lecture-23 (shear force diagram & bending moment diagram)
TRANSCRIPT
![Page 1: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/1.jpg)
Lecture-23 (Shear force
diagram & bending moment
diagram)
Solid Mechanics
(CE 251)
© CE 251 (Solid Mechanics) by Anil Mandariya 1
![Page 2: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/2.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 2
Trigonometric functions:
(1) Line function: y = mx + c
y = dependent variable
x = independent variables
m = slope of line
c = interceptor on y-axis
y = mx
c = 0
m = + ve
y = mx + c
m = +ve
c = +ve
y = mx + c
m = -ve
c = +ve
y = mx + c
m = +ve
c= -ve
y = mx + c
m = -ve
c= -ve
+ x
+y
-x
+y
![Page 3: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/3.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 3
(1) Standard Equations of a Parabola with Vertex at (0, 0)
y = dependent variable
x = independent variables
1.1 y2 = 4ax
Vertex: (0, 0)
Focus: (a, 0)
Directrix: x = a
Symmetric with respect to the x axis
Axis the x axis
a < 0 (open left)
a > 0 (open right)
![Page 4: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/4.jpg)
a < 0 (open down)
a > 0 (open up)
© CE 251 (Solid Mechanics) by Anil Mandariya 4
1.2 x2 = 4ay
Vertex: (0, 0)
Focus: (0, a)
Directrix: y = a
Symmetric with respect to the y axis
Axis the y axis
![Page 5: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/5.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 5
1.3 Convert quadratic equation to standard parabola
a
Aa
bxX
a
c
a
byYwhere
XAY
a
bx
a
c
a
by
a
a
c
a
b
a
bxay
a
c
a
b
a
bx
a
bxay
a
cx
a
bxay
cbxaxy
14;
2
4,
4
)2
()4
(1
4)
2(
)44
(
)(
2
2
2
2
2
2
2
22
2
2
2
22
2
2
a
c
a
by
a
c
a
by
Y
a
bx
a
bx
X
2
2
2
2
2
02
0 and
2
02
0
:are Parabola ofVertex
![Page 6: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/6.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 6
Vertex:
Symmetric with respect to the y axis
Axis the y axis
)2
,2
(2
2
a
c
a
b
a
b
a > 0 (open up)
y
x
0
-x
![Page 7: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/7.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 7
1.4 y = ax3 + bx2 + cx +d
where a ≠ 0, and b, c and d (=0) are constants (can be
zero) and If a > 0 the graph of a cubic looks similar to
one of the following graphs:
yrxqxpx
))()((
IIICase
ypx
3)(
IICase
yx
3
IICase
![Page 8: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/8.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 8
1.4 y = ax3 + bx2 + cx +d
where a ≠ 0, and b, c and d (=0) are constants (can be
zero) and If a < 0 the graph of a cubic looks similar to
one of the following graphs:
yrxqxpx
))()((
IIICase
ypx
3)(
IICase
yx
3
IICase
![Page 9: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/9.jpg)
Sign Conventions SF and BM:
© CE 251 (Solid Mechanics) by Anil Mandariya
9
![Page 10: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/10.jpg)
Sign Conventions of SF & BM
Sagging
Hogging © CE 251 (Solid Mechanics) by Anil
Mandariya 10
![Page 11: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/11.jpg)
Steps to Calculate SF and BM in Beams:
Step-1
Determine the reactions at the supports.
Step-2
Divide the beam span in different segments according
loading and support conditions.
Step-3
Consider left side end of beam as origin.
Step-4
Choose a section corresponding to each segment which
will be at ‘x’ distance from the left end of the beam
(origin). © CE 251 (Solid Mechanics) by Anil Mandariya
11
![Page 12: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/12.jpg)
Step-5
Take a left segment of the beam from section Y-Y.
Step-6
Convert all UDL, UVL in point loads.
Step-7
Choose the sign conventions and write the static
equilibrium equations for the beam segment.
Step-8
Solve static equilibrium equations for the each beam
segment separately according to boundary condition of
segments.
© CE 251 (Solid Mechanics) by Anil Mandariya
12
![Page 13: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/13.jpg)
Step-9
Consider a x and y axis below the beam such that length
of the x-axis will be equal to beam span.
Step-10
Mark the all points according to loading and supports
on the x-axis.
Step-11
Draw the curve separately according to SF and BM
equations for each segments with boundary values.
