CTC / MTC 222 Strength of Materials

Chapter 8Stress Due to Bending

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

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

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

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

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

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

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

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