6. stress caused by combined loadings

8/11/2019 6. Stress Caused by Combined Loadings

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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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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.



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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


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.


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.



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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,


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.


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


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


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.



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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


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


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

6. stress caused by combined loadings

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