© CE 251 (Solid Mechanics) by Anil Mandariya
13
![Page 14: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/14.jpg)
Example 1:
Step-1
Determine the reactions at
the supports.
RB RD
HB
ƩFx = 0 = HB= 0 ..........(1)
ƩFy = 0 = -RB - RD + 20 + 40= 0
RD + RB = 60 kN .............(2)
ƩM = 0 = ƩMB = 0
= - (3 x 40) + (5 x RD) + RB x 0
+ HA x 0 + (2.5 x 20)= 0
RD = 70/5 = 14 kN ..............(3)
From eq. (2) & (3)
RB = 46 kN © CE 251 (Solid Mechanics) by Anil Mandariya
14
![Page 15: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/15.jpg)
Step-2
Divide the beam span in different segments according
loading and support conditions.
Segment-1: AB
Segment-2: BC
Segment-3: CD
© CE 251 (Solid Mechanics) by Anil Mandariya
15
![Page 16: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/16.jpg)
Step-3
Consider left side end of beam as origin.
X
Y
0
Step-4
Choose a section y-y’ corresponding to each segment
which will be at ‘x’ distance from the left end of the
beam (origin).
x
y
y’
© CE 251 (Solid Mechanics) by Anil Mandariya
16
![Page 17: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/17.jpg)
Step-5
Take a left segment of the beam from section Y-Y.
0 x
y
y’ Step-6
Convert all UDL, UVL in point loads.
There is no UDL or UVL.
© CE 251 (Solid Mechanics) by Anil Mandariya
17
![Page 18: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/18.jpg)
Step-7 & 8
Choose the sign conventions and write the static
equilibrium equations for the beam segment & solve them.
0 x
y
y’
Segment-1: AB (0 ≤ x ≤ 2.5)
ƩFx = 0 ..........(1)
ƩFy = 0
= 20 + Vx = 0
Vx = - 20 kN (constant).............(2)
ƩM = 0
= (x * 20) + Mx = 0
Mx = -20x (line equation with –ve
slop and zero intercept on y-axis)..(3)
From eq. (2) & (3)
At x = 0
VA = - 20 kN; MA = 0 kN-m
At x = 2.5 m
VB = - 20 kN; MB = - 50 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya 18
![Page 19: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/19.jpg)
y
y’ x
RB = 46 kN
(x - 2.5)
0
Segment-2: BC (2.5 ≤ x ≤ 5.5)
ƩFx = 0 ..........(1)
ƩFy = 0
= 20 + Vx – 46 = 0
Vx = 26 kN (constant value).............(2)
ƩM = 0
= (x * 20) - (x – 2.5)*46 + Mx = 0
Mx = 26x - 115 (line equation with +ve
slope and –ve intercept on y-axis)..........(3)
From eq. (2) & (3)
At x = 2.5 m
VB = 26.0 kN; MB = - 50.0 kN-m
At x = 5.5 m
VC = 26.0 kN; MC = 28.0 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya 19
![Page 20: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/20.jpg)
Segment-3: CD (5.5 ≤ x ≤ 7.5)
X
Y
0
x
y
y’
(x-2.5-3)
RB = 46 kN
ƩFx = 0 ..........(1)
ƩFy = 0
= 20 – 46 + 40 + Vx = 0
Vx = -14.0 kN (constant value).............(2)
ƩM = 0
= (x * 20) - (x – 2.5)*46 + (x – 2.5 – 3)*40
+ Mx = 0
Mx = -14x + 105 (line eq. with –ve slop
and +ve intercept)................................(3)
From eq. (2) & (3)
At x = 5.5 m
VC = -14.0 kN; MC = 28.0 kN-m
At x = 7.5 m
VD = -14.0 kN; MD = 0 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya
20
![Page 21: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/21.jpg)
Step-9
© CE 251 (Solid Mechanics) by Anil Mandariya
21
Consider a x and y axis below the beam such that length
of the x-axis will be equal to beam span.
x
y
![Page 22: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/22.jpg)
Step-10
© CE 251 (Solid Mechanics) by Anil Mandariya 22
Mark the all points according to loading and supports
on the x-axis.
x
y
B A
C D
![Page 23: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/23.jpg)
Step-11
© CE 251 (Solid Mechanics) by Anil Mandariya
23
Draw the curve separately according to SF and BM
equations for each segments with boundary values.
