stress analysis lecture 4 me 276 spring...

Stress Analysis

Lecture 4

ME 276

Spring 2017-2018

Dr./ Ahmed Mohamed Nagib Elmekawy

Shear and Moment Diagrams

Beam Sign Convention

The positive directions are as follows:

• The internal shear force causes a

clockwise rotation of the beam segment

on which it acts.

• The internal moment causes

compression in the top fibers of the

segment such it bends the segment so

that it “hold water”.

Example 1

Draw the shear and moment diagrams for the beam shown in the attached figure.

)1(2

02

0

xL

wV

VwxwL

Fy


)2(2

022

0

2xLxw

M

Mx

wxxwL

M


8222

22

max

wLLLL

wM

dx

dMV

Note


Example 2

Draw the shear and moment diagrams for

the beam shown in the attached figure.

0 x1 5

)1(kN75.5

075.50

V

VFy

)2(kN.m8075.5

075.5800

1

1

xM

MxM


5 x2 10

)3(kN575.15

0551575.5

0

2

2

xV

Vx

Fy

)4(kN.m5.9275.155.2

02

555

51575.580

0

2

2

2

22

22

xxM

Mx

x

xx

M

The Bending Formula

The Bending Formula

Where

M the resultant internal moment acting at the cross section.

I the moment of inertia of the cross-sectional area about the neutral axis.

c perpendicular distance from the neutral axis to a point farthest away

from the neutral axis. This is where max acts.

max the maximum bending stress in the shaft, which occurs at the outer surface.

y perpendicular distance from the neutral axis to a point where is to be

calculated

yI

M

cI

Mmax

Bending a Cantilever Beam under a Concentrated Load

11

Bending a Cantilever Beam under a Concentrated Load

12

Bending Stress

13

Bending Stress

14

Bending Stress

15

Bending Stress

16

Bending Stress

17

Bending Stress

18

Bending Stress

19

Bending of Curved beam

20

Displacement Stress in x direction

Example 1

A beam has a rectangular cross section and is

subjected to the stress distribution shown in the

figure. Determine the internal moment M at the

section caused by the stress distribution.

The Bending Formula

N.mm10*288

10*840

6020

4

4max

M

Mc

I

M

44

33

mm10*84012

12060

12

bhI

The Bending Formula

Example 2

The simply supported beam has the cross-sectional area shown in the figure.

Determine the absolute maximum bending stress in the beam and draw the

stress distribution over the cross section at this location.

The Bending Formula

463

23

2

mm10*3.3013002012

1

160*250*202025012

12

AdII

MPa7.12

10*3.301

170*10*5.22;

6

6

maxmax

cI

M

The Bending Formula

Example 3

The beam shown in the figure has a

cross-sectional area in the shape of a

channel as shown in the attached figure.

Determine the maximum bending stress

that occurs in the beam at section a-a.

mm09.59

250*20200*15*2

250*20*10200*15*100*2~

A

Ayy

N.mm10*4859

009.59*10002000*2400

0

3

M

M

M NA

The Bending Formula

46

23

23

2

mm10*26.42

1009.59*20*2502025012

1

09.59100*200*152001512

12

AdII

MPa2.16

10*25.42

09.59200*10*859.4;

6

6

maxmax

c

I

M

The Shear Formula

Where

The shear stress in the member at the point located a distance y from the neutral

axis. This stress is assumed to be constant and therefore averaged across the

width t of the member.

V The internal resultant shear force, determined from the method of sections and the

equations of equilibrium.

It

VQ

The Shear Formula

I The moment of inertia of the entire cross-sectional area calculated about the

neutral axis.

t The width of the member’s cross-sectional area, measured at the point where is

to be calculated.

Q = , where A is the area of the top portion of the member’s cross-sectional area,

above the section plane where t is measured, and is the distance from the

neutral axis to the centroid of A.

Ay

y

Example 1

Determine the distribution of the shear stress

over the cross section of the beam shown in the

attached figure.

