The document provides solutions to multiple problems involving calculating support reactions for beams. The problems involve drawing free body diagrams of beams, then applying equations for the sum of forces and sum of moments at the supports to solve for the unknown support reactions. Key steps include considering the absence of any horizontal forces, setting equations for the sum of vertical forces and moments equal to zero, and solving the resulting systems of equations to find the support reactions.

1. Problems on
Q1. Calculate the support reactions for the beam shown in fig
ANS
:
Step 1: Consider
Step 2: Consider
Step 3: Consider
𝑅𝐴 +
Step 4: Consider
Prepared by: Prof. V.V. Nalawade
1
Problems on Support Reactions
Calculate the support reactions for the beam shown in fig
: Consider FBD of beam
: Consider ƩFx
Ʃ𝐹 = 0
As there are no horizontal forces
: Consider ƩFy
Ʃ𝐹 = 0
3 + 4 + 5 − 𝑅𝐴 − 𝑅𝐵 = 0
𝑅𝐴 + 𝑅𝐵 = 3 + 4 + 5
+ 𝑅𝐵 = 12 𝐾𝑁 … … … … … … … … … …
: Consider ƩM@A
Prepared by: Prof. V.V. Nalawade
Support Reactions
Calculate the support reactions for the beam shown in fig.
As there are no horizontal forces
. 𝐸𝑞 1

2. (3
Put the value of RB in Eq
Step 5: Final Answer
Reaction at support A =
Reaction at support B =
Q2. Calculate the support reactions for the beam shown in fig
ANS
:
Step 1: Consider
Step 2: Consider
Prepared by: Prof. V.V. Nalawade
2
ƩM = 0
(3 × 2) + (4 × 3) + (5 × 5) − RB × 6
(3 × 2) + (4 × 3) + (5 × 5) = RB ×
RB × 6 = 6 + 12 + 25
RB =
43
6
= 7.17 KN
Put the value of RB in Eqn
1, we get
RA = 12-7.17 = 4.83 KN
Final Answer
Reaction at support A = RA = 4.83 KN
Reaction at support B = RB = 7.17 KN
Calculate the support reactions for the beam shown in fig
: Consider FBD of beam
: Consider ƩFx
Prepared by: Prof. V.V. Nalawade
= 0
6
4.83 KN
RB = 7.17 KN
Calculate the support reactions for the beam shown in fig.

3. Prepared by: Prof. V.V. Nalawade
3
Ʃ𝐹 = 0
As there are no horizontal forces
Step 3: Consider ƩFy
Ʃ𝐹 = 0
𝑅𝐴 + 𝑅𝐵 = 4 + (2 × 1.5) + 1.5
𝑅𝐴 + 𝑅𝐵 = 8.5 𝐾𝑁 … … … … … … … … … … . 𝐸𝑞 1
Step 4: Consider ƩM@A
ƩM = 0
(4 × 1.5) + (3 × 2.25) + (1.5 × 4.5) = RB × 6
RB × 6 = 6 + 6.75 + 6.75
RB =
19.5
6
= 3.25 KN
Put the value of RB in Eqn
1, we get
RA = 8.5 – 3.25 = 5.25 KN
Step 5: Final Answer
Reaction at support A = RA = 5.25 KN
Reaction at support B = RB = 3.25 KN
Q3. A simply supported beam AB of Span 4.5 m is loaded as shown in
fig. Find the support reactions at A & B.

4. ANS
:
Step 1: Consider
Step 2: Consider
Step 3: Consider
𝑅𝐴 +
Step 4: Consider
Put the value of RB in Eq
Step 5: Final Answer
Reaction at support A =
Reaction at support B =
Prepared by: Prof. V.V. Nalawade
4
: Consider FBD of beam
: Consider ƩFx
Ʃ𝐹 = 0
As there are no horizontal forces
: Consider ƩFy
Ʃ𝐹 = 0
𝑅𝐴 + 𝑅𝐵 = 4.5 + 2.25
+ 𝑅𝐵 = 6.75 𝐾𝑁 … … … … … … … … … …
: Consider ƩM@A
ƩM = 0
(4.5 × 2.25) + (2.25 × 3) = RB × 4.
RB =
16.88
4.5
= 3.75 KN
Put the value of RB in Eqn
1, we get
RA = 6.75 – 3.75 = 3 KN
Final Answer
Reaction at support A = RA = 3
Reaction at support B = RB = 3.75
Prepared by: Prof. V.V. Nalawade
there are no horizontal forces
… . 𝐸𝑞 1
.5
KN
3.75 KN

