Statics Of Rigid Bodies 6276u

Technological Institute of Technology

Sample Problems: Statics of Rigid Bodies Contents: •Structural Analysis •Friction •Moment of Inertia Prepared by:

Engr. RUEL M. GALUTAN

1. Determine the force in member AB,AC,BD,&BC of the crane truss shown in Fig.1

E

C

6' ' 2 1 30.0°

D

B

9'

A 5200 lb

2. For the truss shown in Fig. 2, Find the force acting in member BC & BD.

B

D

F

C

E

G

6'

6'

1000 lb

A 12'

1000 lb 12'

1000 lb 12'

H 12'

3. For the transmission tower shown in Fig. 3, determine the force in member CJ. D

10'

E

780 lb

12' F

C I 12'

G

B J 12'

H

A K 17'

17'

Upper portion of a-a

D

10'

E

780 lb

12' F

C 18’

12'

I

FG

a BC

B

a

BJ

JG J 26’

G

Upper portion of b-b 10'

D

780 lb

E

12' F

C

CJ b

B

BC

I

18’

FJ

b 12'

12 9

G J

26’

4. Compute the force in bars IG and HG of the scissors truss as shown in Fig. 4

12 K F

12 K

12 K

a

G

D 12 K

E

B

48'

12 K

C

I

H

a A

J 10'

10'

10'

10'

10'

10'

5. The pin-connected frame in fig. 5 s a load W=120lb. Find the hor. & ver. components of the pin force at C. Neglect the weight of the .

6'

4'

C

D

4'

F

B 4'

W A

4'

4'

3'

E

6. In the system shown in Fig. 6, Find the reaction at D caused by the 1200 lbs load. Neglect the weight of the .

D

1200 lb 4'

4

8'

B

A

3 5.8'

C

4.2'

CENTROIDS & MOMENT OF INERTIA 4. Given the properties from ASEP handbook, determine the centroids, moment of inertia and radius of gyration about x and y axes of the built-up sections. yo

C15x50

For W shape: d= 413 m A=9484 mm2 bf =179.6 mm tf = 16 mm tw = 9.7 mm

xo

yo

W16x50

xo yo xo xo

yo

For C shape: d= 381 m A=9484 mm2 bf =94.4mm tf = 16.5 mm tw = 18.2 mm xo = 20.27 mm Ix = 168.1 mm4 Iy = 4.58x106 mm 4

x2 =381/2

=190.5 mm

d =381 mm tw =18.2 mm

xo=20.27 mm

d =413mm

431.2 mm

tf =16.5 mm

tf =16.5 mm

C

y2

=410.93 mm

x1 =179.6/2 +16.5 y1

=413/2

y

=106.3 mm

=308.72 mm

=206.5 mm tw =9.7 mm

tf =16 mm

x

bf =179.6 mm

=431.2-20.27

=148.40 mm

Mohr’s Circle for Moment of Inertia -The circle represents all the possible values of I and P with respect to all axes ing through a specified point in an area. Steps: Draw I and P axes. Plot known values (Ix, Iy and Pxy). the points just plotted by a straight line. (Diameter of Mohr’s circle) Draw the circle.

Note: The angle between any two radii on Mohr’s circle is double the actual angle between the two axes of inertia.

C=(Ix+Iy)/2

P ( values of product of inertia)

R2=((Ix-Iy)/2 )2+(Pxy)2

Imax

Imin = C-R (min. moment of inertia)

Ix

Imax = C+R (max. moment of inertia) (Ix-Iy)/2

C=(Ix+Iy)/2

tan 2α = Pxy/(Ix-Iy)/2

Sin 2α = Pxy/R

X-axis

Imin

A (Ix-Iy)/2

Pxy

R

I ( values of moment of inertia)

2α

D

C

O

E

Principal axis

B

Where:

Pyx=-Pxy F y-axis

Iy R = radius of Mohr’s circle 2α = double the actual angle

y

V

U

α α

x

α = location of U&V axes relative to x& y axes. (actual angle) 2α = angle between two radii of Mohr’s circle x & y=usual axes, reference axes, actual axes U,V = inclined axes at an angle of α

4. A certain area has the

P

following properties with respect to x and y axes: Ix=40 in4, Iy=100 in4 and Pxy=40 in4. Determine the maximum moments of inertia and also the angle that the axis of maximum of inertia makes with the x-axis.

