BENDING
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Shear and moment diagrams
Bending Deformation of a Straight Member
The Flexure Formula
Unsymmetric Bending
Composite beams
Reinforced Concrete Beams
1
Shear and bending moment diagram
P
A
RA
w
B
L2
L1
D
MoC
L3
RB
P
RA
RA - P
x1
L3- x1
+
x
RB
M
RAL1 + (RA-P)L2
RAL1
x
RAL1 + (RA-P)L2 - Mo
2
Example 1
Draw the shear and moment diagram for the beam shown.
8 kN
12 kN
3 kN/m
A
C
B
4m
D
4m
4m
3
8 kN
12 kN
3 kN/m
A
RA
C
B
4m
D
4m
4m
RD
+ ΣMA = 0:
-8(4) -12(8) - (3x4)(10) + RD(12) = 0
RD = 20.67 kN
+
ΣFy = 0:
RA - 8 -12 - (3x4) + 20.67 = 0
RA = 11.33 kN
4
12 kN 3(4) = 12 kN
8 kN
A
11.33 kN
C
B
4m
D
4m
20.67 kN
4m
V(kN)
11.33
+11.33(4) = 45.32
3.33
+3.3(4) = 3..32
-8.67
−(1/2)(8.67+20.67)(4)
= −58.64
x
-20.67
M(kN•m)
45.32
58.64
x
5
Example 2
Draw the shear and moment diagram for the beam shown.
10 kN
8 kN•m
3 kN/m
A
C
B
3m
3m
D
2m
6
10 kN
8 kN•m
3 kN/m
A
RA
B
3m
3m
RC
C
D
2m
+ ΣMA = 0:
-10(3) - 8 - (0.5x3x3)(3+(2/3)x3) + RC(6) - (3x2)(7) = 0
RC = 17.08 kN
+
ΣFy = 0:
RA - 10 - (0.5x3x3) + 17.08 - (3x2) = 0
RA = 3.42 kN
7
6 kN
3 kN/m
10 kN
4.5 kN
8 kN•m
3.42 kN
3m
17.08 kN
2m
3m
V(kN)
6
3.42
+3.42(3) = 10.26
x
−
+(1/2)(2)(6)=6
-6.58
-11.08
M(kN•m)
18.26
10.26
x
-6
8
Bending deformation of A Straight Member
Before deformation
M
Horizontal lines
become curved
M
Vertical lines remain
straight, yet rotate
After deformation
9
• Strain ( ε )
y
z
Neutral axis
M
∆x
x
− 2δ max
∆x
=
ε max =
dθ
dθ
y
2
y
δmax
ρ
c
Neutral axis
∆x
Longitudinal axis
ε max =
dθ
2
dθ
)c
2
ρdθ
ε max c
=
ε
y
ε = ε max
− 2(
−c
ρ
ε=
ε=
y
c
−c y
( )
ρ c
−y
ρ
10
y
dθ
z
Neutral axis
M
ε
∆x
δmax
c
y
x
∆x
Longitudinal axis
ε
ε max
=
ε=
ρ
Neutral axis
−y
c
−y
ε max
c
Notes :
ε = f (y )
ε max
= constant
c
11
The Flexure formula
σmax
σ
y
σ max
x
σ
y
y
c
σ = y
c
M
=
σ max
c
For positive bending moment M:
σ = −y
σ max
c
Note :
σ = f (y )
σ max
= constant
c
12
M
σmax
dA
y
dF
NA
σNA = 0
NA
M
Bending Stress diagram
• General formula
• Neutral axis (NA) : A definition
dM = ydF = y (σdA)
∑F
x
y
M = ∫ y ( − σ max )dA
c
A
=−
=−
σ max
c
σ max
c
0 = ∫ σdA
A
y
0 = ∫ ( − σ max )σdA
c
A
2
y
∫ dA
A
Ix = −
σ
y
My
σ =−
I
σ is compression when +M
= 0 = ∫ dF
0=−
Ix
σ max
c
∫ yσdA
A
∫ yσdA = 0
A
yA = 0;
y = 0, A ≠ 0
Pure Bending : σ = 0 at neutral axis
13
• Stress and Strain Distribution for Pure Bending
M
y
NA
z
NA
c
σmax
M
y
σ
σNA = 0
Stress Distribution
•
y
σ
=
σ max c
M
x
εmax
ε
εNA = 0
y
Strain Distribution
•
ε
ε max
=
y
c
14
Beam
P
A
C
B
D
P
Ax
NC
A
C
Ay
VC
VC
MC
MC
D ND
NC C
MD
VD
P
Ax
ND
A
D
Ay
VD
MD
15
Example 3
The simply supported beam has the cross-sectional area shown. Determine
the absolute maximum bending stress in the beam and draw the stress distribution
over the cross section at this location.
