aktu exam 2014 -15

1. AKTU EXAM-(2014-15)
ENGINEERING MECHANICS
Mr. Krishna Kumar

2. QUESTION-1 Two forces of magnitude 20 N and 40 N are acting on
a particle such that the angle between two is 135degree. If both
these forces are acting away from the particle, calculate their
resultant and find its direction.
135 ֺ
Solution-
Given F1 = 20 N
F2= 40 N
Ø = 135
And find R=? , ß=?
R2 = F1
2 + F2
2 +2F1 F2 COS Ø
R = ( F1
2 + F2
2 +2F1 F2 COS Ø)1/2
R = ( 202 + 402 + 2*20*40*COS135)1/2
R = 29.47 N
R
ß
F1
F2

3. F1 sin Ø
F2+ F1 COS Ø
Tan ß =
Tan ß =
20 sin135
40+20cos135
ß = 28.67 degree
R
ß
ß
F1
F2

4. QUESTION-2
Determine the centroid of a uniform
lamina as shown in fig.

5. Solution-
Reference axis is already provided in fig
2
1
a
b
c
d
e
f
(X1,y1)
(X2,y2)

6. Different areas and centroid of their co-ordinates are tabulated form
Sr.
No
.
Figure area(ai ) Co-ordinate of
centroid (xi )
Co-ordinate
of centroid
(yi )
(ai xi ) (ai yi )
1 Rectangle
(abcef)
=100*50
=5000
X1 = 100/2
= 50
Y1 = 50/2
= 25
=5000*50
=250000
=5000*25
=125000
2 Triangle
(cde)
=2*0.5*25
*50
=1250
X2 = 50+25
= 75
Y2 =
50+50/3
=66.66
=1250*75
=93750
=1250*66.6
=83325
sum =6250 =343750 =208325

7. Centroid of the given composite fig .
x =
(a1 x1 + a2 x2 )
X = 55 mm
(a1 y1 + a2 y2 )
Y =
(a1 + a2 )
(a1 + a2 )
Y = 33.33 mm
(55.00,33.33)

8. QUESTION-3 Two spheres each of weight 1kN and
radius 25 cm rest in a horizontal channel of width
90cm as shown in fig . Find the reactions on the points
of contact A,B and C.

9. Solution-
O1
O2
Let O1 and O2 be the centre of the first and second spheres resp. Drop perpendicular O2P
to the horizontal line through O2 Let it make angle Ø with horizontal .
P
O
90
40 25

10. Ø
O2
O1 P
We know
O1P = 40
O1 O2 =50
Cos Ø= O1P/ O1 O2
Ø= 36.87 degree

11. O2
FBD of the spheres and the coordinate directions are shown in the fig. since the
surfaces of the contact are smooth, reaction at 0 is the right angle to tangent at
O. it is in the radical direction.
Ø
1kN
RC
RO
From the equilibrium for sphere-2,
we get,
-Rci – 1000 j + Ro cos 36.87 i + Ro sin 36.87 j =0
(-Rc + Ro cos 36.87) i =0
(– 1000 + Ro sin 36.87) j =0
And
Solving the eq. 1 and eq.2
(1)
(2)
Rc =1333.33N , Ro =1666.66N

12. RAi – 1000 j + RB j - Ro cos 36.87 i - Ro sin 36.87 j =0
Consider the equilibrium of sphere -1
Ro
RB
RA
1kN
Ø
(RA - Ro cos 36.87) i =0
(– 1000 + RB - Ro sin 36.87) j =0
And
(3)
(4)
Solving the eq. 3 and eq.4
RA =1333.33 N , RB = 2000 N , RC= 1333.33N and Ro =1666.66N

13. QUESTION-4
Find the moment of inertia of the shaded area shown in the fig about
axis AB.
A B
P
R
Q
S

14. Solution-
The section is divided into a triangle PQR a Semicircle PSQ having on axis AB and a circle
having its centre on axis AB.
Now,
Moment of inertia of the section about axis AB =
(Moment of inertia of triangle PQR about AB) +
(Moment of inertia of semicircle PSQ about AB ) -
(Moment of inertia of circle about AB)
= (80*803/12) + (π/128) 804 - (π/64)* 404
IAB = 4292979 cm4

15. QUESTION-6
Determine the forces in all members of truss system
shown in fig and indicate the magnitudes and nature
of forces on the diagram of truss.

16. All inclined members have the same inclination to horizontal. Now, length
of an inclined member is

18. At this stage as no other joint is having only two unknowns, no further progress is
possible. Let us find the reactions at the supports considering the whole structure.
Let the reaction be as shown in Fig.

22. Note: When three members are meeting at an unloaded joint and out of them two are
collinear, then the force in third member will be zero. Such situations are illustrated in Fig.

23. A joint is analyzed the forces on the joint are marked on members in fig.

24. QUESTION-7
Draw the SFD and BMD for the overhanging as shown
in fig.
80N

25. Solution-
A B C D E
RAH
RAV RDV
 ΣH = 0,
i.e RAH = 0
 ΣV = 0
i.e
RAV + RDV = 80 + (0.5*40*3)
= 140 (1)

26.  Σ MA = 0,
ACW = +ve
-20+ RDV *5.5 – 80*4 – 2(40*3*0.5)=0
RDV = 83.63 N
and
solving the eq …1
RAV = 56.37 N

35. Thank You