numerical problems based on simply supported beam

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


Forces on left of section 2-2 is RA & 2KN

SF2–2 = 4 – 2 = 2KN (constant value)


SFC = SFD = 2KN


Forces on left of section 3-3 is RA, 2KN, 4KN

SF3–3 = 4 – 2 – 4 =


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


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)


= 4KN (constant value)


2 is RA & 2KN

= 2KN (constant value)


3 is RA, 2KN, 4KN

4 = –2KN (constant value)

MAIT

beam loaded as shown in figure.

...(1)

ΣMA = 0

...(2)

...(3)

total 4 section line, which break the load RA and 2KN (Between


...(4)


...(5)




SFD = SFE = –2KN


Forces on left of section 4-4 is RA

SF4–4 = 4 – 2 – 4 –


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





Consider section 1-1,

Taking moment about section 1-

BM1–1 = 4 X1




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.


1 from point A is X1





-1

MAIT


...(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

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


Taking moment about section 2-2

BM2–2 = 4.X2 – 2.(X2 – 1)

= 2.X2 + 2


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)




BM3–3 = 4.X3 – 2.(X3 – 1) – 4.(X3 – 2)

= –2.X3 + 10



with

X3 = 2 to X3 = 3

At X3 = 2

BMD = 6 ...(12)

At X3 = 3

BME = 4 ...(13)




BM4-4 = 4.X4 – 2.(X4 – 1) – 4.(X4 – 2) - 2.(X4 – 3)

= -4.X4 + 16



with



X4 = 3 to X4 = 4

At X4 = 3; BME = 4 ...(14)

At X4 = 4; BMB = 0 ...(15)


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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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


∑MA = 0

1 X 1 + 4 X 3 + 1 X 5 – RB X 6 = 0

RB = 3KN

From equation (1),

RA = 3KN


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









SF1–1 = 3KN (constant value)


SFA = SFC = 3KN



SF2–2 = 3 – 1 = 2KN (constant value)


SFC = SFD = 2KN


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



: 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







...(4)

& 1KN

1 = 2KN (constant value)


(from point D to the section line i.e. UDL on total distance of (X3 - 2)

MAIT

: Draw the SF and BM diagram for the simply supported beam loaded as shown in

so before solving the

...(5)



SF3–3 = 3 - 1 - 2(X3 - 2) = 6 - 2X3 KN (Equation of straight line)


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.



SF4–4 = 3 – 1 – 4 = – 2KN (constant value)


SFE = SFF = -2KN ...(8)


Forces on left of section 5-5 is RA, 1KN, 4KN, 1KN

SF5-5 = 3 – 1 – 4 – 1 = –3KN (constant value)


SFE = SFB = –3KN ...(9)

Plot the SFD with the help of above shear force values.





BM1–1 = 3.X1


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)




BM2-2 = 3.X2 – 1.(X2 – 1)

= 2.X2 + 1



X2 = 1 to X2 = 2

At X2 = 1

BMC = 3 ...(12)

At X2 = 2

BMD = 5 ...(13)






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


Inclined linear means value of bending moment at b

At X4 = 4

BME = 5

At X4 = 5

BMF = 3




BM5-5 = 3.X5 – 1.(X5 – 1) - 4.(X5

= – 3.X5 + 18


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


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



3 – 2)[(X3 – 2)/2]



= 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)



5 – 3) - 1.(X5 – 5)




: Draw the SF and BM diagram for the simply supported beam loaded as shown in fig

MAIT


...(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.



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


∑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)


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







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)


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.





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)


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





and UDL (from point A to the section line i.e. UDL on total distance

(Equation of straight line)


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

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



SF2-2 = 21.875 – 30 = – 8.125KN (constant value)


SFE = SFD = – 8.125KN ...(6)


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



SFD = SFC = – 8.125KN ...(7)



SF4-4 = 21.875 – 30 -20 = -28.125KN (constant value)


SFC = SFB = –28.125KN ...(8)

Plot the SFD with the help of above shear force values.





BM1-1 = 21.875X1 -20X1(X1/2)



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.



BM2-2 = 21.875X2 – 30(X2 – 0.75)

= –8.125.X2 + 22.5



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





BM3-3 = 21.875X3 – 30(X3 – 0.75) + 30

= –8.125.X2 + 52.5



X3 = 2 to X3 = 3

At X3 = 2

BMD = 36.25 ...(14)

At X3 = 3

BMC = 28.125 ...(15)



BM4-4 = 21.875X4 – 30(X4 – 0.75) + 30 – 20 (X4 – 3)

= –28.125.X4 + 112.5



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


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)


∑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)


Draw the section line, here total 3-section line, which break the load RAV and UDL (Between Point A and B),



No load (Between Point B and C) and UVL (Between Point C and D).

Let






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.


Forces on left of section 2-2 is RAV & 20KN



SFB = SFC = – 1.1KN


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



UVL (Between Point C and D).




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)


& 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

...(6)



= 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.





BM1–1 = 18.9X1 –10X1.X1/2

= 18.9X1 –5 × X12


Parabola means a parabolic curve is formed, value of bending moment at


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




BM2-2 = 18.9X2 – 20(X2 – 1)



At X2 = 2

BMB = 17.8

At X2 = 4

BMC = 15.76

i.e. inclined line 17.8 to 15.76



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



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



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

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)



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


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






SF1-1 = 1000N (constant value)


SFA = SFC = 1000N




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.


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).




MAIT

...(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)




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



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)




BM2-2 = 1000.X2 – 2000.(X2 – 3)

= –1000.X2 + 5000


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



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




3) – 1000


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)


MAIT

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



Draw the section line, here total 2-section line,

which break the point A,D and Point D,B

Let





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


At X1 = 0

SFA = 3000N

At X1 = 3

SFD = 0


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



At X2 = 3

SFA = 0

At X2 = 6



section line,

which break the point A,D and Point D,B



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)


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

at both nearest point of the section is

...(4)

...(5)

...(6)

ormed, value of bending moment at both nearest point of the section is



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.



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)


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



At X2 = 3

BMA = 6000 ...(13)

At X2 = 6

BMD = 0N ...(14)

Plot the BMD with the help of above value.

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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)]



...(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


)/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


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

(Equation of Parabola)


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: A simply supported beam carries distributed load varying uniformly from 125N/m at one end to 250N/m at



= 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,


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)


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


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)


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



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)


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)

numerical problems based on simply supported beam

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