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BCEM


UNIT 5


ENGINEERING MECHANICS

QUESTIONS

 

  • Question 1) Determine moment of Inertia for section about x & y axis as shown in figure.

Description: S 25.png

  • Consider area EFGH

M.I. If this rectangle about xx and yy axis will be,

1=
=

1 = 162226666.7 mm4

1 =
=

1 = 306666.67 mm4

2 =
G2

=

= 106666666.7 mm4

Consider rectangle ABCD area

3 = 461066666.7 mm4

3 = 106666666.7 mm4

=
1 +
2 +
3

= 1084.36
106

1 +
2 +
3

= 213.64

106 mm4

 

  • Consider rectangle PQMN area using parallel axis theorem

2 =
G2= A2h22

h22

+ (400
2)

2 = 461066666.7 mm4

 

Description: S 26.pngQuestion 2): Determine the moment of inertia of shaded area as shown in figure about its centroid as axis.

Answer 2)

As this figure is symmetrical about

Y axis

X = 100 mm

Y =

There ,

Area = rectangle

Area = triangle

Area = circle

Y = 79.95 mm

to find M.I. Of shaded portion , let G is the centroid of shaded area which is at y = 79.95 mm from base.

of shaded portion @ x-x axis passing through its centroid G will be,

x-x) +

x-x) –

x-x axis)

 

1 + 23

= (G1 + A1h12) + G2 + A2h22) - G3 + A3h32)

= + -60] +

+

84329013.21 mm4

of shaded portion about y-y axis passing through its centroid G will be,

y-y) +

y-y axis) –

y-yaxis)

1 + 23

=

= 80000000 + 16666666.67 – 3220623.34

93446043.33 mm4

 

Question 3) Find the support reactions of given beam for loading as shown below.

 

Answer 3)

Draw FBD of given beam & applying conditions of equilibrium

 

∑fx = 0

RHA = 0

∑fxy  = 0

RVA – 20 + RB = 0

RVA + RB = 20    ------(1)

Taking moments at A

∑MA = 0

(20 × 3) – (RB × 5) = 0

60 – 5RB

RB = 12 kN               put in equation (1)

RVA + 12 = 20

RVA = 20-12

RVA = 8 kN

 

Question 4) Find support reaction for given beam

 

Answer 4) Resolving forces horizontally

∑fx = 0

RHA = 0

 

Resolving forces vertically

fy = 0

RVA – 60 + RB = 0

RVA + RB = 60                                 -------(1)

Taking moment at A

∑MA = 0

(60 × 2) – (RB × 4 ) = 0

120 – 4RB = 0

RB =

RB = 30 kN

 

Question 5) . Find the support reactions of a given loading for Beam.

 

Answer 5)

Resolving forces horizontally

∑fX = 0

RHA = 0

Resolving forces vertically

∑fY = 0

RVA – 60 + RB = 0

RVA + RB = 60                    ----------------- (1)

 

Taking moments at A

∑ MA = 0

(RVA × 0) + (60 × 2.67) – 4RB = 0

160.2 – 4RB = 0

    put in equation (1)

We get

RVA + 40 = 60

RVA = 60 – 40

RVA = 20 kN

 

Question 6). Find the support reactions for given beam.

 

Answer 6)

Resolving force horizontally

∑fX = 0

RHA = 0

 

Resolving forces vertically

∑fy = 0

RVA + RB = 0           -------(1)

Taking moments at A

40 – 7RB = 0

RB =

RB = 5.71 kN

Put above value in equation (1)

RVA + RB = 0

RVA + 5.71 = 0

RVA = - 5.71 kN

RVA = 5.71 kN 

 

Question 7)

 

Answer 7)

Resolving forces horizontally

∑fX = 0

RHA = 0

Resolving forces vertically

∑fy = 0

RVA – 240 + RB = 0

RVA + RB = 240              -----------(1)

Taking moment at A

∑ MA = 0

(240 × 3) – 80 – 6RB = 0

640 – 6RB = 0

RB = 106.67 kN

Put this value in equation (1)

RVA + RB = 240

RVA = 240 – 106.67

RVA = 133.33 kN

 

Question 8) Find support reactions for the loading as shown below:

Answer 8)

Resolving forces vertically

∑fy = 0

RVA – 50 – 20 – 120 + RB = 0

RVA + RB = 190   ----(1)

Resolving forces horizontally

∑fX = 0

RHA = 0

 

Taking moments at A

∑ MA = 0

(50 × 2) + (20 × 4) + (120 × 6) – 8RB = 0

100 + 80 + 720 – 8RB = 0

900 – 8RB = 0

RB – 112.5 kN

Put above value in equation (1)

RVA + 112.50 = 190

RVA = 77.5 kN

Question 9) Find the reactions at support A & B.

Answer 9)

  Resolving forces vertically

∑fy = 0

RVA – 30 – 80 + RB = 0

RVA + RB = 110    ------(1)

Resolving forces horizontally

∑fX = 0

RHA = 0

Taking moments at A

∑ MA = 0

(30 × 0.6) + [80 × (4.07 + 0.6)]- (6 × RB) = 0

18 + 373.6 – 6RB = 0

6RB = 391.6

RB= 65.267 kN

Put above value in equation (1)

RVA + RB = 110

RVA + 65.267 = 110

RVA = 110 – 65.267

RVA = 44.733 kN

Question 10) Find support reactions at A & B

Answer 10)

Resolving forces vertically

∑fy = 0

RVA – 8 – 18 – 5 – 20 + RB = 0

RVA + RB = 51   ------(1)

Resolving forces horizontally

∑fX = 0

RHA = 0

 

Taking moments at A

∑ MA = 0

(8 × 2) + (18 × 3.5 ) + (5 × 5 ) + (20 × 8.67) – 10RB = 0

16+ 63 + 25 + 173.4 – 10RB = 0

277.4 – 10RB = 0

RB = 27.74 kN

Put above value in equation (1)

RVA = 51 – 27.74

RVA = 23.26 kN

Question 11) Explain the difference between centre of gravity and centroid?

Answer 11)

Difference Between Center of Gravity and Centroid

Center of Gravity

Centroid

The point where the total weight of the body focuses upon

It is referred to the geometrical center of a body

It is the point where the gravitational force (weight) acts on the body

It is referred to the center of gravity of uniform density objects

It is denoted by g

It is denoted by c

Center of Gravity in a uniform gravitational field is the average of all points, weighted by local density or specific weight

The centroid is a point in a plane area in such a way that the moment of area about any axis throughout that point is 0

It is a physical behavior of the object, a point where all the weight of an object is acting

It is a geometrical behavior. It is the center of measure of the amount of geometry.

 

Question 12: Find the centroid of the triangle whose vertices are A(2, 6), B(4, 9), and C(6,15).

Answer 12) Solution:

Given:

A(x1, y1) = A(2, 6)

B(x2, y2) = B(4,9)

C(x3, y3) = C(6,15)

We know that the formula to find the centroid of a triangle is = ((x1+x2+x3)/3, (y1+y2+y3)/3)

Now, substitute the given values in the formula

Centroid of a triangle = ((2+4+6)/3, (6+9+15)/3)

= (12/3, 30/3)

= (4, 10)

Therefore, the centroid of the triangle for the given vertices A(2, 6), B(4,9), and C(6,15) is (4, 10).