5. Q4. Calculate the support reactions for the beam shown in fig
ANS
:
Step 1: Consider
Step 2: Consider
As there are no horizontal
Step 3: Consider
𝑉𝐴 +
Step 4: Consider
(120 × 3
Prepared by: Prof. V.V. Nalawade
5
Calculate the support reactions for the beam shown in fig
: Consider FBD of beam
: Consider ƩFx
Ʃ𝐹 = 0
𝐻𝐴 = 0
As there are no horizontal forces acting on beam
: Consider ƩFy
Ʃ𝐹 = 0
𝑉𝐴 + 𝑅𝐵 = 120 + 30 + 90
+ 𝑅𝐵 = 240 𝐾𝑁 … … … … … … … … … …
: Consider ƩM@A
ƩM = 0
3) + (30 × 6) + (40) + (90 × 8.67) =
Prepared by: Prof. V.V. Nalawade
Calculate the support reactions for the beam shown in fig.
forces acting on beam
… . 𝐸𝑞 1
) = RB × 10

6. Put the value of RB in Eq
Step 5: Final Answer
Horizontal Reaction at support A = H
Vertical Reaction at support A = V
Reaction at support B =
Q5. Find analytically the support reactions at B and the load P, for
the beam shown in fig. If the reaction of support A is Zero
ANS
:
Step 1: Consider
Step 2: Consider
Prepared by: Prof. V.V. Nalawade
6
Put the value of RB in Eqn
1, we get
VA = 240-136.03 = 103.97 KN
Final Answer
Horizontal Reaction at support A = H
Vertical Reaction at support A = VA = 103.97
Reaction at support B = RB = 136.03
Find analytically the support reactions at B and the load P, for
shown in fig. If the reaction of support A is Zero
: Consider FBD of beam
: Consider ƩFx
Prepared by: Prof. V.V. Nalawade
Horizontal Reaction at support A = HA = 0
103.97 KN
136.03 KN
Find analytically the support reactions at B and the load P, for
shown in fig. If the reaction of support A is Zero

7. Prepared by: Prof. V.V. Nalawade
7
Ʃ𝐹 = 0
𝐻𝐴 = 0
As there are no horizontal forces acting on beam
Step 3: Consider ƩFy
Ʃ𝐹 = 0
𝑉𝐴 + 𝑅𝐵 = 10 + 36 + 𝑃
0 + 𝑅𝐵 − 𝑃 = 46 𝐾𝑁 … … … … … … … … … … 𝑎𝑠 𝑉𝐴 = 0
𝑅𝐵 − 𝑃 = 46 𝐾𝑁 … … … … … … … … … … . 𝐸𝑞 1
Step 4: Consider ƩM@A
ƩM = 0
(10 × 2) − 20 + (36 × 5) − (P × 7) = RB × 6
6RB − 7P = 220… … … … … … … … … … … … 𝐸𝑞 2
Solving Eqn
1 & 2, we get
Putin Eqn
1, we get
Step 5: Final Answer
Reaction at support B = RB = 102 KN
Load = P = 56 KN

8. Q6. Find the support reactions at A and B for the beam loaded as
shown in fig.
ANS
:
Step 1: Consider
Step 2: Consider
𝐻𝐴
Step 3: Consider
𝑉𝐴 + 𝑅𝐵
Step 4: Consider
Prepared by: Prof. V.V. Nalawade
8
Find the support reactions at A and B for the beam loaded as
: Consider FBD of beam
: Consider ƩFx
Ʃ𝐹 = 0
𝐻𝐴 − 𝑅𝐵 𝐶𝑂𝑆 60 = 0 … … … … … … 𝐸𝑞
: Consider ƩFy
Ʃ𝐹 = 0
𝑉𝐴 + 𝑅𝐵𝑠𝑖𝑛 60 = 120 + 90 + 80
𝑅𝐵 𝑠𝑖𝑛60 = 290 𝐾𝑁 … … … … … … … … …
: Consider ƩM@A
ƩM = 0
Prepared by: Prof. V.V. Nalawade
Find the support reactions at A and B for the beam loaded as
𝐸𝑞 1
… … . 𝐸𝑞 2

