numerical problems based on simply supported beam
TRANSCRIPT
Simply Supported Beam
Compiled By: RAMAKANT RANA
Numerical Problems Based on Simply supported beam
Q.1: Draw the SF and BM diagram for the simply supported
Solution:
Let reaction at support A and B be, RA and RB First find the support reaction
For that,
ΣV= 0
RA + RB – 2 – 4 - 2 = 0, RA + RB = 8
Taking moment about point A,
2 x 1 + 4 x 2 + 2 x 3 – RB x 4 = 0
RB = 4KN
From equation (1),
RA = 4KN
Calculation for the Shear force Diagram
Draw the section line, here total 4 section line, which break the load RA and 2KN (Between
Point A and C),
2KN and 4KN (Between Point C and D),
4KN and 2KN (Between Point D and E) and
2KN and RB (Between Point E and B)
Consider left portion of the beam
Consider section 1-1
Force on left of section 1-1 is RA
SF1–1 = 4KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFA = SFC = 4KN
Consider section 2-2
Forces on left of section 2-2 is RA & 2KN
SF2–2 = 4 – 2 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFC = SFD = 2KN
Consider section 3-3
Forces on left of section 3-3 is RA, 2KN, 4KN
SF3–3 = 4 – 2 – 4 =
Simply Supported Beam
ramakantrana.blogspot.com
Numerical Problems Based on Simply supported beam
: Draw the SF and BM diagram for the simply supported beam loaded as shown in figure
Let reaction at support A and B be, RA and RB First find the support reaction
2 = 0, RA + RB = 8
x 4 = 0
Calculation for the Shear force Diagram
total 4 section line, which break the load RA and 2KN (Between
2KN and 4KN (Between Point C and D),
4KN and 2KN (Between Point D and E) and
(Between Point E and B)
Consider left portion of the beam
= 4KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
2 is RA & 2KN
= 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
3 is RA, 2KN, 4KN
4 = –2KN (constant value)
MAIT
Page 1
beam loaded as shown in figure.
...(1)
ΣMA = 0
...(2)
...(3)
total 4 section line, which break the load RA and 2KN (Between
Constant value means value of shear force at both nearest point of the section is equal i.e.
...(4)
Constant value means value of shear force at both nearest point of the section is equal i.e.
...(5)
Simply Supported Beam
Compiled By: RAMAKANT RANA
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFD = SFE = –2KN
Consider section 4-4
Forces on left of section 4-4 is RA
SF4–4 = 4 – 2 – 4 –
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFE = SFB = –4KN
Plot the SFD(shear force diagram)
Calculation for the Bending moment Diagram
Let
Distance of section 1-1 from point A is X
Distance of section 2-2 from point A is X
Distance of section 3-3 from point A is X
Distance of section 4-4 from point A is X
Consider left portion of the beam
Consider section 1-1,
Taking moment about section 1-
BM1–1 = 4 X1
Simply Supported Beam
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Constant value means value of shear force at both nearest point of the section is equal i.e.
2KN
A, 2KN, 4KN, 2KN
2 = – 4KN (constant value)
value means value of shear force at both nearest point of the section is equal i.e.
(shear force diagram) with the help of above shear force values.
Calculation for the Bending moment Diagram
1 from point A is X1
2 from point A is X2
3 from point A is X3
4 from point A is X4
Consider left portion of the beam
-1
MAIT
Page 2
Constant value means value of shear force at both nearest point of the section is equal i.e.
...(6)
value means value of shear force at both nearest point of the section is equal i.e.
...(7)
Simply Supported Beam MAIT
Compiled By: RAMAKANT RANA ramakantrana.blogspot.com Page 3
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies
with
X1 = 0 to X1 = 1
At X1 = 0
BMA = 0 ...(8)
At X1 = 1
BMC = 4 ...(9)
i.e. inclined line 0 to 4
Consider section 2-2,
Taking moment about section 2-2
BM2–2 = 4.X2 – 2.(X2 – 1)
= 2.X2 + 2
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies
with
X2 = 1 to X2 = 2
At X2 = 1
BMC = 4 ...(10)
At X2 = 2
BMD = 6 ...(11)
i.e. inclined line 4 to 6
Consider section 3-3,
Taking moment about section 3-3
BM3–3 = 4.X3 – 2.(X3 – 1) – 4.(X3 – 2)
= –2.X3 + 10
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies
with
X3 = 2 to X3 = 3
At X3 = 2
BMD = 6 ...(12)
At X3 = 3
BME = 4 ...(13)
i.e. inclined line 6 to 4
Consider section 4-4,
Taking moment about section 4-4
BM4-4 = 4.X4 – 2.(X4 – 1) – 4.(X4 – 2) - 2.(X4 – 3)
= -4.X4 + 16
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies
with
Simply Supported Beam MAIT
Compiled By: RAMAKANT RANA ramakantrana.blogspot.com Page 4
X4 = 3 to X4 = 4
At X4 = 3; BME = 4 ...(14)
At X4 = 4; BMB = 0 ...(15)
i.e. inclined line 4 to 0
Plot the BMD(bending moment diag) with the help of above bending moment values.
