mod of cae

KTU PREVIOUS YEAR QUESTIONS
SOLUTIONS
Module-wise / 2016 scheme / 2019 scheme
Krishnendu Sivadas

Course Introduction
Examination pattern and mark distribution (2016 and 2019 scheme)
Solutions to KTU University question papers 2016 and 2019 scheme
• EST 100 - Dec 2019
• BE-100 – May 2019
• BE- 100 – July 2018
Module-wise coverage (2019 syllabus)
KRISHNENDU SIVADAS 2
Krishnendu Sivadas

Examination Pattern – 2019 scheme
EST100
Krishnendu Sivadas

Examination Pattern – 2016 scheme
BE100

PREVIOUS YEAR QUESTIONS
Module 2
Friction – sliding friction - Coulomb’s laws of friction – analysis of single bodies –
wedges, ladder- analysis of connected bodies . Parallel coplanar forces – couple -
resultant of parallel forces – centre of parallel forces – equilibrium of parallel forces
– Simple beam subject to concentrated vertical loads. General coplanar force
system - resultant and equilibrium equations.

Question 1

R1
R2
R3
W
P
𝝁R1
𝝁R3

QUESTION 1 : SOLUTION WEDGE ANALYSIS

Question 2
4 m
A B
3 m
2 m
1m
10 N 6 N 4 N
RA RB

QUESTION 2 : SOLUTION
SIMPLE BEAM
Consider free body diagram of the beam
Equilibrium conditions :
σ 𝐹𝑣 = 0,
σ 𝑀 = 0,
Applying σ 𝐹𝑣 = 0,
RA + RB -10-6-4 =0
RA + RB = 20……..(i)
Applying σ 𝑀 = 0,
Taking moment about A =0,
(10x1) + (6x2) + (4x3) – (RB x4) =0
RB =34/4 = 8.5 N……..(ii)
Therefore RA = 20 - 8.5 = 11.5 N

Question 3
30 N
𝜶
6 N
30 N
𝜶
P
smooth inclined plane
rough inclined plane

ANALYSIS OF SINGLE BODIES
Case 1: Consider the body of weight 30 N placed on a smooth
Inclined plane with angle 𝛼 and load 6 N
Resolving forces along the inclined plane:
6 cos 30 – 30 sin 𝛼 = 0
𝜶 = 9.974
Case 2: Consider the body of weight 30 N placed on a rough
inclined plane with an inclination 𝛼=9.974
Resolving perpendicular to the inclined plane:
R- 30 cos 9.974 = 0
R= 29.54 N
Resolving along inclined plane
P-0.3R-30 sin 9.974 =0
P= (0.3R)+(30sin9.974) =14.06 N
30 N
𝜶
6 N
𝟑𝟎
6 cos 30
6 sin 30
30 N
𝟗. 𝟗𝟕𝟒
P
R
R

Question 4
Krishnendu Sivadas

SIMPLE BEAM
sin𝜃=1/2
𝜃 = sin-1 ½
𝜃 = 26.56
100 sin45
100 cos45
RAV
RAH
150
RD
RD sin26.56
RD cos26.56
1.5 m
1.5 m
1 m
1 m

Consider free body diagram of the beam,
Equilibrium conditions:
σ 𝐹𝑉 = 0,
σ 𝐹𝐻 = 0,
σ 𝑀 = 0,
σ 𝐹𝑉 = 0,
RAV + RD cos26.56-100sin45-150 = 0……..(i)

σ 𝐹𝐻 = 0,
RAH +100 cos45- RD sin26.56 = 0……..(ii)
σ 𝑀 = 0, Taking moment about A,
(100sin45 x 1) + (150 x3.5) – (RD cos26.56 x5) = 0
RD =133.19 kN
Substituting value of RD in eqn (ii), we get;
RAH +100 cos45- RD sin26.56 = 0……..(ii)
RAH = -100 cos45+ RD sin26.56 = -11.15 kN

