ENGR0135 - Statics and Mechanics of Materials 1 (2161)Homework #8

Solution Set

1. One should begin by drawing a free-body diagram for the beam, as shown below. Thereare horizontal and vertical reaction forces due to the pin support at A and a verticalreaction force due to the roller support at B. Distributed loads in a free-body diagramare replaced with their statically equivalent concentrated loads—the distributed loadin this problem was treated at two separate distributed loads.

x

y

A B

18 kN

8 kN

3m

4.6667m

Ax

Ay By

The support reactions are then determined by the equilibrium equations for the free-body diagram.∑

Fx = Ax = 0 =⇒ Ax = 0∑MA = 6By − (3)(18)− (4.6667)(8) = 0 =⇒ By = 15.2222 kN∑Fy = Ay +By − 18− 8 = 0 =⇒ Ay = 10.7778 kN

Thus,Ax = 0 , Ay = 10.8 kN ↑ , By = 15.2 kN ↑

2. There are horizontal and vertical reaction forces and a reaction couple due to the fixedsupport at B.

1500 lb · in

200 lb

1 ft 2 ft

A BBx

By

MB

∑Fx = Bx = 0 =⇒ Bx = 0∑Fy = By − 200 = 0 =⇒ By = 200 lb∑MB = MB − (1500 lb · in) + (36 in)(200 lb) = 0 =⇒ MB = −5700 lb · in

Thus,Bx = 0 , By = 200 lb ↑ , MB = 5700 lb · in �

3. Let T2 be the tension in rope 2 and consider the free-body diagrams shown below.

W

B

C1

2

W

C

T

T

T TT2

T2 T2

From the free-body diagram on the left,∑Fy = T2 + T2 + T − T −W = 0 =⇒ T2 =

1

2W

Then, from the free-body diagram on the right,∑Fy = T + T2 + T −W = 0 =⇒ T =

1

4W

Since W = (500 kg)(9.81 m/s2) = 4905 N,

T = 1226 N

4. Let F1 and F2 be the forces exerted on the cylinder by the ground and curb, respectively,and consider the free-body diagram shown below. Moment equilibrium about the centerof the cylinder shows that the line of action of F2 must pass through the center of thecylinder. Note the angle θ = sin−1 3/6 = 30◦.

F1

F2

θ

50 lb

P

Summing forces in the horizontal x-direction,∑Fx = P − F2 cos θ = 0 =⇒ F2 =

P

cos θ

Then, summing forces in the vertical y-direction,∑Fy = F1 + F2 sin θ − 50 = 0 =⇒ F1 = 50− F2 sin θ = 50− P tan θ

The force exerted by the ground is F1 = 50 lb when the force P = 0 and decreases asthe force P is increased. The cylinder will lift off of the ground when F1 = 0, whichcorresponds to a force

P =50

tan θ= 86.6 lb

5. Consider first a free-body diagram of the entire frame:

6 in 6 in

6 in

6 in

12 in

A

B

C

D

EF

100 lb

Ax

Ay

Fy

Since this frame is statically determinate, the support reactions at A and F can bedetermined from the equilibrium equations for this free-body diagram:∑

Fx = Ax − 100 = 0 =⇒ Ax = 100 lb∑MA = 12Fy − (24)(100) = 0 =⇒ Fy = 200 lb∑Fy = Ay + Fy = 0 =⇒ Ay = −200 lb

Now, take the frame apart and consider free-body diagrams for each of its members,remembering to obey Newton’s third law at each pin joint where the members areattached to each other. The directions of the internal reaction forces are chosenarbitrarily—their actual directions will be determined by their signs in the solutionsbelow.

6 in 6 in

C

EF

6 in

6 inB

E

6 in

6 in

12 in

A

B

C

D100 lb

100 lb

200 lb

200 lbBx

Bx

By

By

Cx

Cx

Cy

Cy

Ex

Ex

EyEy

Note that there are a total of nine remaining equilibrium equations (three for eachfree-body diagram), but only six unknown forces (Bx, By, Cx, Cy, Ex, and Ey). This

means that not all of the equilibrium equations will be required and that there aremultiple ways to determine the unknown forces. One way is shown here.

First consider equilibrium of member CEF :∑MC = 6Ey + (12)(200) = 0 =⇒ Ey = −400 lb∑Fy = −Cy + Ey + 200 = 0 =⇒ Cy = −200 lb

Next consider equilibrium of member BE:∑Fy = −By − Ey = 0 =⇒ By = +400 lb∑MB = −6Ex − 6Ey = 0 =⇒ Ex = +400 lb∑Fx = −Bx − Ex = 0 =⇒ Bx = −400 lb

Finally, again consider equilibrium of member CEF :∑Fx = −Cx + Ex = 0 =⇒ Cx = +400 lb

The results can now be shown using report diagrams:

C

EF

B

E

A

B

C

D100 lb

100 lb

200 lb

200 lb 200 lb

200 lb

400 lb

400 lb400 lb

400 lb

400 lb

400 lb

400 lb

400 lb

400 lb

400 lb

6. (a) Note that members AB and AF are two-force members and consider the followingfree-body diagram. From symmetry, the vertical force at D is half of the 100 lbweight.

12 in

12 in

B

C

D

6 in 6 in

6 in

100 lb

A

B F 45◦45◦45◦

FB

FB

FF

9 in

Cx

Cy

50 lb

Dx

From the free-body diagram for members AB and AF ,∑Fx = −FB cos 45◦ + FF cos 45◦ = 0 =⇒ FF = FB∑Fy = 100− FB sin 45◦ − FF sin 45◦ = 0 =⇒ FB = 70.7107 lb

Then, from the free-body diagram for member BCD,∑Fy = FB sin 45◦ + Cy − 50 = 0 =⇒ Cy = 0∑MC = 12Dx − (9)(50)− (12)(FB cos 45◦)

− (6)(FB sin 45◦) = 0 =⇒ Dx = 112.5 lb∑Fx = Cx +Dx + FB cos 45◦ = 0 =⇒ Cx = −162.5 lb

Using a report diagram to show the forces acting on member BCD,

B

C

D

45◦

50 lb

70.7 lb

112.5 lb

162.5 lb

(b) The shearing stress on the cross section of the 1/4-in.-diameter pin at B is

τ =70.7107 lb

π(0.125 in2)= 1441 psi

7. (a) Let FA and FB be the axial loads in bars A and B and let LA = 100 mm andLB = 200 mm be their lengths. They each have elastic modulus E = 210 GPaand cross-sectional area A = 100 mm2. Then, their changes in length are

δA =FALA

AE, δB =

FBLB

AE

However, the kinematic constraint on the motion of the system is that δA = 2δB,so

FALA

AE= 2

FBLB

AE=⇒ FA = 2

(LB

LA

)FB = 4FB

Now, consider a free-body diagram for member CD:

C

D

50mm

100mm 100mm

P = 10 kN

Dx

FB

Dy

4FB

∑MD = (200)(10)− (50)FB − (100)(4FB) = 0 =⇒ FB = 4.4444 kN

=⇒ FA = 17.7778 kN

Thus, the stresses in bars A and B are

σA =FA

A= 177.8 MPa , σB =

FB

A= 44.4 MPa

(b) The deflection of pin C is twice the change in length of bar A,

δC = 2(17, 777.8 N)(0.1 m)

(0.0001 m2)(210× 109 N/m2)= 1.693× 10−4 m