wk7_8_9_shear moment diagram_bending (080415_150415_220415)

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Week 7

APPLICATIONS

Copyright © 2011 Pearson Education South Asia Pte Ltd

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SHEAR AND MOMENT DIAGRAM

• Shear is obtained by summing forces perpendicular to

the beam’s axis up to the end of the segment.

• Moment is obtained by summing moments about the end of the segment.

• Note the sign conventions are opposite when the summing

processes are carried out with opposite direction.


(from left to right vs from right to left)

EXAMPLE 1

Draw the shear and moment diagrams for the beam

shown in the figure.


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EXAMPLE 1 (cont.)

Solutions

The support reactions are shown in Fig. 6–4c.

Applying the two equations of equilibrium yields


( )

( )

( ) ( )2 2

022

;0

1 2

02

;0

2xLxw

M

Mx

wxxwL

M

xL

wV

VwxwL

Fy

−

=

=+

+

−=∑+

−=

=−−=∑↑+

EXAMPLE 1 (cont.)

Solutions

The point of zero shear can be found from Eq. 1:


2

02

Lx

xL

wV

=

=

−=

8222

22

max

wLLLL

wM =

−

=

From the moment diagram, this value of x represents the point on the beam where the maximum moment occurs.

GRAPHICAL METHOD FOR CONSTRUCTING

SHEAR AND MOMENT DIAGRAMS

Regions of distributed load:


( )dxxwV ∫−=∆

( )dxxVM ∫=∆

Change in moment = area under shear diagram

Change in shear = area under distributed loading

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Regions of concentrated force and moment:

( )FV

VVFV

−=∆

=∆+−−

0

0

0

0

MM

MxVMMM

=∆

=−∆−−∆+

GRAPHICAL METHOD FOR CONSTRUCTING

SHEAR AND MOMENT DIAGRAMS

EXAMPLE 2

Draw the shear and moment diagrams for the beam

shown in Fig. 6–12a.


EXAMPLE 2 (cont.)

• The reactions are shown on the

free-body diagram in Fig. 6–12b.

• The shear at each end is plotted first,

Fig. 6–12c. Since there is no

distributed load on the beam, the shear diagram has zero slope

and is therefore a horizontal line.


Solutions

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EXAMPLE 2 (cont.)

• The moment is zero at each end, Fig. 6–12d. The moment diagram has a constant negative slope of -M0/2L since this is the shear in the beam at each point. Note that the couple moment causes a jump in the moment diagram at the beam’s center, but it does not affect the shear diagram at this point.


Solutions

EXAMPLE 3

Draw the shear and moment diagrams for each of the

beams shown in Figs. 6–13a and 6–14a.


EXAMPLE 3 (cont.)


Solutions

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Week 8

Pure Bending Pure Bending:

Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane

Other Loading Types Eccentric Loading: Axial loading which

does not pass through section centroid produces internal forces equivalent to an axial force and a couple

Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple

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Bending Deformations

• bends uniformly to form a circular arc

• cross-sectional plane passes through arc center and remains planar

• length of top decreases and length of bottom increases

• a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change

• stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it

Beam with a plane of symmetry in pure bending: • member remains symmetric

BENDING DEFORMATION OF A STRAIGHT

MEMBER

Assumptions: 1. Plane section remains plane

2. Length of longitudinal axis remains unchanged

3. Plane section remains perpendicular to the longitudinal axis

4. In-plane distortion of section is negligible


Before deformation After deformation

Horizontal lines become

curved

Vertical lines remains straight,

yet rotate

� Longitudinal strain varies linearly from zero at the neutral axis.

� Hooke’s law applies when material is homogeneous.

� Neutral axis passes through the centroid of the cross-sectional area for linear-elastic material.

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Bending deformation

Strain Due to Bending Consider a beam segment of length L. After deformation, the length of the neutral surface remains L. At other sections,

( )( )

mx

mm

x

c

y

cρ

c

yy

L

yyLL

yL

εε

ερε

ρρθθδ

ε

θρθθρδ

θρ

−=

==

−=−==

−=−−=−′=

−=′

or

linearly) ries(strain va

4 - 24

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2 - 25

Revision: Stress-Strain Diagram of Ductile Materials

415 MPa

275

138

Stress Due to Bending For a linearly elastic material,

For static equilibrium,

∫

∫∫

−=

−===

dAyc

dAc

ydAF

m

mxx

σ

σσ

0

0For static equilibrium,

( ) ( )

I

My

c

y

S

M

I

Mc

c

IdAy

cM

dAc

yydAyM

x

mx

m

mm

mx

−=

−=

==

==

−−=−=

∫

∫∫

σ

σσ

σ

σσ

σσ

ngSubstituti

2

linearly) varies(stressm

mxx

c

y

Ec

yE

σ

εεσ

−=

−==

σ = normal stress in the member M = internal moment I = moment of inertia y = perpendicular distance from the neutral axis

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4- 28

Sample Problem 4.2 SOLUTION:

Based on the cross section geometry,

calculate the location of the section

centroid and moment of inertia.

( )∑ +=∑∑= ′

2dAII

A

AyY x

Apply the elastic flexural formula to

find the maximum tensile and

compressive stresses.

I

Mcm =σ

Calculate the curvature

EI

M=

ρ1

A cast-iron machine part is acted upon

by a 3 kN-m couple. Knowing E = 165

GPa and neglecting the effects of

fillets, determine (a) the maximum

tensile and compressive stresses, (b)

the radius of curvature.

