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Mechanics of Solids (VDB1063)
Stress Caused by Combined Loadings
Lecturer: Dr. Montasir O. Ahmed
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LEARNING OUTCOMES
To evaluate the stressescaused by combined loads
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Lecture Outlines
Stress Analysis
Working Examples
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Copyright 2011 Pearson Education South Asia Pte Ltd
Normal force P leads to:
Shear force V leads to:
Bending moment M leads to:
A
Pstressnormaluniform ,
It
VQondistributistressshear ,
beam)straight(for,I
Myondistributistressallongitudin
Stress Analysis
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Copyright 2011 Pearson Education South Asia Pte Ltd
Resultant stresses by superposition:
Once the normal and shear stress components for each loading
have been calculated, use the principalof
superposition to
determine the resultant normal and shear stress components.
Stress Analysis
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EXAMPLE 2
A force of 15 kN is applied to the edge of the member shown in Fig.
83a. Neglect the weight of the member and determine the state of
stress at pointsB and C.
Copyright 2011 Pearson Education South Asia Pte Ltd
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EXAMPLE 2 (cont)
For equilibrium at the section there must be an axial force of 15 000 N acting
through the centroid and a bending moment of 750 000 Nmm about the
centroidal or principal axis.
The maximum stress is
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
MPa75.3
40100
15000
A
P
MPa25.1110040
12
1 5075000 3max
IMc
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EXAMPLE 2 (cont)
The location of the line of zero stress can be determined by proportional
triangles
Elements of material atB and C are subjected only to normal or uniaxial stress.
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
mm3.33
100
1575
x
xx
(Ans)on)(compressiMPa15
(Ans)(tension)MPa5.7
C
B
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EXAMPLE 4
The member shown in Fig. 85a has a rectangular cross section.
Determine the state of stress that the loading produces at point C.
Copyright 2011 Pearson Education South Asia Pte Ltd
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EXAMPLE 4 (cont)
The resultant internal loadings at the section consist of a normal force, a shear
force, and a bending moment.
Solving,
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
kN89.32kN,21.93kN,45.16 MVN
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EXAMPLE 4 (cont)
The uniform normal-stress distribution acting over the cross section is produced
by the normal force.
At Point C,
In Fig. 85e, the shear stress is zero becauseA/ = 0. thus Q = 0.
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
MPa32.125.005.0
1045.16 3
A
Pc
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EXAMPLE 4 (cont)
Point C is located at y = c = 0.125m from the neutral axis, so the normal stress at
C, Fig. 85f, is
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
MPa16.6325.005.0
125.01089.323
21
3
I
Mcc
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EXAMPLE 4 (cont)
The shear stress is zero.
Adding the normal stresses determined above gives a compressive stress at C
having a value of
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
MPa5.6416.6332.1 IMcc
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EXAMPLE 5
The rectangular block of negligible weight in Fig. 86a is subjected to
a vertical force of 40 kN, which is applied to its corner. Determine the
largest normal stress acting on a section throughABCD.
Copyright 2011 Pearson Education South Asia Pte Ltd
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EXAMPLE 5 (cont)
For uniform normal-stress distribution the stress is
For 8 Kn.m, the maximum stress is
For 16 kN.m, the maximum stress is
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
kPa125
4.08.0
40
A
P
kPa3754.08.0
2.083
121max
x
xx
I
cM
kPa3758.04.0
4.0163
121
max
y
xy
I
cM
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EXAMPLE 5 (cont)
By inspection the normal stress at point C is the largest since each loading
creates a compressive stress there
Copyright 2011 Pearson Education South Asia Pte Ltd
Solutions
(Ans)kPa875375375125 c
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Important Points in this Lecture
When the element is subjected to different types of loads, theprincipal of
superpositioncan be used to predict theresul tant stressas the material
behave in elastic manner
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Next Class
Transformation of Stresses
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Thank You