Bending

Shear and Moment DiagramsShear and Moment DiagramsMembers with support loadings applied perpendicular to their longitudinal axis are calledperpendicular to their longitudinal axis are called beams.Beams classified according to the way they areBeams classified according to the way they are supported.

© 2008 Pearson Education South Asia Pte Ltd© 2008 Pearson Education South Asia Pte Ltd

Chapter 6: BendingChapter 6: BendingMechanics of Material 7Mechanics of Material 7thth EditionEdition

Shear and Moment DiagramsShear and Moment DiagramsShear and moment functions can be plotted in graphs called shear and moment diagramsgraphs called shear and moment diagrams.Positive directions indicate the distributed load acting downward on the beam and clockwise rotation of thedownward on the beam and clockwise rotation of the beam segment on which it acts.



Example 6.1Example 6.1Draw the shear and moment diagrams for the beam shown.

Solution:Solution:From the free-body diagram of the left segment, we apply the equilibrium equations,

∑↑ (1)0 P

∑

∑

==↑+

==↑+

(2) 2

;0

(1) 2

;0

xPMM

PVFy

PP↑∑

2Left segment of the beam extending a distance x within region BC is as follow,

( ) (4) 222

;0

(3) 2

02

;0

L-xPMxPLxPMF

PVVPPF

y

y

=⇒−⎟⎠⎞

⎜⎝⎛ −+=

−=⇒=−−=↑

∑

∑



222 ⎠⎝

Solution:Solution:

The shear diagram represents a plot of Eqs. 1 and 3 g p p q

The moment diagram represents a plot of Eqs. 2 and 4




Solution:Solution:The distributed load is replaced by its resultant force and the reactions.Intensity of the triangular load at the section is found by

LxwwL

wx

w 00 or ==

y g yproportion,

R lt t f th di t ib t d l di i d t i d f th

( )∑ −=⇒=−⎞⎜⎛−=↑+ (1)01;0 22000 xLwVVxxwLwF

Resultant of the distributed loading is determined from the area under the diagram,

( )

( )∑

∑

=+⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛+−=+

⇒⎠

⎜⎝

↑+

(2) 031

21

23 ;0

(1)2

022

;0

002

0 MxxL

xwxLwLwM

xLL

VVxL

Fy



Solution:Solution:

The shear diagram represents a plot of Eqs. 1

The moment diagram represents a plot of Eqs. 2




Solution:Solution:2 regions of x must be considered in order to describe the shear and moment functions for the entire beam.

( ) (2)kNm80755075580;0

(1) kN 75.5075.5 ;0

m, 50 1

+=⇒=+−−=+

=⇒=−=↑+

<≤

∑∑

xMMxM

VVF

x

y

( ) (2) kNm8075.5075.580 ;0 11 +=⇒=+=+∑ xMMxM

( ) ( ) (3)kN5751505515755;0m, 10m 5

22

1

−=⇒=−−−−=↑+

<≤

∑ xVVxFx

( ) ( )

( )

( )0

25551575.580 ;0

(3)kN575.150551575.5 ;0

2

221

22

=+⎟⎠⎞

⎜⎝⎛ −

−+++−−=+

⇒↑+

∑

∑MxxxM

xVVxFy



( ) (4) kNm5.9275.155.2 222 ++−= xxM

Solution:Solution:

The shear diagram represents a plot of Eqs. 1 and 3

The moment diagram represents a plot of Eqs. 2 and 4



Bending Deformation of a Straight MemberBending Deformation of a Straight MemberCross section of a straight beam remains plane when the beam deforms due to bendingwhen the beam deforms due to bending.There will be tensile stress on one side and compressive stress on the other sidecompressive stress on the other side.



Bending Deformation of a Straight MemberBending Deformation of a Straight MemberLongitudinal strain varies linearly from zero at the neutral axisneutral axis.Hooke’s law applies when material is homogeneous.Neutral axis passes through the centroid of theNeutral axis passes through the centroid of the cross-sectional area for linear-elastic material.