C=(Ix+Iy)/2

Imax Ix =40 in4 X-axis

A R

Pxy =40in4

O

D

Imin

E

2α

C

I

B -Pxy=-40in4

(Ix-Iy)/2

F y-axis

Iy =100 in4

Principles of Friction

FA = ʯN

When motion is impending…. W

Tan ɸ = ʯ

direction

bod y

P

A FA

rough surface

NA

RA ɸ

Y

Ɵ

W

X

B

u o R

s h g

e c a f r u

Ɵ

Ɵ N A

RA ɸ

Ɵ Y

W

X

u o R

B

Ɵ

RA ɸ Ɵ

NA

s h g

c a f ur

e

Ɵ Y

W

X

u o R

s h g

B

Ɵ

NA

ɸ R A

Ɵ

c a f ur

e

1. A 200 lb block is in with a plane inclined at 30° to the horizontal. A force P, parallel to and acting up the plane, is applied to the body. If the coefficient of static friction is 0.20, (a) find the value of P to just cause motion to impend up the plane, and (b) find P to just prevent motion down the plane.

Y

W

B

P

30°

X

2. Block 1 weighing 150 N rest on block 2 and is tied with a horizontal string to the wall. Block 2 weighs 50 N. If the coefficient of friction between 1 and 2 is 1/4 and between 2 and the surface is 0.30. Determine the tensile force @ chord & the force necessary to move block 2.

Block 1

150 N

Block 2

50 N

P

3. Two rectangular blocks A &B (Fig.3) having a weight of 3KN & 2KN , respectively, is ed by a solid strut attached to each block. The upper block B is resting on an inclined plane making an angle of 50 with the horizontal while block A is resting on the horizontal plane. The angle that the solid strut makes w/ the horizontal is 30°. The tangent of angle of friction is 11.5°. Find P to prevent motion. 2K

ɸ = 11.5

TAB

FA

RA

N

NB

60

3K N P

B

TAB 30.0°

A ɸ

NA

50

FB

ɸ

RB 50°

38.5

4. A horizontal bar, 10 ft long and of negligible weight, rests on rough inclined planes as shown in fig. If the angle of friction is 15°, How close to B may the 200 lb force be applied before motion impends.

ɸ = 15 FA

100 lb

A

R

N A

45 30

200 lb

x 30.0°

A ɸ

2'

10'

B

N

60°

FB

B

45

ɸ

RB 60

6. As shown in fig. 2, two blocks are separated by a uniform strut attached to each block with frictionless pins. Block A weighs 400 lb, block B weighs 300 lbs, and the strut AB weighs 200 lbs. If ʯ = 0.25 under B, determine the minimum 300 lbs coefficient of friction under A to prevent motion.

ɸB = 14.04

B

Bx

400 lbs

60

Ax Ay

FA RA

30.0°

A

NA

NB

200 Lbs

FB By 60°

60

ɸB

RB 45.96

5. The block A in fig. 6 s a load W = 1000 lbs and is to be raised by forcing the wedge B under it. The angle of friction for all surfaces in is ɸ = 15°. Determine the force P which is necessary to start the wedge under the block. The block and wedge have negligible weight.

A P

B

7. A pulley A is subjected to torque. The coefficient of friction between the belt and pulley is 0.40. Determine the torque (frictional moment), if P=2800 N.

T2 M(torque)

R=0.27

P pulley

β

A

T1 Direction of belt

belt