The Shear Formula

22

42

1

22

2 yh

byh

yh

yAyQ

Applying the shear formula, we have

22

33

22

4

6

12

1

42

1

yh

bh

V

bbh

byh

V

It

VQ

This indicates that the shear stress distribution

over cross section is parabolic with maximum

value of (1.5V/A) at the neutral axis and zero at

top and bottom as shown in the attached figure.

The Shear Formula

It is important to realize that max also acts in the longitudinal direction of the

beam as shown in the attached figure. It is this stress that can cause a timber

beam to fail as shown in the figure. Here horizontal splitting of the wood starts to

occur through the neutral axis at the beam’s ends, since the vertical reactions

subject the beam to large shear stress and wood has a low resistance to shear

along its grains, which are oriented in the longitudinal direction.

Shear Stress due to bending

30

The Shear Formula

Example 2

A steel wide-flange beam has the

dimensions shown in the figure. If is

subjected to a shear of V = 80 kN, plot

the shear-stress distribution acting over

the beam’s cross sectional area.

452

33

mm10*15561102030012

203002

12

20015

I

For point B

34 mm10*6620*300110

mm300

AyQ

t

B

B

The Shear Formula

For point B

34 mm10*66

mm15

BB

B

QQ

t

MPa13.1300*10*1556

10*66*1000*805

4

B

BB

It

VQ

MPa6.2215*10*1556

10*66*1000*805

4

B

BB

It

VQ

For ppint C

33 mm10*735

100*1550300*20110

mm15

AyQ

t

C

C

The Shear Formula

MPa2.2515*10*1556

10*735*1000*805

3

C

CC

It

VQ

From the attached figure, note that the shear

stress occurs in the web and is almost

uniform throughout its depth, varying from

22.6 MPa to 25.2 MPa. It is for this reason

that for design, some codes permit the use of

calculating the average shear stress on the

cross section of the web rather than using the

shear formula; that is,

MPa7.26200*15

1000*80

A

Vavg

The Shear Formula

State of Stress Caused by Combined Loadings

In previous chapters we developed methods for determining the stress distribution in

a member subjected to either an internal axial force, a shear force, a bending moment,

or a torsional moment. Most often, however, the cross section of a member is

subjected to several of these loadings simultaneously. When this occurs, the method

of superposition can be used to determine the resultant stress distribution.

43

Summary

Principal Stresses

44

Example 1For the state of plane stress shown in Fig., determine (a ) theprincipal planes, (b ) the principal stresses, (c ) the maximum shearing stress.

47

Sample Problem 7.1 in Beer’s Book


Example 1

A force of 15 kN is applied to the edge of the member shown in the figure.

Neglect the weight of the member and determine the state of stress at points B

and C.


MPa75.3100*40

15000

A

P

MPa25.11

1004012

1

50*15000

3max

I

Mc

Normal Force

Bending Moment

Superposition


mm3.33

100

155.7

x

xx MPa1575.325.11

MPa5.775.325.11

C

B

Example 2

The member shown in the figure has a rectangular cross section. Determine the state

of stress that the loading produces at point C.


Normal Force

Shear Force

MPa32.1250*50

1000*45.16

A

Pc

0c


Bending Moment

MPa16.63

2505012

1

125*1500*1000*93.21

3

I

McC

Superposition

MPa5.6416.6332.1 C


Example 4

The solid rod shown in the figure has a radius of 7.5 mm. If it is subjected to the

force 500 N, determine the state of stress at point A.

Normal Force

MPa83.25.7

5002

A

PyA


Bending Moment

MPa3.211

1564

5.7*140*500

4

I

McyA

Superposition

MPa13.2143.21183.2 yA

State of Stress Caused by Combined LoadingsExample 5

The solid rod shown in the figure has a radius of 7.5 mm. If it is subjected to the

force 800 N, determine the state of stress at point A.

Shear Force

32mm25.2815.7

2

1

3

5.7*4

AyQ


MPa04.6

5.7*21564

25.281*800

4

It

VQAyz

Bending Moment 0A

Torque

MPa01.169

1532

5.7*1000*112

4

J

TcAyz

Superposition

MPa05.175

01.16904.6

Ayz

stress analysis lecture 4 me 276 spring...

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