Problems on simply supported beams (udl , uvl and couple)

Problems on Simply supported beam
subjected to UDL, UVL and couples
By
Mrs. Venkata Sushma Chinta

Q5. Draw the shear force and B.M. diagrams for a simply supported beam of length
8 m and carrying a uniformly distributed load of10 kN/m for a distance of 4 m as
shown in Fig.

Q6. A beam AB of length L simply supported at the ends A and B and carrying a uniformly
varying load from zero at end A to ‘q’ per unit length at B. Draw shear force and bending
moment diagram.

Estimate the reactions at supports:
𝛴𝑭𝒚 = 0
RA -
𝟏
𝟐
𝒒𝑳+ RB =0
RA + RB =
𝐪𝑳
𝟐
→ (1)
𝛴𝑴𝑨 = 0
-
1
2
𝑞𝐿 (
2𝐿
3
) + RB * L =0
 RB * L = (
𝐪𝑳 𝟐
𝟑
)
RB =
𝐪𝑳
𝟑 → (2 )
Substitute 2 in (1)
RA =
𝐪𝑳
𝟐
- RB
RA =
𝐪𝑳
𝟐
-
𝐪𝑳
𝟑
RA =
𝐪𝑳
𝟔

𝒒
𝑳
=
𝒒 𝒙
𝒙
=> 𝒒 𝒙=
𝒒.𝒙
𝑳

Free body diagram of section1: 0< 𝑥 < L
when x=0 section coincides with A, when x=L section
coincides with B

when x=0 section coincides with A, when x=L section coincides with B
0< 𝑥< L
Section1-1
Shear force
V=
𝐪𝑳
𝟔
−
𝒒.𝒙 𝟐
𝟐𝑳
Bending Moment
𝑴 =
𝐪𝑳
𝟔
. 𝑥 -
𝒒.𝒙 𝟑
𝟔𝑳
𝑥=0
(at A)
VA=
𝐪𝑳
𝟔
MA=
𝐪𝑳
𝟔
. 0 -
𝒒.𝟎 𝟑
𝟔𝑳
= 0
𝑥= L
(at B)
VB=
𝐪𝑳
𝟔
−
𝒒.𝑳 𝟐
𝟐𝑳
=
−
𝒒.𝑳
𝟑
MB=
𝐪𝑳
𝟔
. 𝐿 -
𝒒.𝑳 𝟑
𝟔𝑳
= 0
𝚺𝐅𝐲 = 𝟎
RA-
1
2
.
𝒒.𝒙
𝑳
𝑥 - V =0
V = RA-
1
2
.
𝒒.𝒙
𝑳
𝑥
V =
𝐪𝑳
𝟔
−
𝒒.𝒙 𝟐
𝟐𝑳
𝚺𝐌 𝟏−𝟏 = 𝟎
-
𝐪𝑳
𝟔
. 𝑥 +
1
2
.
𝒒.𝒙
𝑳
𝑥.
𝑥
3
+ M =0
M=
𝐪𝑳
𝟔
. 𝑥 -
𝒒.𝒙 𝟑
𝟔𝑳

when x=0 section coincides with A, when x=L section coincides with B
0< 𝑥< L
Section1-1
Shear force
V=
𝐪𝑳
𝟔
−
𝒒.𝒙 𝟐
𝟐𝑳
Bending Moment
𝑴 =
𝐪𝑳
𝟔
. 𝑥 -
𝒒.𝒙 𝟑
𝟔𝑳
𝑥=0
(at A)
VA=
𝐪𝑳
𝟔
MA=
𝐪𝑳
𝟔
. 0 -
𝒒.𝟎 𝟑
𝟔𝑳
= 0
𝑥= L
(at B)
VB=
𝐪𝑳
𝟔
−
𝒒.𝑳 𝟐
𝟐𝑳
=
−
𝒒.𝑳
𝟑
MB=
𝐪𝑳
𝟔
. 𝐿 -
𝒒.𝑳 𝟑
𝟔𝑳
= 0
𝑥= L/ 3
(at C)
VC=0
MC=
𝐪𝑳
𝟔
.
𝑳
𝟑
-
𝒒.(
𝑳
𝟑
) 𝟑
𝟔𝑳
=
𝒒𝑳 𝟐
𝟗 𝟑
V =
𝐪𝑳
𝟔
−
𝒒.𝒙 𝟐
𝟐𝑳
M=
𝐪𝑳
𝟔
. 𝑥 -
𝒒.𝒙 𝟑
𝟔𝑳