4- 29

Sample Problem 4.2 SOLUTION:

Based on the cross section geometry, calculate

the location of the section centroid and

moment of inertia.

mm 383000

10114 3

=×

=∑∑=

A

AyY

∑ ×==∑

×=×

×=×

3

3

3

32

101143000

104220120030402

109050180090201

mm ,mm ,mm Area,

AyA

Ayy

( ) ( )( ) ( )

49-43

23

12123

121

23

1212

m10868 mm10868

18120040301218002090

×=×=

×+×+×+×=

+=+= ∑∑′

I

dAbhdAII x

4- 30

Sample Problem 4.2 Apply the elastic flexural formula to find the

maximum tensile and compressive stresses.

49

49

m10868

m038.0mkN 3

m10868

m022.0mkN 3

−

−

×

×⋅−=−=

×

×⋅==

=

I

cM

I

cM

I

Mc

BB

AA

m

σ

σ

σ

MPa 0.76+=Aσ

MPa 3.131−=Bσ

Calculate the curvature

( )( )49- m10868GPa 165

mkN 3

1

×

⋅=

=EI

M

ρ

m 7.47

m1095.201 1-3

=

×= −

ρρ

Note the moment is negative

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EXAMPLE 4

The simply supported beam in Fig. 6–26a has the cross-

sectional area shown in Fig. 6–26b. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this

location.


EXAMPLE 4 (cont.)

• The maximum internal moment in the beam, 22.5 kN.m,

occurs at the center.

• By reasons of symmetry, the neutral axis passes through the centroid C at the mid-height of the beam, Fig. 6–

26b.


Solutions

( )( )( ) ( )( )( )[ ] ( )( )[ ]( )

( )( )( ) (Ans) MPa 7.12103.301

17.0105.22 ;

m 103.301

3.002.016.002.025.002.025.02

6

3

46

3

12123

121

2

==−=

=

++=

+∑=

−

−

bB

BI

My

AdII

σσ

EXAMPLE 4 (cont.)

• A three-dimensional view of the stress distribution is shown in Fig. 6–

26d.

• At point B,

Solutions

( )( )( ) MPa 2.11103.301

15.0105.22 ;

6

3

−=−=−=−B

BB

I

Myσσ


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Beam Section Properties The maximum normal stress due to bending,

modulussection

inertia ofmoment section

==

=

==

c

IS

I

S

M

I

Mcmσ

A beam section with a larger section modulus will have a lower maximum stress

Consider a rectangular beam cross section,

Ahbhh

bh

c

IS

613

61

3

121

2====

Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending.

Structural steel beams are designed to have a large section modulus.

Properties of American Standard Shapes

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4 - 42

Properties of American Standard Shapes

S=I/c, elastic section modulus

Norminal depth Mass in kg per meter

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4 - 43

4 - 44

Week 9

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UNSYMMETRICAL BENDING

Moment applied along principal axis


( )

( ) dAyMM

dAzMM

dAFF

AZZR

AyyR

AxR

∫∫∫

−=∑=

=∑=

=∑=

σ

σ

σ

0 ;

0 ;

0 ;

If y and z are the principal axes. ∫ yz dA = 0 (The integral is called the product of inertia)

UNSYMMETRICAL BENDING (cont.)

• Moment arbitrarily applied

• Alternatively, identify the orientation of the principal axes (of which one is the neutral axis)

• Orientation of neutral axis:


y

y

z

z

I

zM

I

yM+−=σ

= +

θα tantany

z

I

I=

= +

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EXAMPLE 5

The rectangular cross section shown in Fig. 6–33a is subjected to a

bending moment of 12 kN.m. Determine the normal stress developed at each corner of the section, and specify the

orientation of the neutral axis.


EXAMPLE 5 (cont.)

• The moment is resolved into its y and z components,

where

• The moments of inertia about the y and z axes are


Solutions

( )

( ) mkN 20.7125

3

mkN 60.9125

4

⋅==

⋅−=−=

z

y

M

M

( )( ) ( )

( )( ) ( ) 433

433

m 10067.14.02.012

1

m 102667.02.04.012

1

−

−

==

==

z

y

I

I

EXAMPLE 5 (cont.)

• For bending stress,

• The resultant normal-stress distribution has been sketched using these values, Fig. 6–33b.


Solutions

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( )

( )( )( ) (Ans) MPa 95.4102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 25.2102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 95.4102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 25.2102667.0

1.0106.9

10067.1

2.0102.7

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

=−−

+−

−=

−=−

+−

−=

−=−

+−=

=−−

+−=

+−=

−−

−−

−−

−−

E

D

C

B

y

z

z

z

I

zM

I

yM

σ

σ

σ

σ

σ

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EXAMPLE 5 (cont.)

• The location z of the neutral axis (NA), Fig. 6–33b,

can be established by proportion.

• We can also establish the orientation

of the NA using Eq. 6–19, which is used

to specify the angle that the axis makes with the z or maximum principal axis.


Solutions

( )m 0625.0

2.0

95.425.2=⇒

−= z

zz

( )( ) ( ) (Ans) 4.791.53tan102667.0

10067.1tan

tantan

3

3

°−=⇒°−=

=

−

−

αα

θαy

z

I

I

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Orientation of neutral axis,

Example • The z section shown is

subjected to bending moment of M=20 kNm.

The principle axes y and

z are oriented as shown, such that they represent

the minimum and

maximum principle moments of inertia,

Iy=0.960(10-3)m4 and

Iz=7.54(10-3)m4,

respectively. Determine

the normal stress at point

P and the orientation of the N.A

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Solve this!

• My=

• Mz=

From the coloured triangles,

• Calculate the coordinate of yp and

zp

Bending stress,

σp= - (Mzyp/Iz) + (Myzp/Iy)

Orientation of neutral axis

wk7_8_9_shear moment diagram_bending (080415_150415_220415)

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