As can be seen in the second diagram of this question.
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Simply Supported Beam
Compiled By: RAMAKANT RANA
Q.2: Draw the SF and BM diagram for the simply supported beam loaded as shown in
Following diagram (this diagram also includes the solution also
numerical try not to see the SFD & BMD).
Solution: Let reaction at support A and B be, RA and R
First find the support reaction.
For finding the support reaction, convert UDL in to point
load and equal to 2 X 2 = 4KN, acting at mid
3m from point A.
For that,
∑V = 0
RA + RB – 1 – 4 – 1 = 0,
RA + RB = 6
Taking moment about point A,
∑MA = 0
1 X 1 + 4 X 3 + 1 X 5 – RB X 6 = 0
RB = 3KN
From equation (1),
RA = 3KN
Calculation for the Shear force Diagram
Draw the section line, here total 5-section line, which break
the load RA and 1KN (Between Point A and C),
1KN and starting of UDL (Between Point C and D), end point
of UDL and 1KN (Between Point E and F) and
1KN and RB (Between Point F and B)
Let
Distance of section 1-1 from point A is X
Distance of section 2-2 from point A is X
Distance of section 3-3 from point A is X
Distance of section 4-4 from point A is X
Distance of section 5-5 from point A is X
Consider left portion of the beam
Consider section 1-1
Force on left of section 1-1 is RA
SF1–1 = 3KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFA = SFC = 3KN
Consider section 2-2
Forces on left of section 2-2 is RA & 1KN
SF2–2 = 3 – 1 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFC = SFD = 2KN
Consider section 3-3
Forces on left of section 3-3 is RA,
1KN and UDL (from point D to the section line i.e. UDL on total distance of (X
Simply Supported Beam
ramakantrana.blogspot.com
: Draw the SF and BM diagram for the simply supported beam loaded as shown in
Following diagram (this diagram also includes the solution also so before solving the
numerical try not to see the SFD & BMD).
and RB
For finding the support reaction, convert UDL in to point
load and equal to 2 X 2 = 4KN, acting at mid point of UDL i.e.
...(1)
X 6 = 0
...(2)
...(3)
Shear force Diagram
section line, which break
and 1KN (Between Point A and C),
1KN and starting of UDL (Between Point C and D), end point
of UDL and 1KN (Between Point E and F) and
1 from point A is X1
2 from point A is X2
3 from point A is X3
4 from point A is X4
5 from point A is X5
Constant value means value of shear force at both nearest point of the section is equal i.e.
...(4)
& 1KN
1 = 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
(from point D to the section line i.e. UDL on total distance of (X3 - 2)
MAIT
Page 5
: Draw the SF and BM diagram for the simply supported beam loaded as shown in
so before solving the
...(5)
Simply Supported Beam MAIT
Compiled By: RAMAKANT RANA ramakantrana.blogspot.com Page 6
SF3–3 = 3 - 1 - 2(X3 - 2) = 6 - 2X3 KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of S.F. at both nearest point of the section is varies with X3 = 2 to
X3 = 4
At X3 = 2
SFD = 2 ...(6)
At X3 = 4
SFE = –2 ...(7)
i.e. inclined line 2 to -2
Since here shear force changes the sign so at any point shear force will be zero and at that point bending moment
is maximum.
For finding the position of zero shear force equate the shear force equation to zero, i.e.
6 – 2X3 = 0; X3 = 3m,
i.e. at 3m from point A bending moment is maximum.
Consider section 4-4
Forces on left of section 4-4 is RA, 1KN, 4KN
SF4–4 = 3 – 1 – 4 = – 2KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFE = SFF = -2KN ...(8)
Consider section 5-5
Forces on left of section 5-5 is RA, 1KN, 4KN, 1KN
SF5-5 = 3 – 1 – 4 – 1 = –3KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFE = SFB = –3KN ...(9)
Plot the SFD with the help of above shear force values.