0< 𝑥< L
Section1-
1
Shear force
V=
𝐪𝑳
𝟔
−
𝒒.𝒙 𝟐
𝟐𝑳
Bending Moment
𝑴 =
𝐪𝑳
𝟔
. 𝑥 -
𝒒.𝒙 𝟑
𝟔𝑳
𝑥=0
(at A)
VA=
𝐪𝑳
𝟔
MA= 0
𝑥= L
(at B)
VB= −
𝒒.𝑳
𝟑
MB=0
𝑥= L/ 3
(at C)
VC=0 MC=
𝒒𝑳 𝟐
𝟗 𝟑

Q6. A simply supported beam of length 5 m carries a uniformly increasing load of 800
N/m run at one end to 1600 N/m run at the other end. Draw the S.F. and B.M. diagrams
for the beam. Also calculate the position and magnitude of maximum bending moment

Estimate the reactions at supports:
𝛴𝑭𝒚 = 0
RA – 4000-2000+ RB =0
RA + RB = 𝟔𝟎𝟎𝟎 → (1)
𝛴𝑴𝑨 = 0
- 4000 ∗ 2.5 – 2000 (
2×5
3
)+ RB * 5 =0
 RB * 5 =16666.7
RB = 𝟑𝟑𝟑𝟑. 𝟑 → (2 )
Substitute 2 in (1)
RA = 𝟔𝟎𝟎𝟎- RB
RA = 𝟔𝟎𝟎𝟎 − 𝟑𝟑𝟑𝟑. 𝟑
RA = 𝟐𝟔𝟔𝟔. 𝟕
Then load on beam due to uniformly distributed load of
800 N/m = 800 x 5 = 4000 N
Load on beam due to triangular loading =
1
2
x 800x 5
=2000 N

Consider any section 1-1at a-distance x from A.
Rate of loading at the section 1-1 =Length CE
=CD +DE
= 800 + 160. 𝑥
C
E
D
D
C
E

2666.7 N
Then load on beam due to uniformly distributed load of 800 N/m = 800 . 𝑥
Load on beam due to triangular loading =
1
2
.160 . 𝑥. 𝑥= 80. 𝑥2
𝚺𝐅𝐲 = 𝟎
RA-800 . 𝑥- 80. 𝑥2 - V =0
V = 2666.7 -800 . 𝑥- 80. 𝑥2
𝚺𝐌 𝟏−𝟏 = 𝟎
- RA . 𝑥 + 800 . 𝑥 .
𝑥
2
+ 80. 𝑥2 .
𝑥
3
+ M =0
M= RA . 𝑥 - 800 . 𝑥 .
𝑥
2
- 80. 𝑥2 .
𝑥
3
M= 2666.7 . 𝑥 - 400 . 𝑥2- 80.
𝑥3
3
Free body diagram of section1: 0< 𝑥 < 5m
when x=0 section coincides with A, when x=5m section coincides with B

2666.7 N
Free body diagram of section1: 0< 𝑥 < 5m
when x=0 section coincides with A, when x=5m section coincides with B
0< 𝑥< 5
Section1-1
Shear force
V = 2666.7 -800 . 𝑥- 80. 𝑥2
Bending Moment
M= 2666.7 . 𝑥 - 400 . 𝑥2- 80.
𝑥3
3
𝑥=0
(at A)
VA=2666.7 −800∗ 0− 80∗ 02 = 2666.7 MA= 2666.7*0 – 400*02- 80*
03
3
= 0
𝑥= 5
(at B)
VB=2666.7 −800 ∗ 5− 80 ∗ 52 =− 3333.3 MB=2666.7*5 – 400*52- 80*
53
3
= 0
𝑥= (at C) VC=0 MC=

2666.7 N
To find maximum bending moment make V=0
V = 2666.7 -800 . 𝑥- 80. 𝑥2
M= 2666.7 . 𝑥 - 400 . 𝑥2- 80.
𝑥3
3

2666.7 N
To find maximum bending moment make V=0
0< 𝑥<5
Section1-1
Shear force
V = 2666.7 -800 . 𝑥- 80. 𝑥2
Bending Moment
M= 2666.7 . 𝑥 - 400 . 𝑥2- 80.
𝑥3
3
𝑥=0
(at A)
VA=2666.7 −800∗ 0− 80∗ 02 =
2667.7
MA= 2666.7*0 – 400*02- 80*
03
3
= 0
𝑥= L
(at B)
VB=2666.7 −800 ∗ 5− 80 ∗ 52 =− 3333.3 MB=2666.7*5 – 400*52- 80*
53
3
= 0
𝑥=2.637 m
(at C)
VC=0 MC=2666.7*2.637 – 400*2.6372- 80*
2.6373
3
=3761.59 N-m
2666.7 -800 . 𝑥- 80. 𝑥2 =0
𝑥= 2.637m

