06 Bending Part 1

8/9/2019 06 Bending Part 1

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! Shear and moment diagrams! Bending Deformation of a Straight Member ! The Flexure Formula! Unsymmetric Bending! Composite beams! Reinforced Concrete Beams

BENDING


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2

x

P

w P

M o

L1 L

2 L3

A B C

D

R A R B

R A R A - P

x

M

R B

+-

x1 L3- x1

R A L1

R A L1 + ( R A- P ) L2

R A L1 + ( R A- P ) L2 - M o

Shear and bending moment diagram


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3

8 kN3 kN/m

A B C

D

4 m 4 m 4 m

12 kN

Example 1

Draw the shear and moment diagram for the beam shown.


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4

8 kN3 kN/m

A B C

D

4 m 4 m 4 m

12 kN

+ ! M A # $%

! F y = 0:+

-8(4) -12(8) - (3x4)(10) + R D(12) = 0

R D = 20.67 kN

R A

- 8 -12 - (3x4) + 20.67 = 0

R A = 11.33 kN

R D R A


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8 kN 3(4) = 12 kN

A B C

D

4 m 4 m 4 m

12 kN 3(4) = 12 kN

&'()*+',-./0*$-./+'1+ # &2,-.1

+11.33(4) = 45.32

+3.3(4) = 3..32

8 kN 12 kN

11.33 kN 20.67 kN

M (kNm)

x

-8.67

45.32 58.64

V (kN)

x

11.33

3.33

-20.67

11.33 kN 20.67 kN


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3 m 3 m 2 m

10 kN8 kNm 3 kN/m

A B C D

Example 2

Draw the shear and moment diagram for the beam shown.


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7

+ ! M A # $%

! F y = 0:+

3 m 3 m 2 m

10 kN8 kNm 3 kN/m

A B C D

R A RC

-10(3) - 8 - (0.5x3x3)(3+(2/3)x3) + RC (6) - (3x2)(7) = 0

RC = 17.08 kN

R A

- 10 - (0.5x3x3) + 17.08 - (3x2) = 0

R A = 3.42 kN


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9

Bending deformation of A Straight Member

Before deformation

M

M

After deformation

Vertical lines remainstraight, yet rotate

Horizontal lines become curved


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10

d 3

Neutral axis

4

5 x

y

x

z

5 x Neutral axis

Longitudinal axis

M

Strain ( 6666 )

c y

yd 23

23 d

7 max

4 6

y

c y

max6 6 #

yc#

6 6 max

4 6

cmax

3 4

3

d

cd )2(2

x !"

# maxmax

2

)(c yc

4 6


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11

d 3

Neutral axis

4

5 x

y

x

z

5 x Neutral axis

Longitudinal axis

M

c y

7 max

max6 6 c y

6

c ymax6 6

constant

)(

: Notes

max

#

#

c

#

y f #


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12

x

y

The Flexure formula

M

c

8 max

y8

For positive bending moment M:

c y

$

$ #

max

c

$

y$ max#

c

$ y$ max

constant

)(

: Note

max ##

c

$ y f $


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13

NA

Bending Stress diagram

NA M

dA y

dF

)( dA y ydF dM 8 ##

dAc y y M

A

)( max9 8

I My

x I cmax8 x I y8

Pure Bending : 8 = 0 at neutral axis

Neutral axis ( NA ) : A definition

9 A

dA yc

2max8

: 9## dF F x 0

9 A

dAc y

8 8 )(0 max

9 A dA yc 8 8 max0

0#9 A

dA y8

0,0;0 ;## A y A y

9# A

dA8 0

8 is compression when + M

M

General formula

8 max

8 NA = 0


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14

Stress and Strain Distribution for Pure Bending

NA M

x

y

z

6 max

6 NA = 0

Strain Distribution

8 max

8 NA = 0

Stress Distribution

M

M NA c y y

8 6

c y#00&>0#

d webd frange

mm7.222)20250()20300(

)20250)(310()20300)(150( #>0>

>0>#:

:# A

A y y


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21

5 kN/m

4 m

17.5 kN30 kNm

4 m 20 kN17.5 kN

V (kN)

x(m)

M (kNm)

x (m)

17.5

-30

40 -20

Internal Loading

17.5

The positive maximum bending M+ = 40 kNm

The maximum negative bending M - = -30 kNm


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22

The Bending Stress in the Maximum Bending in Tension (M = 40 kNm)

40 kNm (8 max )C

(8 max )T

250 mm

20 mm

300 mm

20 mm

222.7 mm

97.3 mm NA B

(C)MPa8.33)m10115(.

)m0973.0)(mkN40()( 431

max #>


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23

The Bending Stress in Maximum Bending in Compression (M = -30 kNm)

250 mm

20 mm

300 mm

20 mm

222.7 mm

97.3 mm NA B

= 25.4 MPa

= 58.1 MPa

)T(MPa4.25)m10115(.

)m0973.0)(mkN30()( 431

max #>


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24

(8 max )T

(8 max )C

NA

= 25.4 MPa

= 58.1 MPa

30 kNm

(8 max )C

(8 max )T

= 33.8 MPa

= 77.5 MPa

40 kNm

The absolute maximum bending stress in tension and also in compression in the beam250 mm

20 mm

300 mm

20 mm

222.7 mm

97.3 mm NA

By comparison,

The maximum bending stress in tension ; (8 max )T = 77.5 MPa ( T )

The maximum bending stress in compression ; (8 max )C = 58.1 MPa ( C )


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25

Example 5

The simply supported beam has the cross-sectional area shown. Determinethe absolute maximum bending stress in the beam and draw the stress distributionover the cross section at this location.

