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Name

MECH 223 – Engineering Statics

Final Exam, May 4th 2015

Question 1 (20 + 5 points)

(a) (8 points) Complete the following table

Force System Free Body Diagram EEs satisfied by

default

Number of

independent EEs

Collinear

∑ 𝑭𝒚 = 𝟎

∑ 𝑴𝒂𝒏𝒚 𝒑𝒐𝒊𝒏𝒕 𝒐𝒏 𝒙 = 𝟎

1

Concurrent at a

Point

∑ 𝑴𝒐 = 𝟎 2

Concurrent with a

Line

∑ 𝑴𝒙 = 𝟎 5

Parallel

∑ 𝑴𝒙 = 𝟎

∑ 𝑭𝒚 = ∑ 𝑭𝒛 = 𝟎

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(b) (9 points) Draw the corresponding

Free Body Diagrams for the three

following cases

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(a) (3 points) In the drawing to the right, the crate is kept in

equilibrium on an inclined rough surface as shown. What are

the two extreme cases and what is the direction of the friction

force in each of these cases (state and explain the cases, no

need to calculate).

Answer: The two extreme cases are (1) the block is about to slide down (friction is up the slope),

and (2) the block is about to be pushed up (friction is down the slope)

(b) (Bonus - 5 points) In each of the following

cases express the cable tension T in terms of

the weight of the crate.

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Question 2 (20 + 5 points) A Polynesian, or duopitch

roof truss is loaded as shown.

(a) (5 points) What is the distance from point A

that the line of action of the resultant of the

external loadings crosses the base of the truss?

Solution:

𝑅 = 200 + 400 + 400 + 400 + 350 + 300 + 300 + 300 + 150 = 2800 𝑙𝑏 ↓

𝑀𝐴 = 𝑅 ∗ 𝑑 = 400 ∗ 6 + 400 ∗ 12 + 400 ∗ 18 + 350 ∗ 24 + 300 ∗ 30 + 300 ∗ 36 + 300 ∗ 42

+ 150 ∗ 48 = 62,400 𝑙𝑏 ∗ 𝑓𝑡

⟹ 𝑑 = 22.3 𝑓𝑡

(b) (10 points) Determine the forces in members FE, FH and FG.

Solution:

From the analysis of the whole truss:

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Now do a cut through FH, FG and EG:

∑ 𝑀𝐺 = 0 = −𝐴 ∗ 24 + 200 ∗ 24 + 400 ∗ 18 + 400

∗ 12 + 400 ∗ 6 − 𝐹𝐹𝐻 ∗4

√42 + 62∗ 6

− 𝐹𝐹𝐻 ∗6

√42 + 62∗ 4.5

⟹ 𝐹𝐹𝐻 =−16800

7.07= −2375.4 𝑙𝑏

⟹ 𝐹𝐹𝐻 = 2375 𝑙𝑏 𝐶

∑ 𝐹𝑦 = 0 = 𝐴 − 200 − 400 − 400 − 400 − 𝐹𝐹𝐻 ∗4

√42 + 62− 𝐹𝐹𝐺 ∗

4.5

√4.52 + 62

⟹ 𝐹𝐹𝐺 =−1217.4

0.6= −2029 𝑙𝑏

⟹ 𝐹𝐹𝐺 = 2029 𝑙𝑏 𝐶

(c) (5 points) If the external force at point B is removed, what the external force at K needs to

be in order for the forces in AB and AC to remain the same as in part (b)? No need to

calculate the actual forces in the above members.

Solution: for AB and AC to remain the same, the reaction at A needs to stay the same. In order

for the reaction at A to remain the same, the moment around N created by the external loading

needs to remain the same.

𝐵 ∗ 𝑑𝑏 + 𝐾 ∗ 𝑑𝑘 = 𝑐𝑜𝑛𝑠𝑡 = 400 ∗ 42 + 300 ∗ 12 = 𝐾 ∗ 12

⟹ 𝐾 = 1700 𝑙𝑏

(d) (Bonus - 5 points) If the external force at point B is removed, by examination, what are

the zero-force members? Explain.

Solution: joint B is a special case => BC=0. Now joint C becomes a special case, leading to

CD=0.

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Question 3 (30 points) The press shown to the right is used

to emboss a small seal at E.

(a) (10 points) Knowing that the coefficient of static

friction between the vertical guide and the embossing

die D is 0.30, determine the force exerted by the die

on the seal.

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(b) (5 points) What is the reaction at A?

Solution: From the ABC member:

∑ 𝐹𝑥 = 0 = 𝐴𝑥 − 𝐹𝐵𝐷 sin 20 ⟹ 𝐴𝑥 = 271 𝑁 →

∑ 𝐹𝑦 = 0 = 𝐴𝑦 + 𝐹𝐵𝐷 cos 20 − 250 ⟹ 𝐴𝑦 = −496 𝑁 ↓

𝐴 = √𝐴𝑥2 + 𝐴𝑦

2 = 565 𝑁 ↘

𝜃 = tan−1𝐴𝑦

𝐴𝑥= 61.3°

(c) (5 points) The machine base shown to the right has a

mass of 75 kg and is fitted with skids at A and B. The

coefficient of static friction between the skids and the

floor is 0.30. If a force P of magnitude 500 N is applied

at corner C explain the two modes of motion possible

for the base.

