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Page 1 of 6 HW1 Due by: 11:59 PM on Monday, May 2, 2022 200 POINTS INSTRUCTIONS: • There are a total of 18 questions. • Everyone needs to work individually and upload the solution to canvas as a PDF....

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Page 1 of 6
HW1
Due by: 11:59 PM on Monday, May 2, 2022
200 POINTS
INSTRUCTIONS:
• There are a total of 18 questions.
• Everyone needs to work individually and upload the solution to canvas as a
PDF.
• Answer each question in an organized and orderly format.
• Write neatly and clearly.
• Making a suitable assumption while solving a problem is fine. However,
please make sure to clearly mention it in the solution.
1. Convert 1,700 lb.ft to appropriate SI units (10 Points).
2. Evaluate each of the following and express with SI units having an appropriate prefix (10
Points):
a. (16,000 mN)(17 GN)
. (1,800 mm)(0.57 MN)2
3. The screw eye in the below figure is subjected to two forces F1 and F2. Determine
the magnitude and direction of the resultant force (10 Points).
100 N
Adr Ma
Page 2 of 6

4. Determine the x and y components of F1 and F2 acting on the boom shown in the
figure below. Express each force as Cartesian vector (10 Points).


5. Determine the magnitude and the coordinate direction angles of the resultant
force acting on the ring as shown in the figure below (10 Points).


6. For each of the figure below, determine the moment of the force about point O
(15 Points).


7. Determine the moment produced by the force F in the Figure below about point
O. Express the result as a Cartesian vector (10 Points).
150 N
{-25j-50k} lb
20 kN 20 kN
20 ft
20 m
Page 3 of 6

8. The sphere in Fig. has a mass of 18 kg and is supported as shown. Draw a free-
ody diagram of the sphere, the cord CE, and the knot at C (10 Points).



9. Determine the tension in cables BA and BC necessary to support the 190-kg
cylinder in the figure below (10 Points).

10. Determine the force in each member of the truss shown below and indicate
whether the members are in tension or compression (10 Points)?

11. Determine the forces acting in all the members of the truss shown in the Figure
and indicate whether the members are in tension or compression (10 Points).
50 degrees
70 degrees
1,500 N
2.7 m
3.2 m
40 degrees
15 kN
5 m
Page 4 of 6

12. Determine the force in each member of the truss shown in the Figure and
indicate whether the members are in tension or compression (15 Points).



13. Determine all the zero-force members of the roof truss shown in the Figure (10
Points).


14. Determine the resultant internal loadings acting on the cross section at C of the
cantilevered bean shown in the Figure (20 Points).
15 kN
16 kN
1,200 N/m
18

32 m
500 N
5 m
Page 5 of 6

15. A uniform A-36 steel bar is shown in Figure below has a diameter of 80 mm and
is subjected to the loading shown. Determine the displacement at D, and the
displacement of point B relative to C (10 Points).
16. A solid shaft and tube shown in Figure are made of a material having an
allowable shear stress of 95 MPa. Determine the maximum torque that can be
applied to each cross section, and show the stress acting on a small element of
material at point A of the shaft, and points B and C of the tube (10 Points).
17. The assembly shown in Figure consists of an aluminum tube AB having a cross
sectional area of 400 mm2. A steel rod having a diameter of 10 mm is attached to
a rigid collar and passes through the tube. If a tensile load of 200 kN is applied to
the rod, determine the displacement of the end C of the rod. Take Est = 200
GPa, Eal = 70 GPa (10 Points).
95 MPa
200 kN
Page 6 of 6

18. The 1.5-in.-diameter shaft shown in Figure below is supported by two bearings
and is subjected to three torques. Determine the shear stress developed at
points A and B, located at section a-a of the shaft (10 Points).
40 kip.in.
    Final Exam
    Due by: 11:59 PM on Monday, May 2, 2022
Answered Same Day May 01, 2022

Solution

Sonam answered on May 02 2022
87 Votes
1) 1 lb-ft = 1.35582 N-m
Therefore, 1700 lb – ft = 2304.89 N-m
2) a. (16,000 mN)(17 GN) = [16000 * 10-3 ][17*109] = 272*109 N2
. (1,800 mm)(0.57 MN)2 = [1800*10-3][0.57*106] = 1026*103 N2
3) net force along horizontal
Fx = 135 cos 27 + 90 sin 18
Fx = 187.19 N
net force along vertical
Fy = 135 sin 27 + 90 cos 18
Fy = 121.51 N
magnitude of net force
F^2 = Fx^2 + Fy^2
F = 223.17 N
direction
x = arctan ( 121.51 / 187.19)
x = 33 deg

4) Given F₁= 150 N. F2 = 147N
Tanx = 5/12
X = 23 deg
Solve for forces in and re direction and represent Fi and Fin Cartesian vecto
· F, Sin 37 = 150 x Sin 37° = 90.27N.
F, cos 37 = 150 x Cos 37° = 119.79 N
So, F1= 90.271 +119.79 Ĵ
· F2 Sin 23° = 147 x Sin 23° = 57.43 N.
F₂ Cos 23° 147 x Cos 23° = 135 31 N.
So, F2 = 135·31 î+57.43 Ĵ·
5) Resultant force= FR = F1 + F₂
(35i+ 49j +63 k) + (-25j-50k)
= 35i +24j+ 13 k
magnitude |FR|= √ (35)² + ( 24 ) ² + (13) ² = 44.38 l
Direction angles.
A= Cos---1 (Ax/A) = Cos¹ (35/44.38) = 37.94°
B= Cos-1 (24 / 44.30) = 57.26°
Y = Cos-1 (13/44.38) = 72.97°
6) A. and B
Shift force 20KN to point A and Create one Clockwise Couple (20x4.5 KN-m) and Compressive force 20 KN
M0 = (20x4.5) KN-m
Mo = 20X4.5 = 90 KN-m = 90X10-3 N-m, Moment of force about point o
C
Moment of force about o
Mo = X1.75X Sin 27°
Mo = 99.3) N-ft (CW)
Mo = 0.3048 X 99.31 N-m
\Mo = 30.269 N-m
7) Position vector of A = 20 k
Position vector of B is: = 31 +17 j
Vectos AB is represented by :
AB = 3i +17j – 20k
unit vector along AB is:
UAB = 3i 17ĵ-20k/[√ 3² + 17² +20²...
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