ENGR0135 - Statics and Mechanics of Materials 1 (2161)Homework #7

Solution Set

1. The shaded area can be divided into component areas in a number of different ways.For example, it could be divided into a large 8 in.×10 in. rectangle with two 3 in.×8 in.rectangular holes. Below, it is divided into a 8 in.×3 in. rectangle (1) and a 2 in.×8 in.rectangle (2).

x

y

1

2

The centroid coordinates and areas for these two component areas are

xC1 = 4 in xC2 = 4 in

yC1 = 1.5 in yC2 = 7 in

A1 = (8 in)(3 in) = 24 in2 A1 = (2 in)(8 in) = 16 in2

The x-coordinate of the centroid of the composite area is

xC =

∑xCiAi∑Ai

=(4 in)(24 in2) + (4 in)(16 in2)

(24 in2) + (16 in2)= 4 in

Actually, one could have deduced this result from the left/right symmetry of the shadedarea, without performing this calculation. The y-coordinate of the centroid of thecomposite area is

yC =

∑yCiAi∑Ai

=(1.5 in)(24 in2) + (7 in)(16 in2)

(24 in2) + (16 in2)= 3.7 in

Thus,xC = 4 in , yC = 3.7 in

2. Let the segments of the rod be labelled as shown below.

x

y

1

2

3

4

5

The centroid coordinates and lengths of these segments are

xC1 = 50 mm xC2 = 25 mm xC3 = 0 mm xC4 = 50 mm xC5 = 100 mm

yC1 = 75 mm yC2 = 100 mm yC3 = 50 mm yC4 = 0 mm yC5 = 75 mm

L1 = 50 mm L2 = 50 mm L3 = 100 mm L4 = 100 mm L5 = 150 mm

The x-coordinate of the centroid of the composite rod is

xC =

∑xCiLi∑Li

=(50)(50) + (25)(50) + (0)(100) + (50)(100) + (100)(150)

50 + 50 + 100 + 100 + 150= 52.8 mm

The y-coordinate of the centroid of the composite rod is

yC =

∑yCiLi∑Li

=(75)(50) + (100)(50) + (50)(100) + (0)(100) + (75)(150)

50 + 50 + 100 + 100 + 150= 55.6 mm

Thus,xC = 52.8 mm , yC = 55.6 mm

3. Label the 20 in×12 in rectangular area 1, the triangular hole 2, and the quarter-circularhole 3. The corresponding centroid coordinates and areas are

xC1 = 0 in xC2 = −8

3in xC3 =

4(8)

3π=

32

3πin

yC1 = 6 in yC2 =8

3in yC3 =

4(8)

3π=

32

3πin

A1 = (20)(12) = 240 in2 A2 =1

2(8)2 = 32 in2 A3 =

π(8)2

4= 16π in2

Treating the holes as “negative area,” the x-coordinate of the centroid of the compositearea is

xC =

∑xCiAi∑Ai

=(0)(240)− (−8/3)(32)− (32/3π)(16π)

240− 32− 16π= −0.541 in

and the y-coordinate of the centroid of the composite area is

yC =

∑yCiAi∑Ai

=(6)(240)− (8/3)(32)− (32/3π)(16π)

240− 32− 16π= 7.51 in

Thus,xC = −0.541 in , yC = 7.51 in

4. Label the cylinder 1 and the hemispherical cap 2. The corresponding centroid z-coordinates, volumes, and masses are

zC1 = 0.1 m zC2 = 0.2 +3(0.15)

8= 0.25625 m

V1 = π(0.15)2(0.2) = 1.4137× 10−2 m3 V2 =2π(0.15)3

3= 7.0686× 10−3 m3

m1 = ρ1V1 = 111.26 kg m2 = ρ2V2 = 32.73 kg

(a) From symmetry, it is observed that the centroid of the composite volume is onthe z-axis. The z-coordinate of the centroid of the composite volume is

zC =

∑zCiVi∑Vi

=(0.1)(1.4137× 10−2) + (0.25625)(7.0686× 10−3)

1.4137× 10−2 + 7.0686× 10−3= 0.1521 m

Thus,xC = yC = 0 , zC = 152.1 mm

(b) Assuming a uniform gravitational field, the center of gravity and the center ofmass are at the same location. From symmetry, it is also observed that the centerof gravity of the composite volume is on the z-axis. Noting that the centers ofgravity for each part of the body are at their centroids, the z-coordinate of thecenter of gravity of the composite volume is

zG =

∑zGimi∑mi

=(0.1)(111.26) + (0.25625)(32.73)

111.26 + 32.73= 0.1355 m

Thus,xG = yG = 0 , zG = 135.5 mm

5. One way to approach this problem is to treat the distributed load as a uniform 3 kN/mdistributed over the entire length of the beam plus a linear distributed load that goesfrom zero at x = 2 m to 4 kN/m at x = 6 m, as shown below. The resultant of each hasa magnitude equal to the area under the distributed load curve and a line of actionthat passes through its centroid.

x

y

A B

18 kN

8 kN

3m

4.6667m

These two resultants can then be replaced by a single resultant, as shown below:

x

y

A B

P

d

The magnitude of the resultant P is the sum of the two component resultants,

P = 18 + 8 = 26 kN

and the distance d from A to its line of action is determined by the requirement thatit have the same moment about A as the two component resultants,

Pd = (3)(18) + (4.6667)(8) =⇒ d = 3.51 m

6. There is a horizontal reaction force, B, at the roller support at B and a reaction forceat the pin support at C that has horizontal and vertical components, Cx and Cy.

AB

C

P

B

Cx

Cy

7. The rough surface exerts a reaction force on the box with normal, Rn, and tangential,Rt, components. The weight, W, is a vertical force whose line of action passes throughthe center of gravity of the box.

P

G

20◦25◦

Rn

Rt

W

8. The hinges are assumed to only transmit forces, because they are properly aligned.The line of action of the weight, W, passes though the center of the door, because thedoor is homogeneous.

x

y

z

A B

C

T

Ax

Ay

Az

Bx

By

Bz

W