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Mechanics of Solids (VDB1063)
Stresses in Beams
Flexural Formula
Lecturer: Dr. Montasir O. Ahmed
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Learning Outcomes
To evaluate the
bending stress by applying the
flexural formula
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Bending Deformation of a Straight Member
Assumptions regarding the way the stress deforms the material:1. Length of longitudinal axis remains unchanged.
2. All cross sections remain plane and perpendicular to the longitudinal axis3. Any deformation of the cross section within its own plane will be neglected (Fig. 6-19b).
Deformation that occur when a
straight beam is subjected to bending
M causes the material within the bottom portion to stretch and within the top portion to compress.Consequently, between these regions there must be a surface ( neutr al sur face ) in which longitudinalfibers of the material will not undergo change in length.
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Bending Deformation of a Straight Member
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The Flexural Formula
Assumptions:
The material behaves in a linear-elastic manner and therefore a linear variation of normal strainmust then be the result of a linear variation in normal stress .
Developing an equation that relates the stress distribution in a beam to the internal resultant bending moment acting on the beams cross section
= -
By applying Hook, we can obtain from
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The Flexural Formula
=
where
max =
max. nor mal str ess in the member, which occur at distance c from the NA.
M = the resul tant internal moment , determined from the method of section and the EOE and
calculated about the NA.
c = the perpendicul ar distance from the NA to a point farthest away from the NA. this is
where max act.
I = the moment of inerti a of the cross section area about the NA. = + 2
= nor mal str ess in the member, which occur at distance y from the NA.
The flexural formula is: OR = -
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EXAMPLE
The simply supported beam in Fig. 6 26a has the cross-sectionalarea shown in Fig. 6 26b. Determine the absolute maximum
bending stress in the beam and draw the stress distribution over thecross section at this location.
10Copyright 2011 Pearson Education South Asia Pte Ltd
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EXAMPLE 4 (cont.)
The maximum internal moment in the beam, 22.5 kN.m, occurs atthe center.
By reasons of symmetry, the neutral axis passes through the
centroid C at the mid-height of the beam, Fig. 6
26b.
11Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
(Ans) MPa7.12103.301
17.0105.22 ;
m103.301
3.002.016.002.025.002.025.02
6
3
46
312123
121
2
b B
B I My
Ad I I
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EXAMPLE 4 (cont.)
A three-dimensional view of the stress distribution is shown in Fig. 6 26d .
At point B,
12
Solutions
MPa2.11103.301
15.0105.22 ; 6
3
B B
B I My
Copyright 2011 Pearson Education South Asia Pte Ltd
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Next Class
Shear in Straight Members
The Shear Formula
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Thank You