Calculation for the Bending moment Diagram
Consider left portion of the beam
Consider section 1-1,
Taking moment about section 1-1
BM1–1 = 3.X1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X1 = 0 to X1 = 1
At X1 = 0
BMA = 0 ...(10)
At X1 = 1
BMC = 3 ...(11)
i.e. inclined line 0 to 3
Consider section 2-2,
Taking moment about section 2-2
BM2-2 = 3.X2 – 1.(X2 – 1)
= 2.X2 + 1
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X2 = 1 to X2 = 2
At X2 = 1
BMC = 3 ...(12)
At X2 = 2
BMD = 5 ...(13)
i.e. inclined line 3 to 5
Simply Supported Beam
Compiled By: RAMAKANT RANA
Consider section 3-3,
Taking moment about section 3-3
BM3-3 = 3.X3 – 1.(X3 – 1) – 2.(X3
= 2.X3 + 1 – (X3 – 2)2
It is Equation of Parabola (Y = mX2 + C),
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is
varies with X3 = 2 to X3 = 4
At X3 = 2
BMD = 5
At X3 = 4
BME = 5
But B.M. is maximum at X3 = 3, which lies between X
So we also find the value of BM at X3 = 3
At X3 = 3
BMmax = 6
i.e. curve makes with in 5 to 6 to 5 region.
Consider section 4-4, taking moment about section 4
BM4-4 = 3.X4 – 1.(X4 – 1) – 4.(X4
= –2.X4 + 13
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at b
At X4 = 4
BME = 5
At X4 = 5
BMF = 3
i.e. inclined line 5 to 3
Consider section 5-5,
Taking moment about section 5-5
BM5-5 = 3.X5 – 1.(X5 – 1) - 4.(X5
= – 3.X5 + 18
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with X
At X5 = 5
BME = 3
At X4 = 6
BMF = 0
i.e. inclined line 3 to 0
Plot the BMD with the help of above bending moment values.
Q.3: Draw the SF and BM diagram for the simply supported beam loaded as shown in fig
Simply Supported Beam
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3 – 2)[(X3 – 2)/2]
It is Equation of Parabola (Y = mX2 + C),
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is
= 3, which lies between X3 = 2 to X3 = 4
= 3
ve makes with in 5 to 6 to 5 region.
4, taking moment about section 4-4
4 – 3)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with X
5 – 3) - 1.(X5 – 5)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with X
Plot the BMD with the help of above bending moment values.
: Draw the SF and BM diagram for the simply supported beam loaded as shown in fig
MAIT
Page 7
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is
...(14)
...(15)
...(16)
oth nearest point of the section is varies with X4 = 4 to X4 = 5
...(17)
...(18)
Inclined linear means value of bending moment at both nearest point of the section is varies with X5 = 5 to X5 = 6
...(19)
...(20)
: Draw the SF and BM diagram for the simply supported beam loaded as shown in figure.
Simply Supported Beam
Compiled By: RAMAKANT RANA
Solution: Let reaction at support A and B be, RA and
find the support reaction.
For finding the support reaction, convert UDL in to
point load and equal to 20 X 1.5 = 30KN, acting at
mid point of UDL i.e. 0.75m from point A.
For that,
∑V = 0
RA + RB - 30 - 20 = 0, RA + RB = 50
Taking moment about point A,
∑MA = 0
30 X 0.75 + 30 + 20 X 3 – RB X 4 = 0
RB = 28.125 KN ...(2)
From equation (1), RA = 21.875KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 4-section line,
which break the load RA and UDL (Between Point A and E),
30KN/m and 20KN (Between Point E and D),
30KN/M and 20KN (Between Point D and C) and
20KN and RB (Between Point C and B)
Let
Distance of section 1-1 from point A is X
Distance of section 2-2 from point A is X
Distance of section 3-3 from point A is X
Distance of section 4-4 from point A is X
Consider left portion of the beam
Consider section 1-1
Force on left of section 1-1 is RA and UDL (from point A to the section line i.e. UDL on total distance
SF1–1 = 21.875 -20X1 KN (Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of shear force at both nearest point of the section is varies with X
At X1 = 0
SFA = 21.875
At X1 = 1.5
SFE = –8.125
i.e. inclined line 21.875 to – 8.125
Since here shear force changes the sign so at any point shear force will be zero and at that point
is maximum.
For finding the position of zero shear force equate the shear
21.875 -20X1 = 0; X1 = 1.09375m, i.e. at 1.09375m from point A bending moment is maximum.