17.5 kN
5 kN/m
30 kN•m
300 mm
4m
4m
250 mm
16
17.5 kN
5 kN/m
• Internal Loading
30 kN•m
17.5 kN
4m
4m
20 kN
V (kN)
17.5
17.5
x(m)
40
M(kN•m)
-20
x (m)
-30
Neutral Axis
(σmax)C
40 kN•m
σNA = 0
(σmax)T
17
0.15 m
0.3 m
0.15 m
Neutral Axis 0.075 m
B
(σmax)C = 10.7 MPa
40 kN•m
5.33 MPa
σNA = 0
5.33 MPa
(σmax)T = 10.7 MPa
0.25 m
• Section Property
I = ∑ ( I + Ad 2 ) =
1
( 0 . 25 m )( 0 . 3 m ) 3 + 0 = 0 . 0005625 m 4
12
• Bending Stress
σ max =
σB =
Mc ( 40 kN • m )( 0 . 15 m )
=
= 10 . 7 MPa (C) ⇐
I
(. 5625 × 10 −3 m 4 )
− My B
− ( 40 kN • m )( 0 . 075 m )
=
= −5 . 33 MPa (C) ⇐
I
(. 5625 × 10 −3 m 4 )
18
Example 4
The simply supported beam has the cross-sectional area shown. Determine
the absolute maximum bending stress in tension and also in compression in the
beam, and draw the stress distribution over the cross section at the mid-span.
17.5 kN
5 kN/m
30 kN•m
250 mm
20 mm
300 mm
4m
4m
20 mm
19
• Section property
250 mm
20 mm
NA
300 mm
y'
y
dfrange
dweb
310 mm
150 mm
20 mm
y =
∑ y A (150 )( 300 × 20 ) + ( 310 )( 250 × 20 )
=
= 222 . 7 mm
∑A
( 300 × 20 ) + ( 250 × 20 )
y ' = 320 − 222.7 = 97.3 mm
I NA = Σ ( I + Ad 2 )
=[
1
1
(20)( 300) 3 + (20 × 300)( 222.7 − 150) 2 ] + [ (250)( 20) 3 + (250 × 20)( 310 − 222.7) 2 ]
12
12
= .1150x109 mm4 = .1150x10-3 m4
20
17.5 kN
5 kN/m
• Internal Loading
30 kN•m
17.5 kN
4m
4m
20 kN
V (kN)
17.5
17.5
x(m)
40
M(kN•m)
-20
x (m)
-30
The positive maximum bending M+ = 40 kN•m
The maximum negative bending M - = -30 kN•m
21
• The Bending Stress in the Maximum Bending in Tension (M = 40 kN•m)
250 mm
20 mm
97.3 mm NA
B
300 mm
40 kN•m
(σmax)C = 33.8 MPa
222.7 mm
(σmax)T = 77.5 MPa
20 mm
(σ max ) C =
Mc 1 ( 40 kN • m )( 0 . 0973 m )
=
= 33 . 8 MPa (C)
−3
4
I
(. 115 × 10 m )
(σ max )T =
Mc 2 ( 40 kN • m )( 0 . 2227 m )
=
= 77 . 5 MPa (T )
−3
4
I
(. 115 × 10 m )
22
• The Bending Stress in Maximum Bending in Compression (M = -30 kN•m)
250 mm
20 mm
97.3 mm NA
B
300 mm
30 kN•m
(σmax)T = 25.4 MPa
222.7 mm
(σmax)C = 58.1 MPa
20 mm
(σ max )T =
Mc 1 ( 30 kN • m )( 0 . 0973 m )
=
= 25 . 4 MPa ( T)
−3
4
I
(. 115 × 10 m )
(σ max ) C =
Mc 2 ( 30 kN • m )( 0 . 2227 m )
=
= 58 . 1 MPa ( C)
I
(. 115 × 10 −3 m 4 )
23
• The absolute maximum bending stress in tension and also in compression in the beam
250 mm
(σmax)C = 33.8 MPa
20 mm
97.3 mm
300 mm
40 kN•m
NA
222.7 mm
(σmax)T = 77.5 MPa
20 mm
(σmax)T = 25.4 MPa
NA
30 kN•m
(σmax)C = 58.1 MPa
By comparison,
The maximum bending stress in tension;
(σmax)T = 77.5 MPa (T)
The maximum bending stress in compression;
(σmax)C = 58.1 MPa (C)
24
Example 5
The simply supported beam has the cross-sectional area shown. Determine
the absolute maximum bending stress in the beam and draw the stress distribution
over the cross section at this location.
