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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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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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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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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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+ ! 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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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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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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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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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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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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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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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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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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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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(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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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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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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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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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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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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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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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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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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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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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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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