9. (120 ×
Put in Eqn
1 & 2
Step 5: Final Answer
Horizontal Reaction at support A = H
Vertical Reaction at support A = V
Reaction at support B =
Q7. Find the support reactions at A and F for the beam loaded as
shown in fig.
ANS
:
Step 1: Consider
FBD of beam AC
FBD of beam DF
Prepared by: Prof. V.V. Nalawade
9
× 5) + (90 × 6) + (80 × 13) = RB sin
& 2, we get
Final Answer
Horizontal Reaction at support A = HA =
Vertical Reaction at support A = VA =
Reaction at support B = RB = 251.72
Find the support reactions at A and F for the beam loaded as
: Consider FBD of beam
FBD of beam AC
DF
Prepared by: Prof. V.V. Nalawade
sin60 × 10
A = 125.86 KN
A = 72 KN
251.72 KN
Find the support reactions at A and F for the beam loaded as

10. Consider FBD of beam DF first,
Step 2: Consider
Step 3: Consider
𝑉𝐹
Step 4: Consider
Put in Eqn
1, we get
Now Consider
Step 5: Consider
Prepared by: Prof. V.V. Nalawade
10
Consider FBD of beam DF first,
: Consider ƩFx
Ʃ𝐹 = 0
𝐻𝐹 = 0
: Consider ƩFy
Ʃ𝐹 = 0
+ 𝑅𝐷 = 120 … … … … … … … … … … . 𝐸𝑞
: Consider ƩM@D
ƩM = 0
(120 × 3) − (VF × 4) = 0
1, we get
Consider the FBD of beam AC,
: Consider ƩFx
Ʃ𝐹 = 0
𝐻𝐴 + 2 cos 59.04 = 0
Prepared by: Prof. V.V. Nalawade
𝐸𝑞 1

11. Prepared by: Prof. V.V. Nalawade
11
𝐻𝐴 = −1.03 𝐾𝑁
i.e. our assumed direction is wrong, Therefore
𝐻𝐴 = 1.03 𝐾𝑁 (←)
Step 6: Consider ƩFy
Ʃ𝐹 = 0
𝑉𝐴 − 𝑅𝐶 = 2 sin 59.04
𝑉𝐴 = 1.71 + 30 = 31.41 𝐾𝑁 (↑)
Step 7: Consider ƩM@A
−M − 2 sin 59.04 × 1 − (30 × 2) = 0
Step 8: Final Answer
Horizontal Reaction at support A = HA = 1.03 KN
Vertical Reaction at support A = VA = 31.41 KN
Moment at support A = MA = 61.71 KN.m
Horizontal Reaction at support F = HF = 0 KN
Vertical Reaction at support F = VF = 90 KN
Reaction at support D = RD = 30 KN
Q8. Two beams AB & CD are arranged as shown in fig. Find the
support reactions at D.

12. ANS
:
Step 1: Consider
FBD of beam AB
FBD of beam
Prepared by: Prof. V.V. Nalawade
12
: Consider FBD of beam
FBD of beam AB
FBD of beam CD
Prepared by: Prof. V.V. Nalawade

13. Step 2: Consider
(
Step 3: Consider
Prepared by: Prof. V.V. Nalawade
13
: Consider FBD of AB & Consider ƩM@A
ƩM = 0
(600 × 4) + 2400 = RB sin 53.13 × 12
Consider FBD of CD & Consider ƩM@C
ƩM = 0
RD sin 36.87 × 10 − 500 × 4.2 = 0
Prepared by: Prof. V.V. Nalawade
M@A
12
M@C
0

14. Q9.
Prepared by: Prof. V.V. Nalawade
14
Prepared by: Prof. V.V. Nalawade

15. Prepared by: Prof. V.V. Nalawade
15
Q10.
Q11.