Consider section 2-2
Forces on left of section 2-2 is RA & 30KN
Simply Supported Beam
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and RB First
reaction, convert UDL in to
point load and equal to 20 X 1.5 = 30KN, acting at
0.75m from point A.
...(1)
Calculation for the Shear force Diagram
section line,
etween Point A and E),
30KN/m and 20KN (Between Point E and D),
30KN/M and 20KN (Between Point D and C) and
1 from point A is X1
2 from point A is X2
3 from point A is X3
4 from point A is X4
and UDL (from point A to the section line i.e. UDL on total distance
(Equation of straight line)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of shear force at both nearest point of the section is varies with X1 = 0
8.125
Since here shear force changes the sign so at any point shear force will be zero and at that point
For finding the position of zero shear force equate the shear force equation to zero, i.e.
= 1.09375m, i.e. at 1.09375m from point A bending moment is maximum.
2 is RA & 30KN
MAIT
Page 8
and UDL (from point A to the section line i.e. UDL on total distance of X1
= 0 to X1 = 1.5
...(4)
...(5)
Since here shear force changes the sign so at any point shear force will be zero and at that point bending moment
Simply Supported Beam MAIT
Compiled By: RAMAKANT RANA ramakantrana.blogspot.com Page 9
SF2-2 = 21.875 – 30 = – 8.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFE = SFD = – 8.125KN ...(6)
Consider section 3-3
Forces on left of section 3-3 is RA & 30KN, since forces are equal that of section 2-2, so the value of shear force at
section 3-3 will be equal that of section 2-2
SF3-3 = 21.875 – 30 = – 8.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFD = SFC = – 8.125KN ...(7)
Consider section 4-4
Forces on left of section 4-4 is RA, 30KN, 20KN
SF4-4 = 21.875 – 30 -20 = -28.125KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFC = SFB = –28.125KN ...(8)
Plot the SFD with the help of above shear force values.
Calculation for the Bending moment Diagram
Consider left portion of the beam
Consider section 1-1,
Taking moment about section 1-1
BM1-1 = 21.875X1 -20X1(X1/2)
It is Equation of Parabola (Y = mX2 + C),
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is
varies with X1 = 0 to X1 = 1.5
At X1 = 0
BMA = 0 ...(9)
At X1 = 1.5
BMC = 10.3125 ...(10)
But B.M. is maximum at X1 = 1.09, which lies between X1 = 0 to X1 = 1.5
So we also find the value of BM at X1 = 1.09
At X1 = 1.09
BMmax = 11.8 ...(11)
i.e. curve makes with in 0 to 11.8 to 10.3125 region.
Consider section 2-2,
Taking moment about section 2-2
BM2-2 = 21.875X2 – 30(X2 – 0.75)
= –8.125.X2 + 22.5
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X2 = 1.5 to X2 = 2
At X2 = 1.5
BME = 10.3125 ...(12)
At X2 = 2
BMD = 6.25 ...(13)
i.e. inclined line 10.3125 to 6.25
Consider section 3-3,
Taking moment about section 3-3
Simply Supported Beam MAIT
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BM3-3 = 21.875X3 – 30(X3 – 0.75) + 30
= –8.125.X2 + 52.5
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X3 = 2 to X3 = 3
At X3 = 2
BMD = 36.25 ...(14)
At X3 = 3
BMC = 28.125 ...(15)
Consider section 4-4,
Taking moment about section 4-4
BM4-4 = 21.875X4 – 30(X4 – 0.75) + 30 – 20 (X4 – 3)
= –28.125.X4 + 112.5
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
X4 = 3 to X4 = 4
At X4 = 3
BMC = 28.125 ...(16)
At X4 = 4
BMB = 0 ...(17)
i.e. inclined line 28.125 to 0
Plot the BMD with the help of above bending moment values.
Q.4: Determine the SF and BM diagrams for the simply supported beam shown in figure No. 4. Also find the
maximum bending moment.
Solution:
Since hinged at point A and D, suppose reaction at support A and D be, RAH, RAV and RDH, RDV first find the support
reaction.
For finding the support reaction, convert UDL and UVL in to point load and,
Point load of UDL equal to
10 X 2 = 20KN,
Acting at mid point of UDL
i.e. 1m from point A.
Point load of UVL equal to
1/2 X 20 X 2 = 20KN, acting at a distance 1/3 of total distance i.e. 1/3m from point D.
