1
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
Collinear
EEs satisfied by
Number of
default
independent EEs
∑ 𝑭𝒚 = 𝟎
1
∑ 𝑴𝒂𝒏𝒚 𝒑𝒐𝒊𝒏𝒕 𝒐𝒏 𝒙 = 𝟎
Concurrent at a
∑ 𝑴𝒐 = 𝟎
2
∑ 𝑴𝒙 = 𝟎
5
∑ 𝑴𝒙 = 𝟎
3
Point
Concurrent with a
Line
Parallel
∑ 𝑭𝒚 = ∑ 𝑭𝒛 = 𝟎
(b) (9 points) Draw the corresponding
Free Body Diagrams for the three
following cases
2
(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.
3
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:
4
Now do a cut through FH, FG and EG:
∑ 𝑀𝐺 = 0 = −𝐴 ∗ 24 + 200 ∗ 24 + 400 ∗ 18 + 400
∗ 12 + 400 ∗ 6 − 𝐹𝐹𝐻 ∗
− 𝐹𝐹𝐻 ∗
⟹
𝐹𝐹𝐻 =
6
√42 + 62
4
√42 + 62
∗6
∗ 4.5
−16800
= −2375.4 𝑙𝑏
7.07
⟹
𝐹𝐹𝐻 = 2375 𝑙𝑏 𝐶
∑ 𝐹𝑦 = 0 = 𝐴 − 200 − 400 − 400 − 400 − 𝐹𝐹𝐻 ∗
⟹
𝐹𝐹𝐺 =
⟹
4
√42 + 62
− 𝐹𝐹𝐺 ∗
4.5
√4.52 + 62
−1217.4
= −2029 𝑙𝑏
0.6
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.
5
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.
6
(b) (5 points) What is the reaction at A?
Solution: From the ABC member:
∑ 𝐹𝑥 = 0 = 𝐴𝑥 − 𝐹𝐵𝐷 sin 20
∑ 𝐹𝑦 = 0 = 𝐴𝑦 + 𝐹𝐵𝐷 cos 20 − 250
⟹
𝐴𝑥 = 271 𝑁 →
⟹
𝐴𝑦 = −496 𝑁 ↓
𝐴 = √𝐴𝑥 2 + 𝐴𝑦 2 = 565 𝑁 ↘
𝜃 = tan−1
𝐴𝑦
= 61.3°
𝐴𝑥
(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).
7
(d) (10 points) Determine the range of values of θ for which the base will not move.
8
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
̅
𝒙
̅
𝒚
̅𝑨
𝒙
̅A
𝒚
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
̅=
𝑿
∑𝒙
̅ 𝒊 𝑨𝒊 475862.8
=
= 𝟕𝟏. 𝟐 𝒎𝒎
∑ 𝑨𝒊
6685.8
̅=
𝒀
∑𝒚
̅ 𝒊 𝑨𝒊 146296.5
=
= 𝟐𝟏. 𝟗 𝒎𝒎
∑ 𝑨𝒊
6685.8
(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 3 1
𝑏ℎ = 20 ∗ 1103 = 8,873,333.3 𝑚𝑚4
3
3
9
𝐼4𝑥′ = 𝐼5𝑥′ =
𝐼2𝑥′ =
1
1
𝑏ℎ3 =
60 ∗ 403 = 320,000 𝑚𝑚4
12
12
1
1
𝑏ℎ3 + 𝑏ℎ ∗ 𝑑2 =
80 ∗ 303 + 80 ∗ 30 ∗ 1252 = 37,680,000 𝑚𝑚4
12
12
1
1
𝐼3𝑥′ = − 𝜋𝑟 4 − 𝜋𝑟 2 ∗ 𝑑 2 = − 𝜋 ∗ 104 − 𝜋 ∗ 102 ∗ 1252 = −4,916,592.5 𝑚𝑚4
4
4
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 = 𝑏⁄𝑎3
𝑘 ′ = 𝑎⁄𝑏 2
𝑦̅ ∫ 𝑑𝐴 = ∫ ̅̅̅̅
𝑦𝑒𝑙 𝑑𝐴
Using a horizontal element: ̅̅̅̅
𝑦𝑒𝑙 = 𝑦,
′
𝑦
𝑑𝐴 = 𝑥 𝑑𝑦 = [(𝑘 )
1⁄
3
− 𝑘′𝑦 2 ] 𝑑𝑦
𝑏
𝑦 1⁄3
3 −1/3 4/3 1
3
1
2
3
∫ 𝑑𝐴 = ∫ [( ) − 𝑘′𝑦 ] 𝑑𝑦 = [ 𝑘
𝑦 − 𝑘′𝑦 ] = 𝑘 −1/3 𝑏 4/3 − 𝑘 ′ 𝑏 3
𝑘
4
3
4
3
0
0
𝑏
3
1
5
= 𝑎𝑏 − 𝑎𝑏 =
𝑎𝑏
4
3
12
𝑏
𝑦 1⁄3
3 −1/3 7/3 1
3
1
2
4
∫ ̅̅̅̅
𝑦𝑒𝑙 𝑑𝐴 = ∫ 𝑦 [( ) − 𝑘′𝑦 ] 𝑑𝑦 = [ 𝑘
𝑦 − 𝑘′𝑦 ] = 𝑘 −1/3 𝑏 7/3 − 𝑘 ′ 𝑏 4
𝑘
7
4
7
4
0
0
𝑏
3
1
5
= 𝑎𝑏 2 − 𝑎𝑏 2 =
𝑎𝑏 2
7
4
28
𝑦̅ =
𝑦𝑒𝑙 𝑑𝐴
5
12
12
3
∫ ̅̅̅̅
=
𝑎𝑏 2 ∗
=
𝑏= 𝑏
28
5𝑎𝑏 28
7
∫ 𝑑𝐴
10
(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).
1
𝑑𝐼𝑥 = 𝑑𝐼1𝑥 + 𝑑𝐼2𝑥 = [𝑦13 − 𝑦23 ]𝑑𝑥
3
3
1
𝑥
𝐼𝑥 = ∫ {[√ ⁄ ] − [𝑘𝑥 3 ]3 } 𝑑𝑥
𝑘′
0 3
𝑎
𝑎
3
3
1
3
2
= ∫ {[√𝑥𝑏 ⁄𝑎] − [𝑏𝑥 ⁄𝑎3 ] } 𝑑𝑥
0 3
Show your work!
Good Luck!
11
Centroids of Common 1D Bodies
Centroids of Common 2D Bodies
12
Centroids of Common 3D Bodies
13
14
Moments of Inertia of Common Cross-Sections