Final Exam Solutions - College of Engineering

1

Name

MECH 223 ¨C 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

¡Æ ?? = ?

1

¡Æ ???? ????? ?? ? = ?

Concurrent at a

¡Æ ?? = ?

2

¡Æ ?? = ?

5

¡Æ ?? = ?

3

Point

Concurrent with a

Line

Parallel

¡Æ ?? = ¡Æ ?? = ?

(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:

? = 200 + 400 + 400 + 400 + 350 + 300 + 300 + 300 + 150 = 2800 ?? ¡ý

?? = ? ? ? = 400 ? 6 + 400 ? 12 + 400 ? 18 + 350 ? 24 + 300 ? 30 + 300 ? 36 + 300 ? 42

+ 150 ? 48 = 62,400 ?? ? ??

?

? = 22.3 ??

(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:

¡Æ ?? = 0 = ?? ? 24 + 200 ? 24 + 400 ? 18 + 400

? 12 + 400 ? 6 ? ??? ?

? ??? ?

?

??? =

6

¡Ì42 + 62

4

¡Ì42 + 62

?6

? 4.5

?16800

= ?2375.4 ??

7.07

?

??? = 2375 ?? ?

¡Æ ?? = 0 = ? ? 200 ? 400 ? 400 ? 400 ? ??? ?

?

??? =

?

4

¡Ì42 + 62

? ??? ?

4.5

¡Ì4.52 + 62

?1217.4

= ?2029 ??

0.6

??? = 2029 ?? ?

(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.

? ? ?? + ? ? ?? = ????? = 400 ? 42 + 300 ? 12 = ? ? 12

?

? = 1700 ??

(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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