Solution: the two extreme cases are (1) the block sliding (small θ), and (2) the block tipping by

rotating around B (large θ).

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(d) (10 points) Determine the range of values of θ for which the base will not move.

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Question 4 (30 + 5 points)

(a) (10 points) For the x-y coordinate system,

determine the location of the centroid of

the composite beam in the drawing to the

right. (Use the tables at the end).

Solution:

A �̅� �̅� �̅� 𝑨 �̅�A

1 2200 70 15 154000 33000

2 2400 70 85 168000 204000

3 -314.2 45 85 -14137.17 -26703.5

4 1200 100 -26.7 120000 -32000

5 1200 40 -26.7 48000 -32000

Total 6685.8 475862.8 146296.5

�̅� =∑ �̅� 𝒊𝑨𝒊

∑ 𝑨𝒊=

475862.8

6685.8= 𝟕𝟏. 𝟐 𝒎𝒎

�̅� =∑ �̅� 𝒊𝑨𝒊

∑ 𝑨𝒊=

146296.5

6685.8= 𝟐𝟏. 𝟗 𝒎𝒎

(b) (10 points) Calculate the moment of inertia of the cross section of the composite beam in

(a) relative to the x’ axis. (Use the tables at the end).

Solution:

𝐼1𝑥′ =1

3𝑏ℎ3 =

1

320 ∗ 1103 = 8,873,333.3 𝑚𝑚4

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𝐼4𝑥′ = 𝐼5𝑥′ =1

12𝑏ℎ3 =

1

1260 ∗ 403 = 320,000 𝑚𝑚4

𝐼2𝑥′ =1

12𝑏ℎ3 + 𝑏ℎ ∗ 𝑑2 =

1

1280 ∗ 303 + 80 ∗ 30 ∗ 1252 = 37,680,000 𝑚𝑚4

𝐼3𝑥′ = −1

4𝜋𝑟4 − 𝜋𝑟2 ∗ 𝑑2 = −

1

4𝜋 ∗ 104 − 𝜋 ∗ 102 ∗ 1252 = −4,916,592.5 𝑚𝑚4

𝐼𝑥′ = 𝐼1𝑥′ + 𝐼2𝑥′ + 𝐼3𝑥′ + 𝐼4𝑥′ + 𝐼5𝑥′ = 41,956,740.8 𝑚𝑚4

(c) (10 points) Calculate by integration the y coordinate of the

centroid of the shaded area in the drawing. Express your

answer in terms of a and b. Solutions by other methods

will carry no credit!

Solution:

𝑘 = 𝑏𝑎3⁄ 𝑘′ = 𝑎

𝑏2⁄

�̅� ∫ 𝑑𝐴 = ∫ 𝑦𝑒𝑙̅̅ ̅̅ 𝑑𝐴

Using a horizontal element: 𝑦𝑒𝑙̅̅ ̅̅ = 𝑦, 𝑑𝐴 = 𝑥′𝑑𝑦 = [(𝑦

𝑘)

13⁄

− 𝑘′𝑦2] 𝑑𝑦

∫ 𝑑𝐴 = ∫ [(𝑦

𝑘)

13⁄

− 𝑘′𝑦2] 𝑑𝑦𝑏

0

= [3

4𝑘−1/3𝑦4/3 −

1

3𝑘′𝑦3]

0

𝑏

=3

4𝑘−1/3𝑏4/3 −

1

3𝑘′𝑏3

=3

4𝑎𝑏 −

1

3𝑎𝑏 =

5

12𝑎𝑏

∫ 𝑦𝑒𝑙̅̅ ̅̅ 𝑑𝐴 = ∫ 𝑦 [(𝑦

𝑘)

13⁄

− 𝑘′𝑦2] 𝑑𝑦𝑏

0

= [3

7𝑘−1/3𝑦7/3 −

1

4𝑘′𝑦4]

0

𝑏

=3

7𝑘−1/3𝑏7/3 −

1

4𝑘′𝑏4

=3

7𝑎𝑏2 −

1

4𝑎𝑏2 =

5

28𝑎𝑏2

�̅� =∫ 𝑦𝑒𝑙̅̅ ̅̅ 𝑑𝐴

∫ 𝑑𝐴=

5

28𝑎𝑏2 ∗

12

5𝑎𝑏=

12

28𝑏 =

3

7𝑏

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(d) (Bonus – 5 points) Using vertical area element derive the integral for the moment of

inertia of the shaded area in part (c) relative to the x axis. Express your answer in terms of

a and b. (No need to solve the integral to produce the final answer for the moment of

inertia).

𝑑𝐼𝑥 = 𝑑𝐼1𝑥 + 𝑑𝐼2𝑥 =1

3[𝑦1

3 − 𝑦23]𝑑𝑥

𝐼𝑥 = ∫1

3{[√𝑥

𝑘′⁄ ]3

− [𝑘𝑥3]3}𝑎

0

𝑑𝑥

= ∫1

3{[√𝑥𝑏2

𝑎⁄ ]

3

− [𝑏𝑥3

𝑎3⁄ ]3

}𝑎

0

𝑑𝑥

Show your work!

Good Luck!

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Centroids of Common 1D Bodies

Centroids of Common 2D Bodies

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Centroids of Common 3D Bodies

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Moments of Inertia of Common Cross-Sections