Segment-1: AB (0 ≤ x ≤ 2.5)
Vx = - 20 kN (constant)
Mx = -20x (line equation with
–ve slop and zero intercept on
y-axis)
At x = 0
VA = - 20 kN; MA = 0 kN-m
At x = 2.5 m
VB = - 20 kN; MB = - 50 kN-m
x
y
B A
C D
-20 kN SFD
![Page 24: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/24.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya
24
Segment-1: AB (0 ≤ x ≤ 2.5)
Vx = - 20 kN (constant)
Mx = -20x (line equation with
–ve slop and zero intercept on
y-axis)
At x = 0
VA = - 20 kN; MA = 0 kN-m
At x = 2.5 m
VB = - 20 kN; MB = - 50 kN-m
x
y
B A
C D
-20 kN
SFD
-50 kN-m
0.0 kN-m
![Page 25: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/25.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya
25
Segment-2: BC (2.5 ≤ x ≤ 5.5)
Vx = 26 kN (constant value)
Mx = 26x - 115 (line equation
with +ve slope and –ve
intercept on y-axis)
At x = 2.5 m
VB = 26.0 kN; MB = - 50.0 kN-m
At x = 5.5 m
VC = 26.0 kN; MC = 28.0 kN-m
Mx = 0
26x – 115 = 0
x = 4.42 m
x
y
B A
C D
-20 kN
SFD
-50 kN-m
0.0 kN-m
26 kN
E x = 4.42 m 28 kN-m
![Page 26: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/26.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya
26
Segment-2: CD (2.5 ≤ x ≤ 7.5)
Vx = -14.0 kN (constant value)
Mx = -14x + 105 (line eq. with –
ve slop and +ve intercept)
At x = 5.5 m
VC = -14.0 kN; MC = 28.0 kN-m
At x = 7.5 m
VD = -14.0 kN; MD = 0 kN-m
x
y
B A
C D
-20 kN
SFD
0.0 kN-m
-50 kN-m
26 kN
E x = 4.42 m 28 kN-m
-14 kN
BMD
![Page 27: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/27.jpg)
Example:2
© CE 251 (Solid Mechanics) by Anil Mandariya
27
![Page 28: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/28.jpg)
Step-1
Determine the reactions at the supports.
Couple: arm = 2 m and
force = 8 kN
Mo = 2 x 8 = 16 kN-m
UDL: 2 Nos.
q1 = q2 = q = 4 kN/m on 2 m
Resultant point load = 4 x 2 = 8 kN
Will act at centroid of the UDL
= 2/2
= 1 m from A and B © CE 251 (Solid Mechanics) by Anil Mandariya
28
![Page 29: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/29.jpg)
1.0 m 3.0 m 1.0 m
8.0 kN 8.0 kN
3.0 m
HA
RA RB
ƩFx = 0 = HA= 0 ..........(1)
ƩFy = 0 = -RA - RB + 8.0 + 8.0 = 0
RA + RB = 16.0 kN .............(2)
ƩM = 0 = ƩMB = 0
= RB x 0 + (1 x 8) + - (Mo = 16) + (7 x 8) - (8 x RA) + HA x 0 = 0
RA = 48/8 = 6.0 kN ..............(3)
From eq. (2) & (3)
RB = 10.0 kN
So Reactions are HA= 0; RA = 6.0 kN; RB = 10.0 kN
© CE 251 (Solid Mechanics) by Anil Mandariya
29
![Page 30: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/30.jpg)
Step-2
Divide the beam span in different segments according
loading and support conditions.
C E D
Segment-1: AC
Segment-2: CD
Segment-3: DE
Segment-4: EB © CE 251 (Solid Mechanics) by Anil Mandariya 30
![Page 31: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/31.jpg)
Step-3
Consider left side end of beam as origin.
C E D
Y
X
O
© CE 251 (Solid Mechanics) by Anil Mandariya
31
![Page 32: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/32.jpg)
Step-4
Choose a section y-y’ corresponding to each segment
which will be at ‘x’ distance from the left end of the
beam (origin). 1st choose for segment AC and so on.
C E D
Y
X
O
x
y
y’
© CE 251 (Solid Mechanics) by Anil Mandariya
32
![Page 33: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/33.jpg)
Step-5
Take a left segment of the beam from section y-y'.
C E D
Y
X
O
x
y
y’
0 kN
6.0 kN
Step-6
Convert all UDL, UVL in point loads.
