A visual description of Flexural Stress
Fig. 1: Flexural Stress in a fishing pole (Source: bjamin)

Bending Stress (aka flexural stress, aka torque) is the stress caused by a moment or a couple?. A great example of bending stress can be seen in Figure 1. Since the load caused by the fishing line is cantilevered off the end of the pole and since the cross section of a fishing pole is relatively small, a fishing pole will have high flexural stresses. This observation can be made objectively by noticing the large deflection which occurs in the pole.

Some common facts concerning bending stress are:

While Shear Stress is simply the shear force divided by the beams area, Bending Stress introduces new variables based on the dimensions of the beam to solve for:

f_b = -\frac {My}{I} = \frac{M}{S}

This will be explained in greater detail below.

Understanding Bending (Flexural) Stress
  1. How to calculate the bending stress
  2. How to calculate the allowable bending capacity
  3. Example #1
  4. Example #2



How to Calculate the Bending Stress in a member:


Once the Moment is obtained, the bending stress acting on a beam (fb) can be found by:

f_b = -\frac {My}{I} = \frac{M}{S}

where:

fb = the bending stress acting on the member (ksi)
M = The moment acting on the beam (lb-in, kip-in, kip-ft, etc.)
I = the Moment of Inertia of the member (in4)
y = the distance from the neutral axis to the extreme fiber of the cross section (in)
S = the Section Modulus? of the cross section = \frac{I}{c} (in3)

Note: For standard shapes, the Moment of Inertia (I) and Section Modulus (S) can be found in the various tables of shape properties available in each code.


Calculating the Allowable Bending (Flexural) Capacity in a Member

Calculating the flexural capacity for a member is all based on the type of material you're working with (and also often dependent on the shape you're using). For example concrete (solved by finding the allowable moment (not bending stress)), steel, wood, and every other material will calculate it out differently. For the sake of the examples below we will use steel & wood.

For a laterally supported Steel I-shaped beam:

Most common simplified bending equation:

f_b = {M\over Z_x} \le F_b \approx {0.6*F_y}

where:

fb = The computed stress in the beam in bending
M = The maximum moment acting on the beam
Zx = The Plastic Section Modulus in the x or strong axis. Zx is similar to the Section Modulus of a member (it is usually a minimum of 10% greater than the Section Modulus) (in3)
Fb = The allowable stress of the beam in bending
Fy = The Yield Strength of the Steel (e.g. 36 ksi, 46 ksi, 50 ksi)
Ωb = The Safety Factor for Elements in Bending = 1.67

Compare this to Wood where:

Most common simplified bending equation:

f_b = {M\over S_x} \le F_b' = F_b*C_d*C_M*C_t*C_L*C_F*C_{fu}*C_i*C_r

All the variables can be found here in greater detail. As you can see there is a world of difference between the two materials.

Example #1

You have a simply supported wooden 4x10 beam spanning 12' and supporting a live load of 40 pounds per foot. What is the flexural stress acting on the beam?

1) Find the total weight acting on the beam (note: a 4x10 beam is actually 3 1/2" x 9 1/4")

TL = DL + LL = w = \frac{3.5 * 9.25}{144} * 36 pcf + 40 plf ~ 48 plf

2) Using the appropriate moment diagram, solve for the maximum moment acting on the beam.

M_{max} = \frac{wl^2}{8} = \frac{48plf(12 ft^2)}{8} = 864 lb-ft = 10.4 kip-in

3) Solve for the Section Modulus? of the beam:

S = \frac{bh^2}{6} = \frac{3.5*9.25^2}{6} = 49.9 in^3

4) Solve for your flexural stress (fb):

f_b = \frac{M}{S} = \frac {10.4 kip-in}{49.9 in^3} = 0.208 ksi = 208 psi

Note: Keep watch of your units (your final result should be psi or ksi (or Pa or kPa for those of us on the metric system)


Example #2


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