1

AER 520Chapter 6 Bending

1.Shear and moment diagram2.Graphic method3.Bending deformation 4.Flexure formula5.Unsymmetric bending

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6.1 Shear and Moment DiagramsBeam: slender, support loading are perpendicular to the longitudinal axis

3

Determine the internal forces (normal force, shear force and bending moment) on a section passing

through the beam at point C.

Using balance equations: Force balance and moment balance two free body diagrams

4

Sign Convention

• Shear ~ positive direction is denoted by an internal force that causes a clockwise rotation on which it acts.

• Moment ~ positive direction is denoted by and internal moment that causes a compression or pushing on the upper part of the member

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Shear and Moment Diagrams

• For beams we can draw the shear and moment diagrams as a function of position along the longitudinal axis of the beam.

• Internal normal force will not be considered as most loadings are vertical on a beam and are primarily concerned about shear and bending failures.

6

Support reaction forces: Symmetric structure

Internal shear force and moment: first half

02

02

y

A

PF V

PM M x

Example 6.1

Sign convention: V: clockwise rotation M: Compression on the upper part

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Internal shear force and moment: the left segment including region BC

0 / 2 0

0 02

y

A

F V P P

LM M Vx P

/ 2

2

V P

PM L x

Example 6.1

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Support reaction forces:

0

0

0 0

0 0

A C

C A

M F L M

M F L M

Example 6.2

0

0

C

A

MF

LM

FL

Shear and moment in region AB:

0

0

0 0

0 0

y

MF V

LM

M M xL

0

0

MV

LM

M xL

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Example 6.2

Shear and moment in region BC:

0

00

0 0

0 0

y

MF V

LM

M M x ML

0

0 1

MV

Lx

M ML

A up jump at the middle

dMV

dx

10

Problem 6-12

Point B: pin joint, only vertical force

Support reaction forces: AB

0 6*4 *6 0

0 6 0

A B

y B A

M F

F F F

4kip

10kipB

A

F

F

Support reaction forces: BC

0 8 0y B CF F F 4kipCF

6 ?kip

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Problem 6-12Free-body diagram:

V

M

6kip

Shear and moment functions:

6kip

6 kip.ft

V

M x

dMV

dx

12

Example 6.5

FA FC

0 *18 36*9 36*12 0

0 *18 36*9 36*6 0

A C

C A

M F

M F

42kip

30kipC

A

F

F

Shear and moment functions:2

30 29

xV x

2

2 * * 30* 02 9 3

x x xM x x

3230

27

xM x x

Support reaction forces:

EXAMPLE: Differences

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AER 520

Chapter 6 Bending

6.2 Graphical Method for shear and moment diagrams

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Regions of distributed load

0 0

0 0

y

O

F V V w x x V

M M M w x x k x M V x

2

V w x x

M V x w x k x

slope of = -distributed

the shear load intensity

dVw x

dx

slope of = the shear

the moment

dMV

dx

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Regions of concentrated force and moment

0 0yF V V F V

For concentrated force: (downward)

V F

For concentrated moment: (clockwise)

00 0OM M M M M

0M M

Same direction of F

0same sign of M

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Shear and moment diagram

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Procedure for analysis

• Support reaction forces and moments• Shear diagram:

/ or - ( )dV dx w x V w x dx V F

• moment diagram:

/ ( )dM dx V or M V x dx 0M M

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Support reaction forces:

0

0

0 0

0 02

y O

O

F F w L

LM M w L

Example 6.9

0

20

2

O

O

F w L

w LM

Shear diagram:

0

0

; 0

/O eF w L F

dV dx w

Moment diagram:2

0 ; 02O e

w LM M

M Vdx

V is positive and decrease to zero;shear diagram: a straight linemoment diagram: parabolic

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Support reaction forces:

4kipA BF F

Problem 6-4

Shear diagram:

(0) 4; (20) 4

/ 0

V V

dV dx

(0) 0; (20) 0

( )

M M

M V x dx

Moment diagram:

There are four jumps (downward at –2kip)

A B

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Support reaction forces:

0 *45 2*45 / 2*45 / 3 0

0 *45 2*45 / 2*2*45 / 3 0

E O

O E

M F

M F

Example 6.11

15

30O

E

F

F

Shear diagram:

0

15; 30

/O EF F

dV dx w

V: slope from 0 to –2

To find the cross point where V=0

115 2 0 26.0

2 45

xx x ft

O E

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Example 6.11

0; 0

/O EM M

dM dx V

Moment diagram:

x=0, slope=15;x=45, slope=-30

The maximum M is at x=26 (V=0)

max

1 26 2615*26 2 26 260

2 45 3M

M Vdx

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Support reaction forces:

1 68 6 2*6* 2*4*2 /10 4.4kip

2 3

1 68 4 2*6* 6 4 2*4*12 /10 17.6kip

2 3

A

C

F

F

Example 6.13

Shear diagram:

4.4; 0;A DV V

B: downward jump 8kip (FA)

BC: further downward at a slope -w

C: upward jump 17.6kip (FC)

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Example 6.13

0; 0

( )

A DM M

M V x dx

Moment diagram:

Cross point of the moment:

AB: linear increase at slope 4.4

BC: decrease in the slope of V

dMV

dx

B and C: turning points

B CM M

parabolic

parabolic

cubic

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Support reaction forces:

2*8*5 /10 8kip

2*8*5 /10 8kip

A

B

F

F

Problem 6-24

4kip/ft upward2

BB

Fw

Shear diagram:

8; 0;

/A BV V

dV dx w

Moment diagram:

0; 0;

/

( )

A BM M

dM dx V

or M V x dx

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Support reaction forces:

2*8*5 /10 8kip

2*8*5 /10 8kip

A

B

F

F

Problem 6-24

4kip/ft upward2

BB

Fw

Shear diagram:

8; 0;

/A BV V

dV dx w

Moment diagram:

0; 0;

/

( )

A BM M

dM dx V

or M V x dx

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6.3 Bending Deformation

Symmetric cross-section

Homogeneous material

Subject to a bending moment

Top compress; bottom stretch

Neutral surface: there is no change in length

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Bending Deformation

Neutral surfaceCompress

Stretch

30

0

0

lim

lim

s

s

s s

sy

y

Normal strain calculation

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Maximum Strain

max

c

y

max y

c

max

y

c

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6.4 The Flexure Formula

Applying Hooke’s Law– σ=Eε

max

c

y

max

c

yNormal strain relation:

A linear variation in strain results in a linear variation in stress.

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From the force equilibrium:

0

0

x

A

F

ydA

It means that the neutral axis is the horizontal centroidal axis

ZM M

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From the force equilibrium:

0xF

It means that the neutral axis is the horizontal centroidal axis

ZM M

0A

dA max

c

y

0A

ydA

A

M ydF max

A A

yy dA y dA

c

2max

A

M y dAc

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Flexure Formula

I

Mcmax

2

A

I y dA

I: the moment of inertia of the cross sectional area about the neutral axis

My

I

T

J

J: the polar moment of inertia of the cross sectional area about the center

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Moment of Inertia Review

Centroid ydA2Moment of Inertia y dA

2Polar Moment of Inertia dA

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Parallel Axis Theorem

2'

2'

x x y

y y x

I I Ad

I I Ad

x

y

x

y

yd

xd(m 4)

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Determine the area moment of inertia for the rectangle shown about the axis x’ and about the axis xb.

22 2

2

33 2

2

1

3 12

h

xh

h

h

I y dA y bdy

bhby

2 2

0

33

0

1

3 3

b

h

x

h

I y dA y bdy

bhby

23 32

' 12 2 3bx x y

bh h bhI I Ad bh

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Composite Shapes

2( )x x yI I Ad

2'

33

2

( )

0.25 6 12[ 6 0.25

12 12

0.25*6 3.125 ]

x x yI I Ad

x

Take away two small blocks

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Example 6.15Internal moment:

15*3 5 3 *1.5 22.5kNMM

Section property:

2'

3 2

3

( )

12 0.25 0.02 0.25 0.02 0.16

12

10.02 0.3

12

x x yI I Ad

Bending stress:

3

max 6

22.5 10 *0.1712.7MPa

301.3 10

Mc

I

max 12.7MPaD

max

0.1512.7 11.2MPa

0.17B

B

y

c

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Example 6.16Neutral axis NA:

0.01 0.25*0.02 2 0.1 0.2*0.015

0.25*0.02 2* 0.2*0.015

0.05909

i i

i

y Ay

A

m

Section property:

2'

3 2

3 2

6 4

( )

10.25 0.02 0.25 0.02 0.04909

12

12 0.015 0.2 0.015 0.2 0.04091

12

42.26 10 m

x x yI I Ad

Internal moment about the NA:

2.4*2 1*0.05909 4.859kNMM

Maximum bending stress:

3

max 6

4.859 10 0.140916.2MPa

42.26 10

Mc

I

?