UDL: 1 Nos.; q = 4 kN/m & length = x m
Resultant load of UDL = 4x kN
Will act at centroid of UDL: x/2 m
4x kN
x/2 m
© CE 251 (Solid Mechanics) by Anil Mandariya
33
![Page 34: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/34.jpg)
Step-7 & 8
Choose the sign conventions and write the static
equilibrium equations for the beam segment & solve them.
Segment-1: AC (0 ≤ x ≤ 2.0 m)
4x kN
x/2 m
ƩFx = 0 ............................................(1)
ƩFy = 0 = - 6.0 + 4x + Vx = 0
Vx = (6- 4x) kN ............................................(2)
ƩM = 0
= - 6*x + (x/2 * 4x) + Mx = 0
Mx = - 2x2 + 6x ...........................................(3)
From eq. (2) & (3)
At x = 0 m
VA = 6.0 kN; MA = 0.0 kN-m
At x = 2.0 m
VC = -2.0 kN; MC = 4.0 kN-m © CE 251 (Solid Mechanics) by Anil
Mandariya 34
![Page 35: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/35.jpg)
Segment-2: CD (2.0 ≤ x ≤ 4.0 m)
x
y
y’
6.0 kN x
y
y’
6.0 kN
8.0 kN
1.0 m
ƩFx = 0 ..........................(1)
ƩFy = 0 = - 6.0 + 8 + Vx = 0
Vx = -2.0 kN .............................. (2)
ƩM = 0
= - 6*x + (x – 1) * 8 + Mx = 0
Mx = - 2x + 8 .............................(3)
From eq. (2) & (3)
At x = 2.0 m
VC = -2.0 kN; MC = 4.0 kN-m
At x = 4.0 m
VD = -2.0 kN; MD = 0.0 kN-m
© CE 251 (Solid Mechanics) by Anil Mandariya 35
![Page 36: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/36.jpg)
Segment-3: DE (4.0 ≤ x ≤ 6.0 m)
x
C D
y
y’ 6.0 kN
2.0 m
8.0 kN
1.0 m
ƩFx = 0 .....................................(1)
ƩFy = 0 = - 6.0 + 8 + Vx = 0
Vx = -2.0 kN ....................................... (2)
ƩM = 0
= - 6*x + (x – 1) * 8 – Mo (= 16) + Mx = 0
Mx = - 2x + 24 ....................................(3)
From eq. (2) & (3)
At x = 4.0 m
VD = -2.0 kN; MD = 16.0 kN-m
At x = 6.0 m
VE = -2.0 kN; ME = 12.0 kN-m
© CE 251 (Solid Mechanics) by Anil Mandariya 36
![Page 37: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/37.jpg)
Segment-4: EB (6.0 ≤ x ≤ 8.0 m)
2.0 m
x
C D
y
y’ 6.0 kN
2.0 m
8.0 kN
1.0 m
4.0 kN/m
E
© CE 251 (Solid Mechanics) by Anil Mandariya
37
![Page 38: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/38.jpg)
ƩFx = 0 .....................................(1)
ƩFy = 0 = - 6.0 + 8 + 4*(x – 6) + Vx = 0
Vx = 22 – 4x kN ................................ (2)
ƩM = 0
= - 6*x + (x – 1) * 8 – Mo (= 16) + 4*(x – 6)*(x – 6)/2 + Mx = 0
Mx = - 2x2 + 22x - 48 ....................................(3)
x
C D
y
y’ 6.0 kN
2.0 m
8.0 kN
1.0 m
4.0 kN/m
2.0 m
4*(x – 6)
(x – 6)/2 m
E
From eq. (2) & (3)
At x = 6.0 m
VE = -2.0 kN;
ME = 12.0 kN-m
At x = 8.0 m
VB = -10.0 kN;
ME = 0.0 kN-m
© CE 251 (Solid Mechanics) by Anil Mandariya 38
![Page 39: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/39.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 39
C E D
x
x
y
y
A
A B D E C
C D E B
Consider a x and y axis below the beam such that length of the x-
axis will be equal to beam span and mark the all points according
to loading and supports on the x-axis.
Step-9
& 10
![Page 40: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/40.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 40
Draw the curve separately according to SF and BM equations for
each segments with boundary values.
Step-11
Segment-1: AC (0 ≤ x ≤ 2.0 m)
Vx = (6- 4x) kN (line eq. with –ve slop and +ve intercept on y-axis and SF =0 at
x = 1.5 m).
