2. Since the bending moment is
negative, meaning the beam is bent concave facing downward, and since the
location is below the beam centroid, then this stress is compressive.