Substituting value of RD in eqn (ii), we get;
RAV + RD cos26.56-100sin45-150 = 0……..(i)
RAV = - RD cos26.56+100sin45+150 = 101.57 kN
Therefore resultant reaction at A = σ 𝑅𝐴𝐻
2
+ σ 𝑅𝐴𝐻
2
= 11.152 + 101.572 = 102.18 kN
Inclination = tan-1 (σ 𝑅𝐴𝑉/ σ 𝑅𝐴𝐻)
= 83.73 with horizontal or 90-83.73 =6.26 with vertical
Reaction at D= RD =133.19 kN

Question 5
Question 5

LADDER FRICTION
Free body diagram:
Given data:
Length of ladder = 4 m
Weight of ladder W=200 N
𝜇w= 0.25
𝜇f= 0.35
Inclination of ladder with floor = 60
Load at top = 1000 N
To find:
a) Horizontal force P to prevent
slipping
b)Minimum inclination to prevent
slipping if P is not applied.
200 N
1000 N
P
Rf
Rw
4 m
60
0.25Rw
0.35Rf

Consider the equilibrium of ladder:
σ 𝐹𝑥 = 0,
P +0.35 Rf – Rw = 0…..(i)
σ 𝐹𝑦 = 0,
Rf +0.25Rw – 200 -1000 =0
Rf +0.25Rw =1200…..(ii)
(200x2cos60) +(1000 x 4cos60) – (0.25Rw x 4cos60)
-(Rw x 4sin 60) = 0
200 + 2000 – 0.5 Rw -3.464 Rw = 0
2200 = (0.5+3.464) Rw
Rw = 554.98 N
From (ii), Rf = 1200 -0.25 Rw = 1061.26 N
From (i), P = Rw -0.35 Rf = 183.5 N
200 N
1000 N
P
Rf
Rw
4 m
60
0.25Rw
0.35Rf

σ 𝐹𝑥 = 0,
0.35 Rf – Rw = 0
Rw = 0.35 Rf …..(i)
σ 𝐹𝑦 = 0,
Rf +0.25Rw – 200 -1000 =0
Rf +0.25Rw =1200
Rf +0.25x0.35 Rf =1200
Rf=1103.45 N
Rw = 0.35 x 1103.45 = 386.2 N
(200x2cos𝜃) +(1000 x 4cos𝜃) – (0.25Rw x 4cos𝜃)
-(Rw x 4sin 𝜃) = 0
cos𝜃(400 + 4000- 0.25x4x386.2) = 4x386.2sin 𝜃
𝜽=68.95
200 N
1000 N
Rf
Rw
4 m
𝜃
0.25Rw
0.35Rf

Question 6
Krishnendu Sivadas

Question 7

SIMPLE BEAM
RA RB
30x6=180 kN 45 kN 90 kN
1.5 m
2.5 m
2 m
3 m
3 m

Consider the free body diagram of the beam AB
σ 𝐹𝑉 = 0,
σ 𝑀 = 0,
σ 𝐹𝑉 = 0,
RA+RB -180-45-90 =0
RA+RB =315 kN…..(i)

σ 𝑀 = 0, Taking moment about A
(180 x 3) + (45x8) +(90x10.5) – (RB x12) =0
RB =153.75 kN
From eqn. (i)
RA=315 - RB = 315-153.75 =161.25 kN

Question 8

LADDER FRICTION
Free body diagram:
Given data:
Length of ladder = 3 m
Weight of ladder W=180 N
𝜇w= 0.25
𝜇f= 0.35
Inclination of ladder with floor = 60
Load at top = 900 N
To find:
Minimum horizontal force P to
prevent slipping
180 N
900 N
P
Rf
Rw
3 m
60
0.25Rw
0.35Rf

