1. CTC / MTC 222 Strength of Materials
Chapter 8
Stress Due to Bending

2. Chapter Objectives
Compute the maximum bending stress in a beam using the flexure formula.
Compute the bending stress at any point in a beam cross section.
Determine the appropriate design stress and design beams to carry a given load safely

3. The Flexure Formula
Positive moment – compression on top, bent concave upward
Negative moment – compression on bottom, bent concave downward
Maximum Stress due to bending (Flexure Formula)
σmax = M c / I
Where M = bending moment, I = moment of inertia, and c = distance from centroidal axis of beam to outermost fiber
For a non-symmetric section distance to the top fiber, ct , is different than distance to bottom fiber cb
σtop = M ct / I
σbot = M cb / I
Conditions for application of the flexure formula
Listed in Section 8-3, p. 308

4. The Flexure Formula Conditions for application of flexure formula
Beam is straight, or nearly straight
Cross-section is uniform
Beam is not subject to torsion
Beam is relatively long and narrow in proportion to its depth
Beam is made from a homogeneous material with equal moduli of elasticity in tension and compression
Stress is below the proportional limit
No part of beam fails from instability
Location where stress is computed is not close to point of application of a concentrated load

5. Stress Distribution
For a positive moment, top of beam is in compression and bottom is in tension
Stress at centroidal axis (neutral axis) is zero
Stress distribution within cross section is linear
Magnitude of stress is directly proportional to distance from neutral axis
Stress at any location in cross section due to bending
σ = M y / I , where y = distance from neutral axis of beam to location in cross section

6. Section Modulus, S Maximum Stress due to bending
σmax = M c / I
Both I and c are geometric properties of the section
Define section modulus, S = I / c
Then σmax = M c / I = M / S
Units for S – in3 , mm3
Use consistent units
Example: if stress, σ, is to be in ksi (kips / in2 ), moment, M, must be in units of kip – inches
For a non-symmetric section S is different for the top and the bottom of the section
Stop = I / ctop
Sbot = I / cbot

7. Stress Concentration Factors
Changes in cross-section of a member can cause stress concentrations
Stress concentration factor KT
Depends on geometry of the member
Can be measured experimentally, or by computerized analyses
KT = σmax / σnom
Solving for and σmax using σnom = M / S
σmax = M KT / S
See Sections 3-9, 8-9 and Appendix A-21

8. Flexure Center
For the flexure formula to apply, the applied load must not cause the beam to twist
The loads must pass through the flexural center or shear center, of the beam
If section has an axis of symmetry, the shear center is on that axis
If section doesn’t have an axis of symmetry, the shear center may be outside the limits of the section itself
Example - Channel

9. Calculation of Design Stress
Design stress (or allowable stress), σd
Sometimes based on yield strength: σd = Sy / N
Sometimes based on ultimate strength σd = Su / N
Ductile Materials - >5% elongation before failure
Static loads - σd = Sy / N , N = 2
Repeated loads - σd = Su / N , N = 8
Impact or shock load - σd = Su / N , N = 12
Brittle Materials - <5% elongation before failure
Static loads - σd = Su / N , N = 6
Repeated loads - σd = Su / N , N = 10
Impact or shock load - σd = Su / N , N = 15
Design stresses for specific materials are specified in design codes
Examples: AISC or Aluminum Association