1. The beam is simply supported over 8m with UDL of 10kN/m over the first 4m from the left support. 2. The left support reaction is 40kN and right support is 20kN. 3. The shear force diagram shows V reducing linearly from 40kN to 0kN over 0-4m and then constant at -20kN from 4-8m. 4. The bending moment diagram shows M increasing parabolically from 0 to 80kNm over 0-4m and then linearly decreasing from 80kNm to 0 over 4-8m.

1. Problems on Simply supported Beam
Mrs. Venkata Sushma Chinta

2. Q1. A simply supported beam of length 6 m, carries point load of 3 kN and 6 kN at distances
of 2 m and 4 m from the left end. Draw the shear force and bending moment diagrams for
the beam.

3. Estimate the reactions at supports:
𝛴𝑭𝒚 = 0
RA -3 - 6 + RB =0
RA + RB = 9 → (1)
𝛴𝑴𝑨 = 0
-3 * 2 – 6 * 4 + RB * 6 =0
 RB * 6 = 3 * 2 + 6 * 4
RB = 5 kN → (2 )
Substitute 2 in (1)
RA = 9 - RB
RA = 9 – 5
RA = 4 kN

5. Free body diagram of section1: 0< 𝑥< 2 m
when x=0 section coincides with A, when x=a section coincides with C
0< 𝑥<2
Section1-1
Shear force
(𝑘𝑁)
V= 4
Bending
Moment (kN-m)
M= 4 * 𝑥
𝑥=0 (at A) VA= 4 MA= 4 ∗ 0= 0
𝑥= 2 m (at C) VC=4 MC=4 * 2= 8
𝚺𝐅𝐲 = 𝟎
RA-V =0
V =RA = 4 kN
𝚺𝐌 𝟏−𝟏 = 𝟎
- RA * 𝑥 + M =0
M= RA * 𝑥 = 4. 𝑥

6. Free body diagram of section2:
when x=2 m section coincides with C, when x=4 m section coincides with D
2< 𝑥< 4
Section2-2
Shear force
V= 1 kN
Bending Moment(kN-m)
M= 𝑥 + 6
𝑥=2 (at C) VC= 1 MC= 2+ 6= 8
𝑥= 4 (at D) VD= 1 MD=4 + 6 = 10
𝚺𝐅𝐲 = 𝟎 RA- 3 -V =0
V = RA- 3
V = 4- 3 = 1kN
𝚺𝐌 𝟐−𝟐 = 𝟎
-RA . 𝑥 +3 * (𝑥 -2) + M =0
M= 4 . 𝑥 – 3 * (𝑥 -2)
M= 4 . 𝑥 – 3 * 𝑥 + 6
M= 𝑥 + 6
2 m < 𝑥< 4 m

7. 𝚺𝐅𝐲 = 𝟎
Free body diagram of section 3: 4 m < 𝑥< 6 m
when x=4 m section coincides with D, when
x=6 m section coincides with B
RA – 3 – 6 – V =0
V = RA- 9
V = 4- 9 = - 5 kN
𝚺𝐌 𝟐−𝟐 = 𝟎 -RA . 𝑥 +3 * (𝑥 -2) + 6 * (𝑥 -4) + M =0
M= 4 . 𝑥 – 3 * (𝑥 -2) – 6 * (𝑥 -4)
M= - 5 * 𝑥 + 30
4 < 𝑥< 6
Section 3-3
Shear force
V= -5 kN
Bending Moment(kN-m)
M= - 5 * 𝑥 + 30
𝑥=4 (at D) VD= -5 MD= - 5 * 4 + 30= 10
𝑥=6 (at B) VB= -5 MB=- 5 * 6 + 30 = 0

8. 𝚺𝐅𝐲 = 𝟎
Free body diagram of section 3: 0 m < 𝑥< 2m
when x=4 m section coincides with D, when
x=6 m section coincides with B
RB +V =0
V = - RB
V =- 5 kN
𝚺𝐌 𝟐−𝟐 = 𝟎 RB . 𝑥 - M=0
M= RB . 𝑥
M= 5 * 𝑥
4 < 𝑥< 6
Section 3-3
Shear force
V= -5 kN
Bending Moment(kN-m)
M= 5 * 𝑥
𝑥=2 (at D) VD= -5 MD= 5 * 2= 10
𝑥=0 (at B) VB= -5 MB= 5 * 0= 0
=5k N