30 kN•m
4m
17.5 kN
5 kN/m
500 mm
300 mm
250 mm
2.25 m
4m
250 mm
25
• Internal Loading
30 kN•m
17.5 kN
5 kN/m
4m
B
A
C
2.25 m
17.5 kN
D
4m
20 kN
V (kN)
17.5
17.5
x(m)
M(kN•m)
9.38
40
-20
x (m)
-30
26
(σmax)T = 9.33 MPa
0.15 m
NA
0.3 m
30 kN•m
0.15 m
(σmax)C = 9.33 MPa
0.25 m
• Section Property
I = ∑ ( I + Ad 2 ) =
1
( 0 . 25 m )( 0 . 3 m ) 3 + 0 = 0 . 0005625 m 4
12
• Bending Stress at A
σ max =
Mc ( 30 kN • m )( 0 . 15 m )
=
= 9 . 33 MPa
I
(. 5625 × 10 −3 m 4 )
27
(σmax)C = 3.84 MPa
0.25 m
NA
0.5 m
40 kN•m
σNA = 0
0.25 m
(σmax)T = 3.84 MPa
0.25 m
• Section Property
I = ∑ ( I + Ad 2 ) =
1
( 0 . 25 m )( 0 . 5 m ) 3 + 0 = 0 . 002604 m 4
12
• Bending Stress at D
σ max =
Mc ( 40 kN • m )( 0 . 25 m )
=
= 3 . 84 MPa
−3
4
I
( 2 . 604 × 10 m )
Maximum bending stress occurs at D
σ max = 9 . 33 MPa
28
30 kN•m
4m
17.5 kN
5 kN/m
B
A
C
2.25 m
17.5 kN
M(kN•m)
9.38
D
4m
20 kN
40
x (m)
-30
0.15 m
0.3 m
(σmax)T = 9.33 MPa
NA
30 kN•m
0.15 m
0.25 m
(σmax)C = 9.33 MPa
Maximum bending stress occurs at A
σ max = 9 . 33 MPa
29
Unsymmetric Bending
• Moment Applied Along Principal Axis
y
y
εmax
z
ε
y
dA
y
z
dF = σdA
C
M
x
c
x
Normal Strain Distribution
(Profile View)
y
σmax
σ
y
Bending Stress Distribution
(Profile View)
c M
x
30
• Moment Arbitrarily Applied
y
Mz = M cos θ
y
z
x
M
z
θ
y
x
My = M sin θ
σ =−
Mz y M yz
+
Iz
Iy
z
x
31
[(σx)max - (σ´x)max]
y
(σx)max
[(σx)max + (σ´x)max]
z
α
(σx)max
N
z
[(σx)max + (σ´x)max]
Mz
A
x
[(σx)max - (σ´x)max]
y
My
(σ´x)max
(σ´x)max
32
• Orientation of the Neutral Axis (α
α)
[(σx)max - (σ´x)max]
α
y
y
[(σx)max + (σ´x)max]
A
F
N
z
A
[(σx)max + (σ´x)max]
B
F
D
E
x
A
A
α
O
H
C
G
z
x
[(σx)max - (σ´x)max]
C
B
G
D
α = tan −1 (
FH
)
HO
33
Example 6
The rectangular cross section shown is subjected to a bending moment of
M = 12 kN•m. Determine the normal stress developed at each corner of the
section, and specify the orientation of the neutral axis.
x
0.2 m
0.2 m
0.1 m
B
E
M = 12 kN•m.
D5
3
4
z
C
0.1 m
y
34
x
0.2 m
0.2 m
0.1 m
B
E
(4/5)(12) = 9.60 kN•m.
M = 12 kN•m.
D5
3
C
4
z
(3/5)(12) = 7.20 kN•m.