The Flexure FormulaThe Flexure FormulaResultant moment on the cross section is equal to the moment produced by the linear normal stressthe moment produced by the linear normal stress distribution about the neutral axis.

MyI

My−=σ

σ = normal stress in the memberM lt t i t l tM = resultant internal momentI = moment of inertiay = perpendicular distance from the neutral axis

By the right-hand rule, negative sign is compressive since it acts in the negative x direction.



Example 6.15Example 6.15The simply supported beam has the cross-sectional area as shown. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this locationthe cross section at this location.

Solution:Solution:The maximum internal moment in the beam is kNm 5.22=M



Solution:Solution:By symmetry, the centroid C and thus the neutral axis pass through the mid-height of the beam, and the moment of inertia is

( )( )( )( ) ( )( )( ) ( )( )323

2

3.002.012116.0002.025.002.025.0

1212 ⎥⎦

⎤⎢⎣⎡+⎥⎦

⎤⎢⎣⎡ +=

+=∑ AdII

( ) 46 m 103.301 −=⎦⎣⎦⎣

Applying the flexure formula where c = 170 mmApplying the flexure formula where c = 170 mm,

( )( ) (Ans) MPa 7.12103301

17.05.22 ; 6maxmax === −σσI

Mc( )103.301 6I



Example 6.17Example 6.17The beam has a cross-sectional area in the shape of a channel. Determine the maximum bending stress that occurs in the beam at section a–a.

Solution:The resultant internal moment must be computed about the beam’s neutral axisThe resultant internal moment must be computed about the beam s neutral axis at section a–a. Since the neutral axis passes through the centroid,

( )( )( ) ( )( )( )250020010015020102∑ Ay ( )( )( ) ( )( )( )( )( ) ( )( )

mm09.59m05909.0 25.002.0015.02.02

25.002.001.0015.02.01.02

==

++

==∑∑

AAy

y



Solution:Applying the moment equation of equilibrium about the neutral axis, we have

( ) ( ) kNm8594005909001242;0 =⇒=+=+∑ MMM ( ) ( ) kNm859.4005909.00.124.2 ;0 =⇒=−+=+∑ MMM NA

The moment of inertia about the neutral axis is

1 ⎤⎡ ( )( ) ( )( )( )

( )( ) ( )( )( )23

23

05909.01.02.0015.02.0015.01212

01.005909.002.025.002.025.0121

⎥⎦⎤

⎢⎣⎡ −++

⎥⎦⎤

⎢⎣⎡ −+=I

( )( ) ( )( )( )

( ) 46 m 1026.42 12

−=

⎥⎦⎢⎣

The maximum bending stress occurs at points farthest away from the neutral axis.

( )( ) (Ans) MPa 2.16102642

05909.02.0859.46max =

−== −I

Mcσ



( )1026.42 6max I

Stress ConcentrationsStress ConcentrationsThe maximum normal stress at each of discontinuities occurs at smallest cross sectionaldiscontinuities occurs at smallest cross sectionalarea.Once K is obtained the maximum bending stress isOnce K is obtained, the maximum bending stress is determined using

IMcK=maxσ



Example 6.26Example 6.26The transition in the cross-sectional area of the steel bar is achieved using shoulder fillets. If the bar is subjected to a bending moment of 5 kNm, determine the maximum normal stress developed in the steel The yield stress is σ = 500 MPamaximum normal stress developed in the steel. The yield stress is σY = 500 MPa.

Solution:From the geometry of the barFrom the geometry of the bar,

5.180

120 2.08016

====hw

hr

Thus K is 1.45 and we have

( ) ( )( )( )( )

MPa 3401

04.0545.13

max =⎤⎡

==I

McKσ( )( )08.002.0

121 3

⎥⎦⎤

⎢⎣⎡I

This result indicates that the steel remains elastic since the stress is below the yield stress (500 MPa).



y ( )

Bending

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