For that,
∑V = 0
RAV + RDV – 20 – 20 = 0, RA + RB = 40 ...(1)
Taking moment about point A,
∑MA = 0
20 X 1 + 20 X 5.33 – RDV X 6 = 0
RDV = 21.1 KN ...(2)
From equation (1), RAV = 18.9KN ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 3-section line, which break the load RAV and UDL (Between Point A and B),
Simply Supported Beam
Compiled By: RAMAKANT RANA
No load (Between Point B and C) and UVL (Between Point C and D).
Let
Distance of section 1-1 from point A is X
Distance of section 2-2 from point A is X
Distance of section 3-3 from point A is X
Consider left portion of the beam
Consider section 1-1
Force on left of section 1-1 is RAV and UDL
(from point A to the section line i.e. UDL on
total distance of X1
SF1-1 = 18.9 -10X1 KN
(Equation of straight line)
It is Equation of straight line (Y = mX + C),
inclined linear.
Inclined linear means value of shear force at
both nearest point of the section is varies
with X1 = 0 to X1 = 2
At X = 0
SFA = 18.9 ...(4)
At X1 = 2
SFB = –1.1 ...(5)
i.e. inclined line 18.9 to - 1.1
Since here shear force changes the sign so at
any point shear force will be zero and at that
point bending moment is maximum.
For finding the position of zero shear force
equate the shear force equation to zero, i.e.
18.9 –10X1 = 0; X1 = 1.89m,
i.e. at 1.89m from point A bending moment is maximum.
Consider section 2-2
Forces on left of section 2-2 is RAV & 20KN
SF2-2 = 18.9 – 20 = – 1.1KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFB = SFC = – 1.1KN
Consider section 3-3
Forces on left of section 3-3 is RAV & 20KN and UVL of 20KN/m over (X
First calculate the total load of UVL over length of (X
Consider triangle CDE and CGF
DE/GF = CD/CG
Since DE = 20
20/GF = 2/(X3 – 4)
GF = 10(X3 – 4)
Now load of triangle
CGF = 1/2 X CG X GF = 1/2 X (X3 – 4) X 10(X
See figure
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UVL (Between Point C and D).
1 from point A is X1
2 from point A is X2
3 from point A is X3
and UDL
(from point A to the section line i.e. UDL on
It is Equation of straight line (Y = mX + C),
Inclined linear means value of shear force at
both nearest point of the section is varies
...(4)
...(5)
Since here shear force changes the sign so at
point shear force will be zero and at that
For finding the position of zero shear force
equate the shear force equation to zero, i.e.
at 1.89m from point A bending moment is maximum.
& 20KN
1.1KN (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
& 20KN and UVL of 20KN/m over (X3 – 4) m length,
First calculate the total load of UVL over length of (X3 – 4)
4) X 10(X3 – 4)
MAIT
Page 11
...(6)
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= 5(X3 – 4)2, at a distance of (X3 – 4)/3 from G
SF3-3 = 18.9 – 20 –5(X3 - 4)2
= – 1.1 –5(X3
Parabola means a parabolic curve is formed, value of
moment at both nearest point of the section is varies with X
to X3 = 6
At X3 = 4
SFC = –1.1KN
SFD = –21.1KN
Plot the SFD(shear force diagram)
above shear force values.
Calculation for the Bending moment Diagram
Consider left portion of the beam
Consider section 1-1,
Taking moment about section 1-1
BM1–1 = 18.9X1 –10X1.X1/2
= 18.9X1 –5 × X12
It is Equation of Parabola (Y = mX2 + C),
Parabola means a parabolic curve is formed, value of bending moment at
varies with X1 = 0 to X1 = 2
At X1 = 0
BMA = 0
At X1 = 2
BMB = 17.8
But B.M. is maximum at X1 = 1.89, which lies between X
So we also find the value of BM at X
At X1 = 1.89
BMmax = 17.86
i.e. curve makes with in 0 to 17.86 to 17.8 region.
Consider section 2-2,
Taking moment about section 2-2
BM2-2 = 18.9X2 – 20(X2 – 1)
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of bending moment at both nearest point of the section is varies with
At X2 = 2
BMB = 17.8
At X2 = 4
BMC = 15.76
i.e. inclined line 17.8 to 15.76
Consider section 3-3,
Taking moment about section 3-3
BM3-3 = 18.9X3 – 20(X3 – 1) – 5(X3 – 4) 2
. (X
It is cubic Equation which varies with X3
At X3 = 4
BMC = 15.76
Simply Supported Beam
ramakantrana.blogspot.com
4)/3 from G ...(7)
3 – 4) 2
(Parabola)
Parabola means a parabolic curve is formed, value of bending
section is varies with X3 = 4
...(8)
...(9)
(shear force diagram) with the help of
Diagram
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
, which lies between X1 = 0 to X1 = 2
So we also find the value of BM at X1 = 1.89
i.e. curve makes with in 0 to 17.86 to 17.8 region.