σ 𝐹𝑥 = 0,
P +0.35 Rf – Rw = 0…..(i)
σ 𝐹𝑦 = 0,
Rf +0.25Rw – 180 -900 =0
Rf +0.25Rw =1080…..(ii)
(180x1.5cos60) +(900 x 3cos60) – (0.25Rw x 3cos60)
-(Rw x 3sin 60) = 0
135 + 1350 – 0.375 Rw -2.59 Rw = 0
1485 = (0.375+2.59) Rw
Rw = 500.84 N
From (ii), Rf = 1080 -0.25 Rw = 954.79 N
From (i), P = Rw -0.35 Rf = 166.6 N
180 N
900 N
P
Rf
Rw
3 m
60
0.25Rw
0.35Rf

Question 9
Krishnendu Sivadas

QUESTION 9 : SOLUTION BEAM SUPPORTS

Question 10

Characteristics of dry friction
 Co-efficient of friction (𝜇)
 Angle of friction (𝜙)
 Angle of repose (𝛼)
 Cone of friction
Co-efficient of friction
Limiting friction is proportional to the normal reaction at the contact surface
𝐹max ∝ 𝑅𝑁
𝐹max = 𝜇𝑅𝑁
𝜇 = co-efficient of friction at contact surface
Angle of friction:
It is the angle between the normal reaction at the contact surface and the resultant of normal reaction and
limiting friction. Denoted by 𝜙
QUESTION 10 : SOLUTION COULOMB FRICTION
𝑅𝑁 =
𝐹max
𝜇
tan 𝜙 =
𝐹
= 𝜇𝑅𝑁
= 𝜇
𝑅𝑁 𝑅𝑁

Angle of Repose:
It is the maximum inclination of a plane on which a body can rest/repose without applying external force
Cone of Friction
If the resultant reaction is rotated about normal reaction force, it will form a cone which is known as cone
of friction.

Question 11

SIMPLE BEAM
30sin45
30cos45
20 x 2 =40kN
3 m 2 m 1 m 2 m
40kNm
RAH
RAV RB

Consider the free body diagram of the beam;
σ 𝐹𝑉 = 0,
σ 𝐹𝐻 = 0,
σ 𝑀 = 0,
σ 𝐹𝑉 = 0,
RAV + RB -30sin45-40 = 0……..(i)

σ 𝐹𝐻 = 0,
RAH – 30cos45 = 0
RAH = 30cos45 = 21.213 kN
(40) + (30sin45 x5) –(RBx 6)+(40 x 7) = 0
RB =71.01 kN
Substituting value of RD in eqn (i), we get;
RAV + RB -30sin45-40 = 0……..(i)
RAV = - RB +30sin45+40 = -9.79kN

Therefore resultant reaction at A = σ 𝑅𝐴𝐻
2
+ σ 𝑅𝐴𝐻
2
= 21.212 + 9.792 = 23.36 kN
Inclination = tan-1 (σ 𝑅𝐴𝑉/ σ 𝑅𝐴𝐻)
= 24.77 with horizontal or 90-24.77 =65.223 with vertical

Question 12

F
F
F
F
RN
RN
RN
RN
R
R
R
R
QUESTION 12 : SOLUTION WEDGE FRICTION

Consider the block to the right
σ 𝐹𝑦 = 0,
R1 cos 15 – R2 sin15 -100 =0……(i)
σ 𝐹𝑥 = 0,
R2 cos15- R1 sin 15 -2.5 = 0…..(ii) Solving we get, R1 =112.28 kN and R2 =32.67kN
R
R
R
R
15
15
15
15
15

Consider the block to the right
σ 𝐹𝑥 = 0,
R3 cos 30 = R2 cos15
R3 = (32.67 cos15)/cos 30 = 36.44 kN
σ 𝐹𝑦 = 0,
R2 sin15+ R3 sin 30 –P = 0
P= R2 sin15+ R3 sin 30 = 32.67 sin15+ 36.44 sin 30 =26.67 kN
R
R
R
R
15
15
15
15
15

THANK YOU
Krishnendu Sivadas

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