500 mm

250 mm

300 mm

250 mm

5 kN/m

4 m

17.5 kN30 kNm

2.25 m

4 m


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28

40 kNm(8 max )C

8 NA

= 0

(8 max )T

0.5 m

0.25 m

NA

Section Property

432 m002604.00)m5.0)(m25.0(121

)( #0#0# : Ad I I

Bending Stress at D

0.25 m

0.25 m

MPa84.3)m10604.2(

)m25.0)(mkN40(43max #>


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29

Maximum bending stress occurs at A

MPa33.9max #$

5 kN/m

4 m

17.5 kN30 kNm

2.25 m

4 m

20 kN17.5 kN

M (kNm)

x (m)

-30

40

9.38

D A B

C

30 kNm

0.3 m

0.25 m

NA0.15 m

0.15 m

(8 max )T

(8 max )C

= 9.33 MPa

= 9.33 MPa

0.15 m


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30

x

Bending Stress Distribution(Profile View)

y

M

Unsymmetric Bending

Moment Applied Along Principal Axis

z

y

C

x

M

y

z

dAdF = 8 dA

6 max

c

y8

8 max

c

x

Normal Strain Distribution(Profile View)

y

y

6


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31

Moment Arbitrarily Applied

x

y

z M 3

x

y

z

M z = M cos 3

x

y

z

M y = M sin 3

y

y

z

z

I

z M

I y M 0


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32

x

y

z

N

A

?

z M z

(8 x)max

(8 x)max

y

M y

(8 x)max

(8 x)max

[(8 x)max + (8 x)max ] [(8 x)max - (8 x)max ]

[(8 x)max - (8 x)max ] [(8 x)max + (8 x)max ]


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B

C D

A

H O

Orientation of the Neutral Axis ( ???? )

x

y

z

N

A?

[(8 x)max + (8 x)max ] [(8 x)max - (8 x)max ]

[(8 x)max - (8 x)max ] [(8 x)max + (8 x)max ]

x

y

z

B C

E D

A

A F

G

)(tan 1 HO FH ?

F

G

?


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x


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35

8 B = ( ) + ( )

8 C = ( ) + ( )

8 D = ( ) + ( )

8 E = ( ) + ( )

Section Properties

433 m)10(2667.0)m2.0)(m4.0(12

1 # y I

433 m)10(067.1)m4.0)(m2.0(121 # z I

Bending Stress I

My#8 :

MPa35.1m)10(067.1

)m2.0(m N)10(2.7:mkN20.7 43

3

@#


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36

B C

D

z

y

E

400 mm

C

D

c

B

0.2 m4.95 MPa

2.25 MPa2.25 MPa

4.95 MPa

Orientation of Neutral Axis

m625.0;)25.295.4(

2

25.2#

0# FD FD

E F D

4.95 MPa

-2.25 MPa2 m

137.5 mm

62.5 mm

o4.79))5.625.137

400(tan 1 #

&? ?

z

N

A

m375.1;)25.295.4(

2

95.4#

0# GC GC

G B

C

2.25 MPa

-4.95 MPa2 m

N A

F

G


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37

Example 7

A T-beam is subjected to the bending moment of 15 kNm as shown. Determinethe maximum normal stress in the beam and the orientation of the neutral axis.

y

z

z

M = 15 kNm

30o

30 mm

100 mm

80 mm40 mm 80 mm

z Section Properties


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38

y

z

M = 15 kNm

30o

30 mm

100 mm

80 mm

40 mm 80 mm

M y = 15 cos 30 o = 13 kNm

M z = 15 sin 30 o = 7.5 kNm B

C

4633 m)10(53.20)m20.0)(m03.0(121)m04.0)(m10.0(

121 # z I

4623

23

m)10(92.13])m089.0m115.0)(m03.0m2.0()m03.0)(m20.0(121[

])m05.0m089.0)(m10.0m04.0()m10.0)(m04.0(121

[

&>00

&>0# y I

Section Properties

z

m0890.0)m2.0m03.0()m04.0m1.0(

)m2.0m03.0)(m115.0()m04.0m1.0)(m05.0( #>0>

>0>## ::

A

A z z

z Maximum Normal Stress

My#8:


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39

4646

m)10(92.13m)10(53.20 && 0# B$

4646 m)10(92.13m)10(53.20 && 0#C $

M z = 7.5 kNm M y = 13 kNm

4646 m)10(92.13m)10(53.20 && 0# D$

0.041 m0.041 m

y

z

0.089 m M y = 13 kNm

M z = 7.5 kNm B

C

46 m)10(53.20 z I 46

m)10(92.13 &

# y I M y = 13 kNm

M z = 7.5 kNm

0.089 m

Maximum Normal Stress, #8 :

(+7.5 kNm)= 74.8 MPa ( T )

(0.02 m) (-13 kNm) (0.089 m)= -90.4 MPa ( C )

(-7.5 kNm)

(-7.5 kNm)

(0.10 m)

(0.10 m)

(+13 kNm)

(+13 kNm)

(0.041 m)

(0.041 m)= 1.76 MPa ( T )

D

0.1 m

0.02 m

0.1 m

0.02 m

0.08 m0.08 m

MPa874#$ Orientation of Neutral Axis


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0.041 m

y

z

z

0.089 m

0.1 m

M y = 13 kNm

M z = 7.5 kNm B

C

M y = 13 kNm

M z = 7.5 kNm

MPa8.74# B$

0.1 m

Orientation of Neutral Axis

D

B D

y

z

N

A

?

MPa76.1# D$

B D0.2 m

74.8 MPa

1.76 MPa

e

m)10(738.2,8.74

2.076.1

3# eee

o%% 68,m0410.0

)m002738.0m10.0(tan #0#

0.041 m

0.1 m e = 0.00273 m

Documents

06 Bending Part 1