0< 𝑥< 5
Section
1-1
Shear
force
Bending
Moment
𝑥=0
(at A)
VA=2667.7 MA= 0
𝑥= L
(at B)
VB= − 3333.3 MB=0
𝑥=2.637
m
(at C)
VC=0 MC=3761.59
N-m

SHEAR FORCE AND BENDING MOMENT DIAGRAMS FOR BEAMS SUBJECTED TO COUPLES
Q7. A simply supported beam AB of length 6 m is hinged at A and B. It is subjected to a
clockwise couple of24 kN-m at a distance of 2 m from the left end A. Draw the S.F. and B.M.
diagrams.

Problems on simply supported beams (udl , uvl and couple)

Free body diagram of section1: 0< 𝑥 < 2
when x=0 section coincides with A, when x=2 section
coincides with C
0< 𝑥< 2
Section
1-1
Shear force
V = -4
Bending Moment
M= -4𝑥
𝑥=0
(at A)
VA= -4 MA=-4(0)=0
𝑥= 2
(at C)
VC= -4 MC=-4(2)= -8

Free body diagram of section1: 2< 𝑥 < 6
when x=2 section coincides with C, when x=6 section coincides with
B
2< 𝑥< 6
Section
2-2
Shear force
V = -4
Bending Moment
M= 24-4𝑥
𝑥=2
(at C)
VC= -4 MC=24-4 2 = 16
𝑥= 6
(at B)
VB= -4 MB=24-4 6 = 0

0< 𝑥< 2
Section
1-1
Shear force
V = -4
Bending Moment
M= -4𝑥
𝑥=0
(at A)
VA= -4 MA=-4(0)=0
𝑥= 2
(at C)
VC= -4 MC=-4(2)= -8
2< 𝑥< 6
Section
2-2
Shear force
V = -4
Bending Moment
M= 24-4𝑥
𝑥=2
(at C)
VC= -4 MC=24-4 2 = 16
𝑥= 6
(at B)
VB= -4 MB=-24-4 6 = 0

Q8. A beam 10 m long and simply supported at each end, has a uniformly distributed load of
1000 N/m extending from the left end up to the center of the beam. There is also an anti-
clockwise couple of 15000 Nm at a distance of 2.5 m from the right end. Draw the S.F. and
B.M. diagrams

0< 𝑥< 5
Section1
-1
Shear force
V =5250-1000𝑥
Bending Moment
M= 5250𝑥 – 500 𝑥2
𝑥=0
(at A)
VA=5250-1000∗ 0 =
5250
MA=5250∗ 0 – 500 ∗
02 =0
𝑥= 5
(at C)
VC=5250-1000∗ 5 =
250
MC=5250∗ 5– 500 ∗
52 =13750

2.5< 𝑥< 5
Section2-2
Shear force
V =250
Bending Moment
M= 15000- 250𝑥
𝑥=2.5
(at D)
VD=250 MD=15000- 250∗ 2.5 = 14735
𝑥= 5
(at C)
VC=250 MC=15000- 250∗ 5 = 13750

0< 𝑥<
2.5
Section
3-3
Shear force
V =250
Bending Moment
M= -250𝑥
𝑥=2.5
(at D)
VD=250 MD=-250*2.5=-625
𝑥= 0
(at B)
VB=250 MB=-250*0= 0

Section
1-1
Shear
force
Bending
Moment
𝑥=0
(at A)
VA=5250 MA=0
𝑥= 5
(at C)
VC=250 MC=13750
Section2-
2
Shear
force
Bending
Moment
𝑥=2.5
(at D)
VD=250 MD= 14735
𝑥= 5
(at C)
VC=250 MC=13750
0< 𝑥<
2.5
Section
3-3
Shear force
V =250
Bending Moment
𝑥=2.5
(at D)
VD=250 MD=-625
𝑥= 0
(at B)
VB=250 MB= 0

Problems on simply supported beams (udl , uvl and couple)

Problems on simply supported beams (udl , uvl and couple)

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