0.1 m
N
0.2 m
Iy =
1
( 0 . 4 m )( 0 . 2 m) 3 = 0 . 2667 (10 −3 ) m 4
12
Iz =
1
( 0 . 2 m )( 0 . 4 m ) 3 = 1 . 067 (10 −3 ) m 4
12
• Bending Stress : σ =
y
2.25 MPa
z B
• Section Properties
My
I
M = 7 . 20 kN • m :
7 . 2 (10 3 ) N • m ( 0 . 2 m )
σ =
= ±1 . 35 MPa
1 . 067 (10 −3 )m 4
M = 9 . 60 kN • m :
9 . 60 (10 3 ) N • m ( 0 . 1 m )
σ =
= ±3 . 60 MPa
0 . 2667 (10 −3 ) m 4
4.95 MPa
A
E
2.25 MPa
D
M = 7.2 kN•m M = 9.6 kN•m
σB = ( - 1.35 MPa ) + ( + 3.60 MPa ) = 2.25 MPa (T)
C
4.95 MPa
σC = ( - 1.35 MPa ) + ( - 3.60 MPa ) = -4.95 MPa (C)
σD = ( + 1.35 MPa ) + ( - 3.60 MPa ) = -2.25 MPa (C)
σE = ( + 1.35 MPa ) + ( + 3.60 MPa ) = 4.95 MPa (T) 35
• Orientation of Neutral Axis
4.95 MPa
4.95 MPa
c A
2.25 MPa
D
2.25 MPa
2m
C
z B
N
F
E
D
-2.25 MPa
FD
2
=
; FD = 0 . 625 m
2 . 25 ( 4 . 95 + 2 . 25 )
4.95 MPa
0.2 m
2.25 MPa
B
G
2m
y
F
E
C
-4.95 MPa
62.5 mm
GC
2
=
; GC = 1 . 375 m
4 . 95 ( 4 . 95 + 2 . 25 )
D
400 mm
NA
z
B
G
α
C
137.5 mm
α = tan −1 (
400
) = 79.4 o
137.5 − 62.5)
36
Example 7
A T-beam is subjected to the bending moment of 15 kN•m as shown. Determine
the maximum normal stress in the beam and the orientation of the neutral axis.
z
30 mm
M = 15 kN•m
100 mm
30o
y
z
80 mm
40 mm 80 mm
37
z
• Section Properties
Mz = 15 sin 30o = 7.5 kN•m
B
30 mm
M = 15 kN•m
100 mm
30o
z
C
z
z
y
My = 15 cos 30o = 13 kN•m
80 mm
40 mm 80 mm
z A ( 0 . 05 m )( 0 . 1 m × 0 . 04 m ) + ( 0 . 115 m )( 0 . 03 m × 0 . 2 m )
∑
=
=
= 0 . 0890 m
×
+
×
A
(
0
.
1
m
0
.
04
m
)
(
0
.
03
m
0
.
2
m
)
∑
Iz =
1
1
( 0 . 10 m )( 0 . 04 m ) 3 +
( 0 . 03 m )( 0 . 20 m ) 3 = 20 . 53 (10 −6 ) m 4
12
12
Iy =[
1
( 0 . 04 m )( 0 . 10 m ) 3 + ( 0 . 04 m × 0 . 10 m )( 0 . 089 m − 0 . 05 m) 2 ]
12
1
+ [ ( 0 . 20 m )( 0 . 03 m ) 3 + ( 0 . 2 m × 0 . 03 m )( 0 . 115 m − 0 . 089 m ) 2 ] = 13 . 92 (10 −6 ) m 4 38
12
• Maximum Normal Stress, : σ =
z
My
I
Mz = 7.5 kN•m
B
0.041 m
D
I z = 20 . 53 (10 −6 ) m 4 0.089 m
y
My = 13 kN•m
I y = 13 . 92 (10 −6 ) m 4
C
0.02 m
z
Mz = 7.5 kN•m
0.1 m
0.08 m
My = 13 kN•m
σB =
(+7.5 kN•m) (0.10 m)
20 . 53 (10 −6 ) m 4
+
(+13 kN•m) (0.041 m)
13 . 92 (10 −6 ) m 4
= 74.8 MPa (T)
σC =
(-7.5 kN•m) (0.02 m)
20 . 53 (10 −6 ) m 4
+
(-13 kN•m)(0.089 m)
13 . 92 (10 −6 ) m 4
= -90.4 MPa (C)
σD =
(-7.5 kN•m) (0.10 m)
20 . 53 (10 −6 ) m 4
+
(+13 kN•m) (0.041 m)
13 . 92 (10 −6 ) m 4
= 1.76 MPa (T)
39
• Orientation of Neutral Axis
z
z
σ D = 1 . 76 MPa
Mz = 7.5 kN•m
B
0.1 m
B
α
D
0.041 m
e = 0.00273 m
A
D
0.041 m
y
0.089 m
y
My = 13 kN•m
C
z
σ B = 74 . 8 MPa
0.1 m
N
74.8 MPa
1.76 MPa
0.1 m
B
D
0.2 m
0.2 + e
e
=
,
1 . 76
74 . 8
tan α =
e
e = 2 . 738 (10 −3 ) m
( 0 . 10 m + 0 . 002738 m )
,
0 . 0410 m
α = 68 o
40