Mx = - 2x2 + 6x (quadratic eq. of parabola with vertex (1.5, 4.5))
At x = 0 m
VA = 6.0 kN; MA = 0.0 kN-m
At x = 2.0 m
VC = -2.0 kN; MC = 4.0 kN-m
y
x
0
-x
Select this part
of parabola for
BMD according
to boundary
value
![Page 41: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/41.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 41
C E D
x
x
y
y
A
A B D E C
C D E B
6 kN
-2 kN 1.5 m
4 kN-m
2 m
![Page 42: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/42.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 42
Segment-2: CD (2.0 ≤ x ≤ 4.0 m)
Vx = -2.0 kN (constant value, horizontal axis parallel to x-
axis)
Mx = - 2x + 8 (line eq. with –ve slop and +ve intercept on y-
axis)
At x = 2.0 m
VC = -2.0 kN; MC = 4.0 kN-m
At x = 4.0 m
VD = -2.0 kN; MD = 0.0 kN-m
![Page 43: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/43.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 43
6 kN
-2 kN 1.5 m
2 m C E D
x
x
y
y
A
A B D E C
C D E B
4 kN-m
-2 kN
2 m 2 m
![Page 44: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/44.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 44
Segment-3: DE (4.0 ≤ x ≤ 6.0 m)
Vx = -2.0 kN (constant value, horizontal axis parallel to x-
axis)
Mx = - 2x + 24 (line eq. with –ve slop and +ve intercept on
y-axis)
At x = 4.0 m
VC = -2.0 kN; MC = 16.0 kN-m
At x = 6.0 m
VD = -2.0 kN; MD = 12.0 kN-m
![Page 45: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/45.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 45
6 kN
-2 kN 1.5 m
2 m C E D
x
x
y
y
A
A B D E C
C D E B
4 kN-m
-2 kN
16 kN-m
12 kN-m
-2 kN
2 m 2 m 2 m
![Page 46: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/46.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 46
Segment-4: EB (6.0 ≤ x ≤ 8.0 m)
Vx = 22 – 4x kN (line eq. with –ve slop and +ve intercept
on y-axis)
Mx = - 2x2 + 22x - 48 (quadratic eq. of parabola with vertex
(5.5, 36.5))
At x = 6.0 m
VE = -2.0 kN;
ME = 12.0 kN-m
At x = 8.0 m
VB = -10.0 kN;
ME = 0.0 kN-m
y
x
Select this part
of parabola for
BMD according
to boundary
value
![Page 47: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/47.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 47
6 kN
-2 kN 1.5 m
2 m C E D
x
x
y
y
A
A B D E C
C D E B
4 kN-m
-2 kN
16 kN-m
12 kN-m
-2 kN
2 m 2 m 2 m
-10 kN
![Page 48: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/48.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 48
SFD & BMD for different types of beam and
loading condition:
(1) Simply supported beam:
![Page 49: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/49.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 49
![Page 50: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/50.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 50
(2) Cantilever beam:
![Page 51: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/51.jpg)
© CE 251 (Solid Mechanics) by Anil Mandariya 51
![Page 52: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/52.jpg)
BM will be maximum at:
1. At sections where a concentrated load acts and the
shear force changes its sign.
2. At sections where shear force = 0
3. At supports providing vertical reactions.
4. At sections where a couple are applied.
Point of contra flexure:
A point where BM changes its sign.
Anil Mandariya 52
Important points for BM:
![Page 53: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/53.jpg)
The slop of the SFD at any section of the beam is equal
to the intensity of the distributed load over
corresponding section.
Presence of the concentrated loads are indicated by
abrupt changes in SFD. The abrupt change (change in
ordinate) of SF at a section is equal to the
concentrated load at the section.
The slop of the BMD at any section is equal to the SF
at the corresponding section.
At concentrated load point, the slop of BMD is abruptly
chnges.
Anil Mandariya 53
Properties of SFD & BMD:
![Page 54: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/54.jpg)
Total load on a beam between two sections =
difference between the SF at these two sections
The area of SFD between two section = difference
between BM at these two sections.
Anil Mandariya 54
![Page 55: Lecture-23 (Shear Force Diagram & Bending Moment Diagram)](https://reader037.vdocuments.site/reader037/viewer/2022103105/577ccd081a28ab9e788b50a0/html5/thumbnails/55.jpg)
Thanks
© CE 251 (Solid Mechanics) by Anil Mandariya
55