It is Equation of straight line (Y = mX + C), inclined linear.
linear means value of bending moment at both nearest point of the section is varies with
. (X3 – 4)/3
3 = 4 to X3 = 6
MAIT
Page 12
both nearest point of the section is
...(10)
...(11)
...(12)
linear means value of bending moment at both nearest point of the section is varies with X2 = 2 to X2 = 4
...(13)
...(14)
...(15)
Simply Supported Beam
Compiled By: RAMAKANT RANA
At X3 = 6
BMD = 0
Plot the BMD with the help of above bending
Q.5: Draw the SF and BM diagrams for a simply supported beam 5m long carrying a load of 200N through a
bracket welded to the beam loaded as shown in following figure.
Solution:
The diagram is of force couple system, let us apply at C
2000N. Now the vertically upward load of 2000N at C and vertically downward load of 2000N
anticlockwise couple at C whose moment is 2000 X 0.5 = 1000Nm
And we are left with a vertically downward load of 2000N acting at C.
Let reaction at support A and B be, RA and R
Taking moment about point A;
2000 X 3 – 1000 – RB X 5 = 0
RB = 1000N
RV = 0, RA + RB – 2000 = 0
RA = 1000N
Calculation for the Shear force Diagram
Draw the section line, here total 2 section line, which break the load
RA and 2000N (Between Point A and C),
2000N and RB (Between Point C and B).
Let
Distance of section 1-1 from point A is X
Distance of section 2-2 from point A is X
Consider left portion of the beam
Consider section 1-1
Force on left of section 1-1 is RA
SF1-1 = 1000N (constant value)
Constant value means value of shear force at both nearest point of the section is equal i.e.
SFA = SFC = 1000N
Simply Supported Beam
ramakantrana.blogspot.com
Plot the BMD with the help of above bending moment values.
Draw the SF and BM diagrams for a simply supported beam 5m long carrying a load of 200N through a
bracket welded to the beam loaded as shown in following figure.
The diagram is of force couple system, let us apply at C two equal and opposite forces each equal and
2000N. Now the vertically upward load of 2000N at C and vertically downward load of 2000N at D forms an
anticlockwise couple at C whose moment is 2000 X 0.5 = 1000Nm
downward load of 2000N acting at C.
and RB first find the support reaction.
Calculation for the Shear force Diagram
Draw the section line, here total 2 section line, which break the load
and 2000N (Between Point A and C),
(Between Point C and B).
1 from point A is X1
2 from point A is X2
Constant value means value of shear force at both nearest point of the section is equal i.e.
MAIT
Page 13
...(16)
Draw the SF and BM diagrams for a simply supported beam 5m long carrying a load of 200N through a
two equal and opposite forces each equal and parallel to
at D forms an
...(1)
...(2)
...(3)
Simply Supported Beam
Compiled By: RAMAKANT RANA
Consider section 2-2
Forces on left of section 2-2 is RA &
2000N
SF2-2 = 1000 – 2000 = –1000
(constant value)
Constant value means value of shear
force at both nearest point of the
section is equal i.e.
SFC = SFB = –1000N ...(4)
Plot the SFD with the help of above
shear force values.
Calculation for the bending moment
Diagram
Consider section 1-1,
Taking moment about section 1-1
BM1-1 = 1000.X1
It is Equation of straight line
(Y = mX + C), inclined linear.
Inclined linear means value of
bending moment at both nearest
point of the section is varies with X1
= 0 to X1 = 3
At X1 = 0
BMA = 0 ...(5)
At X1 = 3
BMC = 3000 ...(6)
i.e. inclined line 0 to 3000
Consider section 2-2,
Taking moment about section 2-2
BM2-2 = 1000.X2 – 2000.(X2 – 3)
= –1000.X2 + 5000
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
At X2 = 3
BMC = 2000
At X2 = 5
BMB = 0
i.e. inclined line 2000 to 0
Plot the BMD with the help of above bending moment values.
Q.6: A simply supported beam 6m long is subjected to a triangular load of 6000N as shown
the S.F. and B.M. diagrams for the beam.
