Problems on support reaction.pdf

Engineering mechanics subject Equilibrium chapter problems solved on BEAM REACTION.

1. Problems on
Q1. Calculate the support reactions for the beam shown in fig
ANS
:
Step 1: Consider

Step 2: Consider
Step 3: Consider

????????????+
Step 4: Consider
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1
Problems on Support Reactions
Calculate the support reactions for the beam shown in fig
: Consider FBD of beam
: Consider ƩFx
Ʃ??????
?= 0
As there are no horizontal forces
: Consider ƩFy
Ʃ??????
?= 0
3+4+5−????????????−???????????? = 0
????????????+???????????? = 3+4+5
+???????????? = 12 ????????????…………………………
: Consider ƩM@A
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Support Reactions
Calculate the support reactions for the beam shown in fig.


As there are no horizontal forces
.????????????
?
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
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2
ƩM
E=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 Eq
n
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
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=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

Ʃ??????
?=0
As there are no horizontal forces
Step 3: Consider ƩFy
Ʃ??????
?= 0
????????????+????????????= 4+(2×1.5)+1.5
????????????+????????????= 8.5 ????????????…………………………. ????????????
?
1
Step 4: Consider ƩM@A
ƩM
E= 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 Eq
n
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
????????????+
Step 4: Consider
Put the value of RB in Eq
Step 5: Final Answer
Reaction at support A =
Reaction at support B =
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4
: Consider FBD of beam
: Consider ƩFx
Ʃ??????
?= 0
As there are no horizontal forces
: Consider ƩFy
Ʃ??????
?= 0
????????????+???????????? = 4.5+2.25
+???????????? = 6.75 ????????????…………………………
: Consider ƩM@A
ƩM
E= 0
(4.5×2.25)+(2.25×3)= RB×4.
RB =
16.88
4.5
= 3.75 KN
Put the value of RB in Eq
n
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



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there are no horizontal forces
….????????????
?
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
????????????+
Step 4: Consider
(120×3
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5
Calculate the support reactions for the beam shown in fig
: Consider FBD of beam
: Consider ƩFx
Ʃ??????
?= 0
???????????? = 0
As there are no horizontal forces acting on beam
: Consider ƩFy
Ʃ??????
?= 0
???????????? +???????????? = 120+30+90
+???????????? = 240 ????????????…………………………
: Consider ƩM@A
ƩM
E= 0
3)+(30×6)+(40)+(90×8.67)=
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Calculate the support reactions for the beam shown in fig.


forces acting on beam
….????????????
?
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
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6

Put the value of RB in Eq
n
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
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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

Ʃ??????
?=0
????????????=0
As there are no horizontal forces acting on beam
Step 3: Consider ƩFy
Ʃ??????
?= 0
????????????+????????????= 10+36+??????
0+????????????−??????= 46 ????????????………………………… ???????????? ????????????= 0
????????????−??????= 46 ????????????…………………………. ????????????
?
1

Step 4: Consider ƩM@A
ƩM
E= 0
(10×2)−20+(36×5)−(P×7) = RB ×6
6RB−7P = 220……………………………… ????????????
?
2
Solving Eq
n
1 & 2, we get

Putin Eq
n
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
????????????
Step 3: Consider
???????????? +????????????

Step 4: Consider
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8
Find the support reactions at A and B for the beam loaded as
: Consider FBD of beam
: Consider ƩFx
Ʃ??????
?= 0
????????????−???????????? ?????????????????? 60 = 0………………????????????
: Consider ƩFy
Ʃ??????
?= 0
???????????? +?????????????????????????????? 60 = 120+90+80
???????????? ??????????????????60 = 290 ????????????………………………
: Consider ƩM@A
ƩM
E=0
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Find the support reactions at A and B for the beam loaded as


????????????
?
1

…….????????????
?
2

9. (120×
Put in Eq
n
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
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×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
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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
????????????

Step 4: Consider
Put in Eq
n
1, we get
Now Consider
Step 5: Consider



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10

Consider FBD of beam DF first,
: Consider ƩFx
Ʃ??????
?= 0
???????????? = 0
: Consider ƩFy
Ʃ??????
?= 0
+???????????? = 120………………………….????????????
: Consider ƩM@D
ƩM
H= 0
(120×3)−(VF×4)= 0

1, we get

Consider the FBD of beam AC,
: Consider ƩFx
Ʃ??????
?= 0
????????????+2cos59.04 = 0
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????????????
?
1

11. Prepared by: Prof. V.V. Nalawade

11

????????????=−1.03 ????????????
i.e. our assumed direction is wrong, Therefore
????????????= 1.03 ???????????? (←)
Step 6: Consider ƩFy
Ʃ??????
?= 0
????????????−????????????= 2sin59.04
????????????= 1.71+30 = 31.41 ???????????? (↑)

Step 7: Consider ƩM@A
−M
E−2sin59.04 ×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

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12
: Consider FBD of beam
FBD of beam AB
FBD of beam CD
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13. Step 2: Consider
(
Step 3: Consider

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13
: Consider FBD of AB & Consider ƩM@A
ƩM
E=0
(600×4)+2400 = RBsin53.13 ×12

Consider FBD of CD & Consider ƩM@C
ƩM
G= 0
RDsin36.87 ×10−500×4.2 = 0

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M@A
12
M@C
0

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

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Q10.

Q11.