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Determine the maximum tensile and compressive bending stress in the part

Neutral axis NA:

i i

i

y Ay

A

Section property:

2'( )x x yI I Ad

Bending stress:

Tensile bending stress at the bottom;Compressive bending stress at the top.

My

I

0.0175m

6 40.3633 10 m

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Flexure Formula1. Internal moment reaction support force shear force moment diagram2. Section property neutral axis parallel axis theorem3. Bending stress coordinate of the position

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AER 520

Stress Analysis

6.5 Unsymmetric bending

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Symmetric cross-section

46

Unsymmetric BendingPrincipal axes: y and z

z

My

I

The moments of inertia are the maximumor minimum; see A.4 for detail

Moment applied to the principal axis:

Y axis: y

Mz

I

Z axis:

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Unsymmetric Bending

cos

sinz

y

M M

M M

Moment arbitrarily applied:

1. Moment divided to two moments:

2. Normal stress calculation:

yz

z y

M zM y

I I

3. Orientation of the neutral axis NA:

0yz

z y

M zM y

I I

tan tanz

y

Iy

z I

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Example 6.181. Moment divided to two moments

312* 7.2kNM

54

12* 9.6kNM5

z

y

M

M

2. Section properties:

3 3 410.2 0.4 1.067 10 m

12zI

3 3 410.4 0.2 0.2667 10 m

12yI

3. Bending stress:

yz

z y

M zM y

I I

3 3

3 3

7.2 10 0.2 9.6 10 0.12.25MPa

1.067 10 0.2667 10B

3 3

3 3

7.2 10 0.2 9.6 10 0.14.95MPa

1.067 10 0.2667 10C

Z axis is the NA

y axis is the NA

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Example 6.18

2.25

4.95D

E

MPa

MPa

4. Orientation of neutral axis:

tan tanz

y

I

I

79.4

306.9 ; 53.1

From +z toward +y

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Calculate the bending stress at A and B

1. The centroid location:

376*12 *6 2 150*12 *75

36.6mm376*12 2 150*12

z

2. Section properties:

800cos( 60 ) 400Nm

800sin( 60 ) 692.82Nmz

y

M

M

3. Bending stress:

yz

z y

M zM y

I I

3 6

692.82 0.15 0.0366400*0.24.38MPa

0.18869 10 16.3374 10A

3 6

400 0.2 692.82 0.03661.13MPa

0.18869 10 16.3374 10B

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Example 6.191. Internal moment components:

15cos 60 7.5kN M

15sin 60 12.99kN Mz

y

M

M

2. Section properties:

0.05 0.1 0.04 0.115 0.03 0.2

0.1 0.04 0.03 0.2

.089

zAz

A

m

3 3

6 4

1 10.1 0.04 0.03 0.2

12 12

20.53 10 m

zI

3 2

3 2

6 4

10.04 0.1 0.04 0.1 0.89 0.5

121

0.3 0.03 0.3 0.03 0.115 0.8912

13.92 10 m

yI

Z axis is the NA

Y axis is the NA

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Example 6.193. Bending stress:

yz

z y

M zM y

I I

4. Orientation of neutral axis:

tan tan ; 60z

y

I

I

6 6

7.5 0.1 12.99 0.04174.8

20.53 10 13.92 10B MPa

6 6

7.5 0.02 12.99 0.08990.4

20.53 10 13.92 10C MPa

68.6

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Determine the force P so that max

180A MPa

1. Internal moment components

0; 2 cos30 0z zM M P 0; 2 sin 30 0y yM M P

1.732

1.0z

y

M P

M P

2. Section properties:

3

3 6 4

12 0.01 0.2

12

10.015 0.01 13.3458 10

12

yI

m

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2. Section properties:

3 3

6 4

1 10.2 0.17 0.19 0.15

12 12

28.4458 10

zI

m

3. Maximum bending stress:

yz

z y

M zM y

I I

1.732

1.0z

y

M P

M P

Choose y positive and z negative: point A

6

6 6

1.732 0.085 1.0 0.1180 10

28.4458 10 13.3458 10

P P

14208 14.2P N kN

Determine the maximum bending stress in the shaft with 30mm diameter.