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# Physics for

Engineers
Week 3
Last week:
More truss analysis
Stress and Strain
Elasticity and Plasticity

This week:
Shear, Bearing stress
Shear Force and Bending Moments
An axial load of 100 kN is applied to the
material shown in Figure Q1(a).
Determine the stress in section A and
the extension in section C. Take E=
209 GN/m2.

50x50m
m

Diameter = Diameter =
20mm 30mm
Shear Stress

## A shear stress is produced whenever the

applied forces cause one section of the object
to slide past the adjacent section (usually due

## Hole Pin carrying a

A bearing strass failure causes the hole to
elongate, or tear out completely

## Where: t = plate thickness, d = diameter of

pin (txd = contact area)
Shear Force and Bending
Moment
The bending Diagrams
moment diagram indicates the
bending moment withstood by the beam section
along the length of the beam.
The bending moment at any section is the
algebraic sum of the moments of the forces
about the section, taken on either side.

## The shear force diagram indicates the shear

force withstood by the beam section along the
length of the beam.
At any section in a beam carrying transverse
These shearing
diagrams force isthe
illustrate defined as the
variation in BM and
algebraic
SF sum
along the of theof
length forces taken
a beam, andon either side
therefore
of the section.
clearly show the positions on the beam that are
subjected to the greatest shear or bending
Sign Convention
(BM)

## Clockwise moments to the left and

anticlockwise to the right are positive. (a)
shows a positive bending moment system
resulting in sagging of the beam at X-X and
(b) illustrates a negative B.M. system with its
associated hogging beam.
Sign
Convention SF

## Forces upwards to the left of a section or

downwards to the right of the section are
positive. (a) shows a positive S.F. system at X-X
and (b) shows a negative S.F. system.
Method

F
1. Calculate support reactions
2. Take beam section by section,
calculating the SF and BM at each
section

S.F. at A = 10
S.F. at B = -10
S.F. at C = 20
S.F. at D = -20
S.F. at E = -30
S.F. at F = 30
3. Draw diagrams
Try these example
questions:
Point of Contraflexure / Point of
Inflection
is the point where the curvature of the
beam changes sign:

## *This occurs where the

BM = 0*
BMx = wLx wx(x) = wLx - wx

dBM/dx = wL - wx

SF = wL - wx

So dBM/dx = SF

## Max/Min of a curve occurs when dy/dx = 0,

therefore when BM is max/min, dBM/dx = 0, and
so SF must be zero
CANTILEVERS (not required
for exams)
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