9. 0< 𝑥<2
Section1-1
Shear
force (𝑘𝑁)
V= 4
Bending
Moment
M= 4 * 𝑥
𝑥=0 (at A) VA= 4 MA= 0
𝑥= 2 m (at C) VC=4 MC= 8
2< 𝑥< 4
Section2-2
Shear
force
V= 1 kN
Bending
Moment
M= 𝑥 + 6
𝑥=2 (at C) VC= 1 MC= 8
𝑥= 4 (at D) VD= 1 MD= 10
4 < 𝑥< 6
Section 3-3
Shear force
V= -5 kN
Bending Moment
M= - 5 * 𝑥 + 30
𝑥=4 (at D) VD= -5 MD=10
𝑥=6 (at B) VB= -5 MB= 0
4 kN 5 kN

10. Q2. Draw the shear force and bending moment diagram for a simply
supported beam of length 9 m and carrying a uniformly distributed load of 10
kN/m for a distance of 6 m from the left end. Also calculate the maximum B.M.
on the section.

12. Estimate the reactions at supports:
𝛴𝑭𝒚 = 0
RA - 10 * 6 + RB =0
RA + RB = 60 → (1)
𝛴𝑴𝑨 = 0
- 10 *6 * 3 + RB * 9 =0
 RB * 9 = 180
RB = 20 kN → (2 )
Substitute 2 in (1)
RA = 60 - RB
RA = 60 – 20
RA = 40 kN

13. Free body diagram of section1: 0< 𝑥< 6 m
when x=0 section coincides with A, when x=6 section coincides with C
0< 𝑥<2
Section1-1
Shear force (𝑘𝑁)
V=40 -10 *𝑥
Bending Moment (kN-m)
M= 40*𝑥 - 10 *
𝑥2
2
𝑥=0 (at A) VA=40 -10 *0 = 40 MA= 40*0 - 10 *
02
2
= 0
𝑥= 6 m (at C) VC=40 −10 ∗6 = −20 MC=40*6 - 10 *
62
2
= 60
𝑥= 4 m (at D) V=0 B.M is maximum
MD=40*4 - 10 *
42
2
= 80
𝚺𝐅𝐲 = 𝟎
RA-10 *𝑥 - V =0
V = RA-10 *𝑥
V = 40 -10 *𝑥
𝚺𝐌 𝟏−𝟏 = 𝟎
- RA * 𝑥 + 10. * 𝑥.
𝑥
2
+ M =0
M= 40 * 𝑥 - 10* 𝑥 ∗
𝑥
2
= 40*𝑥 - 10 *
𝑥2
2
40 kN
V = 0
40 -10 *𝑥 = 0
10 *𝑥 = 40
𝑥 = 4 m

14. Free body diagram of section2: 0< 𝑥< 3 m
when x=0 section coincides with B, when x=3 section coincides with C
0< 𝑥< 3
Section2-2
Shear force (𝑘𝑁)
V=-20
Bending
Moment (kN-m)
M= 20 * 𝑥
𝑥=0 (at B) VB=-20 MB= 20*0 = 0
𝑥= 3 m (at C) VC=−20 MC=20*3 = 60
𝚺𝐅𝐲 = 𝟎
RB + V =0
V = - RB
V = -20 kN
𝚺𝐌 𝟐−𝟐 = 𝟎
− M + RB * 𝑥 =0
M= 20 * 𝑥

15. 0< 𝑥<2
Section1-1
Shear force
(𝑘𝑁)
V=40 -10 *𝑥
Bending
Moment (kN-m)
M= 40*𝑥 - 10 *
𝑥2
2
𝑥=0 (at A) VA=40 MA= 0
𝑥= 6 m (at C) VC=−20 MC= 60
𝑥= 4 m (at D) V=0 B.M is maximum
MD= 80
0< 𝑥< 3
Section2-2
Shear force
(𝑘𝑁)
V=-20
Bending
Moment (kN-m)
M= 20 * 𝑥
𝑥=0 (at B) VB= -20 MB= 0
𝑥= 3 m (at
C)
VC=−20 MC= 60

16. Q3. 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.