Solution:
Let
Suppose reaction at support A and B be, R
Due to symmetry,
RA = RB = 6000/2 = 3000N
Calculation for the Shear force Diagram
Simply Supported Beam
ramakantrana.blogspot.com
3) – 1000
It is Equation of straight line (Y = mX + C), inclined linear.
Inclined linear means value of Bending moment at both nearest point of the section is varies with
Plot the BMD with the help of above bending moment values. The SFD and BMD is shown in figure
A simply supported beam 6m long is subjected to a triangular load of 6000N as shown in fig
the S.F. and B.M. diagrams for the beam.
Suppose reaction at support A and B be, RA and RB first find the support reaction.
...(1)
Calculation for the Shear force Diagram
MAIT
Page 14
Inclined linear means value of Bending moment at both nearest point of the section is varies with X2 = 3 to X2 = 5
...(7)
...(8)
D and BMD is shown in figure No.5
in figure below. Draw
Simply Supported Beam
Compiled By: RAMAKANT RANA
Draw the section line, here total 2-section line,
which break the point A,D and Point D,B
Let
Distance of section 1-1 from point A is X
Distance of section 2-2 from point A is X
Consider left portion of the beam
Consider section 1-1
Forces on left of section 1-1 is RA and UVL of
6000N/m over X1 m length,
Since, Total load = 6000 = 1/2 X AB X CD
1/2 X 6 X CD = 6000, CD = 2000N
First calculate the total load of UVL over length of
X1
Consider triangle ADC and AFE
DC/EF = AD/AF
Since DC = 2000
2000/EF = 3/X1
EF = (2000X1)/3
Now load of triangle AEF = 1/2 X EF × AF
= (1/2 X 2000X
=(1000.X12)/3
distance of X1/3 from F
...(3)
SF1-1 = 3000 – (1000X12)/3 (Parabola)
Parabola means a parabolic curve is formed, value of bending moment
varies with X1 = 0 to X1 = 3
At X1 = 0
SFA = 3000N
At X1 = 3
SFD = 0
Consider section 2-2
Forces on left of section 2-2 is RA and UVL of 2000N/m(At CD) and UVL over (X
First calculate the total load of UVL over length of (X
Consider triangle CDB and BGH
DC/GH = DB/BG
Since DC = 2000
2000/GH = 3/(6 - X2)
GH = 2000(6-X2)/3
Now load of triangle BGH = 1/2 X GH X BG
= [1/2 X 2000(6-X2)/3] X (6
= 1000(6 – X2)2/3, at a distance of X
Load of CDB = 1/2 X 3 X 2000 = 3000
Now load of CDGH = load of CDB - load of BGH
= 3000 – 1000(6 – X2)
SF2-2 = 3000 – 3000 – [3000 – 1000(6
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
varies with X2 = 3 to X2 = 6
At X2 = 3
SFA = 0
At X2 = 6
Simply Supported Beam
ramakantrana.blogspot.com
section line,
which break the point A,D and Point D,B
1 from point A is X1
2 from point A is X2
and UVL of
Total load = 6000 = 1/2 X AB X CD
...(2)
First calculate the total load of UVL over length of
Now load of triangle AEF = 1/2 X EF × AF
= (1/2 X 2000X1)/3 × (X1)
)/3 a
/3 from F
(1000X12)/3 (Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
and UVL of 2000N/m(At CD) and UVL over (X2 – 3) m length,
First calculate the total load of UVL over length of (X2 – 3)
Now load of triangle BGH = 1/2 X GH X BG
)/3] X (6-X2)
)2/3, at a distance of X1/3 from F
load of BGH
)2/3 ...(7)
1000(6 – X2) 2
/3] (Parabola)
ormed, value of bending moment at both nearest point of the
...(8)
MAIT
Page 15
at both nearest point of the section is
...(4)
...(5)
...(6)
ormed, value of bending moment at both nearest point of the section is
Simply Supported Beam
Compiled By: RAMAKANT RANA
SFD = –3000N
Plot the SFD with the help of above value as
shown in fig.
Since SF change its sign at X2 = 3, that means at a
distance of 3m from point A bending moment is
maximum.
