Solution of Engineering Mechanics For UCER Students 1995976039
Solution of Engineering Mechanics For UCER Students 1995976039
Solution of Engineering Mechanics For UCER Students 1995976039
UNIT-I
LONG QUESTIONS
= sin ( ) = 25.37°
=90+25 =115
= 90+25.377 = 115.377
= 360-(115+115.377) = 129.623
= =
() () (
)
Putting the values we get,
P =175.95 kg
2.
R=l00√2 kN
So HB = 15 kN
AC = 72-30-15 = 27 mm
BC = √45 − 27 =36 mm
Hi 115 kN
HB 15 kN
100√2 kN
R
3. Two cylinders A and B weighing 4 kN and 3 kN, respectively, rest on smooth inclined plane as,
shown in figure, They are connected by a bar of negligible weight hinged to each cylinder at its
geometric centre by smooth pins. Find the force P to be applied to the smaller cylinder at 45· to the
vertical to hold the system in the given position.
Ans.
R = 4522N
R= 4906.7 N
For equilibrium of point B, we have.
∑ f =0,
R cos 20° - Re cos 45° - P cos 45° = 0
4522 cos 20° - (Re + P) cos 45° = 0
or Re + P = 6010.31 ... (1)
∑ f =0
Rc cos 45° - 3000 - R cos 70° - P cos 45° = O.
or Rc - P = 6430.85 ... (2)
Adding (1) and (2), we get
Rc = 6220.58 N
P = 210.27 N Ans,
4. Three cylinders A, Band C each weighing 100 N and diameter 80 mm are placed in a channel of 180
10m width as shown in Fig. 1. Determine the pressure exerted by' the cylinder A and B at the point of
contact.
RC = 315.37 N
Sin θ = 600/100
Or θ = 36.86°
Or 2θ = 73.72 °
By applying Lami’s Theorem,
=
( θ)
(θ)
Or T = 377.86 N
6. uniform bar AB of length Land weight W lien in it vertical plane with its end resting
on two smooth on OA and find angle for-the equilibrium of bar as shown in Fig.
From the geometry we obtain
the various angles
The reaction R1 , R2 and W of the rod should cut at same point for the equilibrium. By applying Lami’s theorem at
point O.
= =
() () ()
Solving we get
R1 = W/2
R2 = 1.73 W
Taking moment at the centre of the rod:
Clockwise moment = Anti Clockwise moment. .
O1A1 = L/2 cos
O1B1 = L/2 Sin
From above, we get
Tan = 1/√3
Or, = 30°
7 . Blocks A and B, f weight 150 N and 200 N, respectively rest on an inclined plane as shown in the figure. The
coefficient of friction between the two blocks is 0.3 and 'between blocks A and inclined plane is 0.4. Find the value
of e for which either one or both the blocks start slipping. At that instant, what is the friction force between B and
A? Between A and inclined plane?
Ans:
Let x b'e the distance man climbs on ladder for equilibrium from point B.
∑ f =0 , Ra - µRb = 0 Ra = 0.2 Rb .. ,(1)
∑ f =0, Rb + µRa - 300 - 750 = 0
Rb = 1009.6 N, Ra = 201.92 N
For finding out distance x take moment about B.
∑ M =0, 750 x cos 60° + 300 x 3 cos 60° - µRa 6 sin 60° -
Ra x 6 cos 60° = 0
x = 1.92 rn
9. Block 2 rests 011 block 1 and is attached· by horizontal rope AB on the wall as shown in fig. What force P is
necessary to cause motion of the block 1 to impend? The coefficient of friction between the blocks is 1/4 and
between the floor and block 1 is 1/3. Mass of the blocks 1 and 2 are 14 kg and 9 kg respectively.
Ans:
M1= 14kg,
m2 =9kg
FBD of block 2
∑ = 0 T = ×
∑ = 0
= 9× 9.8
FBD of block 1
∑ f = 0 ; Pcos45°= µ × N + µ N2 (3)
∑ f = 0 ; Psin45°+N1=N2+14×9.8 (4)
By solving all. equations
P=103.183N
UNIT-II
1. Determine the magnitude and nature of forces in all members of the truss shown in Fig.
Ans:
For equilibrium of truss consider it fully as free body
Ra + Rb = 4
∑M = 0
4 × AD = Rb × AB
Putting the value we get,
Rb = 3 kN
Ra = 1 kN
FBD of Joint A
Ans:
First consider whole truss as a free body
∑ f =0, Rex = 15 kN
∑ f =0, Rey +Ra = 0
∑ M =0,, Ra x4.5 = 1.5 X 6
Ra= 20 kN
R = -20 kN
Joint C
∑ f =0, fcd sin 36.87 - 15 = 0
∑ f =0, fcb + fcd cos 36.87 = 0
fcd = 25 kN (T),.
Joint B
∑ f =0
fab+ Ra + fad cos 36.87 = 0
- 20 + 20 +fad cos 36.87 = 0
fad= 0
∑ f =0
fac + fad sin 36.87 = 0
fac= 0
Joint D
Fdc = fde =25 kN (T)
3.. Determine the forces and their nature in each member of truss loaded as shown in fig.
