ctc / mtc 222 strength of materials chapter 8 stress due to bending
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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
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