Calculation for the Bending moment Diagram
Consider section 1-1
BM1-1 = 3000X1 – [(1000X12)/3]
(Cubic)
Cubic means a parabolic curve is formed, value of
bending moment at both nearest point of the
section is varies with X1 = 0 to X1 = 3
At X1 = 0
BMA = 0 ...(10)
At X1 = 3
BMD = 6000 ...(11)
Consider section 2-2
Point of CG of any trapezium is = h/3[(b + 2a)/(a +
b)]
i.e. Distance of C.G of the trapezium CDGH is
given by,
= 1/3 X DG X [(GH + 2CD)/(GH + CD)]
= 1/3.(X2-3).{[2000(6-X2)/3] + 2 X 2000)}/{[ 2000(6
= {(X2 – 3)(12 – X2)}/{3(9 – X2)}
BM2-2 = 3000X2-3000(X2-2)-[3000-1000(6
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
varies with X2 = 3 to X2 = 6
At X2 = 3
BMA = 6000 ...(13)
At X2 = 6
BMD = 0N ...(14)
Plot the BMD with the help of above value.
______________________________________________________________________________
Q.7: A simply supported beam carries distributed load varying uniformly from 125N/m at one end to 250N/m at
the other. Draw the SF and BM diagram and determine the maximum B.M.
Solution:
Total load = Area of the load diagram ABEC
= Rectangle ABED + Triangle DEC
= (AB X BE) + (1/2 X DE X DC) = (9 X 125) +
[1/2 X 9 X (250-125)]
Simply Supported Beam
ramakantrana.blogspot.com
...(9)
Plot the SFD with the help of above value as
3, that means at a
distance of 3m from point A bending moment is
Calculation for the Bending moment Diagram
)/3] X1/3
Cubic means a parabolic curve is formed, value of
nearest point of the
Point of CG of any trapezium is = h/3[(b + 2a)/(a +
i.e. Distance of C.G of the trapezium CDGH is
X DG X [(GH + 2CD)/(GH + CD)]
X2)/3] + 2 X 2000)}/{[ 2000(6-X2)/3]+[ 2000]}
X2)} ...(12)
1000(6 – X2)2/3]{+ (X2 – 3)(12 – X2)}/{3(9 – X2)} (Equation of
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the
Plot the BMD with the help of above value.
______________________________________________________________________
: A simply supported beam carries distributed load varying uniformly from 125N/m at one end to 250N/m at
the other. Draw the SF and BM diagram and determine the maximum B.M.
Total load = Area of the load diagram ABEC
+ (1/2 X DE X DC) = (9 X 125) +
MAIT
Page 16
(Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is
______________________________________________________________________
: A simply supported beam carries distributed load varying uniformly from 125N/m at one end to 250N/m at
Simply Supported Beam MAIT
Compiled By: RAMAKANT RANA ramakantrana.blogspot.com Page 17
= 1125N + 562.5N ...(1)
Centroid of the load of 1125N (rectangular load) is at a distance of 9/2 = 4.5m from AD and the centroid of the load
of 562.5N (Triangular load) is at a distance of 1/3 X DE = 1/3 X 9 = 3m from point A.
Let support reaction at A and B be RA and RB. For finding the support reaction,
Taking moment about point A,
1125 X 4.5 + 562.5 X 3 - RB X 9 = 0
RB = 750N ...(2)
Now, RV = 0
RA + RB = 1125 + 562.5 = 1687.5
RA = 937.5N ...(3)
Calculation for the Shear force Diagram
Draw the section line, here total 1-section line, which break the point A and B
Let
Distance of section 1-1 from point B is X
Consider right portion of the beam
Consider section 1-1
Forces on right of section 1-1 is RB and Load of PBEF and Load of EFH
SF1-1 = RB - load on the area PBEF - load on the area EFH
= RB - X.125 - 1/2.X.FH
In the equiangular triangles DEC and FEH
DC/DE = FH/FE
or,
125/9 = FH/X
FH = 125X/9
S.F. between B and A = 750 - 125X - 125X2/18 (Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is
varies with X = 0 to X = 9
At X = 0
SFB = 750N ...(4)
At X = 9
SFA = –937.5N ...(5)
Since the value of SF changes its sign, which is between the point A and B we get max. BM For the point of zero
shear,
750 – 125X – 125X2/18 = 0
On solving we get, X = 4.75m
That is BM is max. at X = 4.75 from point B
Calculation for the Bending moment Diagram
Consider section 1-1
BM1–1 = 750X – PB.BE.X/2 – 1/2.FE.FH.1/3.FE
= 750X – X.125.(X/2) – 1/2.X.(125X/9)(X/3)
= 750x – 125x2/2 – 125X2/54 (Equation of Parabola)
Parabola means a parabolic curve is formed, value of bending moment at both nearest point of the section is
varies with X = 0 to X = 9
At X = 0
BMB = 0 ...(6)
At X = 4.75
BMmax = 1904N-m ...(7)
At X = 9
BMA = 0 ...(8)