Ans.
The reaction at the supports can be determined by considering equilibrium of the engine
truss.
From the figure.
AB sin 30° = BD
AB =2 BD =2X
BD.=BC=X
∑ f =0
RA + Rc= 57-5 KN
∑ M =0
7.5 x 2Xcos 30 + 50 x 2X=Rc × 3X
Or, Rc =37.08 KN
RA = 20.42KN
FAD = - 20.42/sin 30 = - 40.84 kN (C)
FAB = . +40.84× √3/2= 35·36 kN (T)
FCD = -74.16 kN (C)
Fbc = + 74.16 × √3 / 2
= 64.221 kN
FBD = 50 / sin 60 = 57.71 kN (T)
4.A truss is loaded as shown in Fig. 6. Find the reactions and forces in the members of truss.
Ans:
∑ f =0,
F sin 60 = -1
F = -1.155 kN
∑ fh =0,
F = −F sin 60
F = 0.577 kN
Consider Point C
∑ f =0,
FCB sin 60° + FAC cos 30° =.- 2
0.866 FCB + 0.866 FAC = - 2
∑ fh =0,
FAC sin 30° - FCB cos60° = Fe})
0.5 FAC – 0.5 FCB = 0.577 ... (ii)
From equations (i) and (ii)
FAC = - 0.577 kN (comp.) Ans.
FCB = - 1.731 kN (comp.) Ans.
Consider point 'A'
5.Draw the shear force and bending moment diagrams for the simple supported beam as shown in Fig.
Ans:
6. For the simply supported beam as shown in figure, draw shear force and bending moment diagrams after finding
the equations for shear force and bending moment.
For BMD- when the Beam is simply supported at end point B.M. will be zero.
BMA = BMB=0
M= '8.33 x – 5/2x2 0 ≤ x ≤2
M = 8.33x -10(x -1) -1 – 25 (x-2)2 2 ≤ x ≤4
with help of BM load we will draw the BMD show in fig
7. Draw the shear force and bending moment diagrams for the simple supported beam as shown in fig.
Ans.
∑ f =0
Ra + Rb = 15 + 5 =20 KM
Taking moment about point A
∑ M =0
15 x 1·5 + 10 + 5×8 - Rb x10 = 0
Rb =7.25 kN
Ra = 12·75 kN
UNIT-I
SHORT QUESTION
1. . Principle of Transmissibility of a Force. It states that the condition of equilibrium or of motion of rigid body will
remain unchanged if the point of application of a force acting on the rigid body is transmitted to act at any other
point along its line of action. A force F acting on the rigid body at' point B having the line of action AB, can be
replaced by the same force F but acting .at the point A provided this new point A lies along the line of action AB of
the force.
In other words, the force F acting at point A can be transmitted to act any other point B along its line of action
without changing its effect on the rigid ·body. Next consider a prismatic bar AB which is acted upon by two equal
and opposite coaxial forces P and P 'as shown in Fig.
Varignon's Theorem: It is also known as principle of moments: it states that "The algebraic sum of the moments of a
system of coplanar forces about a moment centre in their plane is equal to the moment of their resultant force about
the same moment centre.
Fl x Xl + F2 x X2 + F3 x X3 = R x X
F1, F2, F3 are the forces and R is the resultant.
5.Angle of Friction (4)): It is defined as the angle made by the resultant of the normal reaction and the limiting force
of friction (f) with the normal reaction R.
tan ∅ = F/R =µ .
Angle of repose: It is the maximum inclination of an inclined at which body starts moving due to it's own weight. In
limiting condition this angle is equal to angle of friction.
7.'Laws of static friction : If the two surfaces, which' are in contact are at rest, the force experienced by
one surface is called static friction' following are the coulomb’s laws for static (Dry) friction:
1. Frictional force is always acting .in opposite direction in which body tends to move.
2. Frictional force is directly proportional to normal reaction.
3. Frictional force is equal to the net force acting in direction of motion.
4.Frictional force is independent of the surface area in contact.
5. Frictional force depends upon the nature or roughness of surface.
UNIT-II
SHORT QUESTION
1.
Beam
Staticaly Staticaly
determinate indeterminate
beam beam
Simply supported beam: When a beam is made to freely rest on supports which may be knife edges or
rollers is called simply supported beam.
Point Load
Cantilever Beam: In cantilever beam one end is fixed or built in and other end is free to deflect.
4. The point at which bending moment changes it sign is called point of contraflexture. At this point bending
moment is zero point of contraflexture and point inflection is same.
5.Assumptions:
(b) Method of Sections: It is for finding out forces in some specified members. Take a section plane
which will cut only three members where forces are unknown including specified members. Draw the
FBD for either side and apply condition of equilibrium for finding out unknowns
Method of section is preffered when forces in some of the member are required.and also preferred
when method of joint fails to proceed.