ME2114-2014-15 solutions

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NATIONAL UNIVERSITY OF SINGAPORE

ME2114 - MECHANICS OF MATERIALS II

(Semester 2 : A Y20 14/20 15)

Time Allowed : 2 Hours

INSTRUCTIONS TO STUDENTS:

1.

Please write your Student Number only in the box above. Do not write your name.

2.

This assessment paper contains FOUR (4) questions and comprises TWENTY -ONE (21) printed pages.

3.

Students are required to answer ALL FOUR (4) questions.

4.

Students should write their answers in the space provided in this answer booklet.

5.

This is a CLOSED-BOOK ASSESSMENT. Students are NOT allowed to bring in any reference materials into the examination hall.

6.

Programmable calculators are NOT allowed for this assessment. Question Number

Marks Obtained

Maximum Marks

1

25

2

25

3

25

4

25

Total

100

YA u~ .:l

"---"

QUESTION 1 A solid shaft which has a radius of 30 mm and length 2m is made of an elastic perfectly plastic material. It is loaded by a slowly increasing torque Tp until a plastic zone has occurred partially in the shaft. Derive an expression for the torque Tp. Assuming the yield stress in shear Ty =150 MPa, G = 80 GPa, determine: (a)

the torque Ty at the on set of first yield,

(b)

the angle of twist if the torque is increased to 1.2 Ty,

(c)

the residual shearing stress at the outer surface after the torque of 1.2 Ty is removed. (25 marks)

DO NOT

WRIT E ON T HIS

MARGI N

J.n V~

I

QUESTION 2 A rigid horizontal bar pivoted at the left-hand end is loaded by a point load Pat the right-hand end as shown in Figure 2a. It is supported by a column, which is pinned at both ends and is restrained at mid-height from in-plane (x-y plane) displacement, but free to deflect out-of-plane (y-z plane). The column has a height of 300 mm and a rectangular cross-section of 5 x 15 mm as shown in Figure 2b. Assuming a factor of safety of 1.5, determine the maximum allowable load P that the column can support. Assume a Young' s modulus of200 GPa and a compressive yield stress of 300 MPa. State if buckling will occur in the x-y plane (in-plane) or y-z plane (out-of-plane). The Euler formula used must be derived from first principles. The solution to the differential equation:

2

d v +A-2v = 0

dx2

is given by: v

= C1 sin llx + C2 cosllx (25 marks) p

i

100

a

dimensions in mm

H

y

a

150

300

~}

X cross -section at aa

150

Fig. 2b

z

Fig. 2a

-------·--------4-~---------------------·

c Shear- stress distribution

The torque in the bar above is given by :

Tp =

J:

r (2nr)dr -r

(Elastic)

Tp =

+

LR

ry

2

2nr dr

(Plastic)

s: [r~r]

2

2nr dr +

r

Ty

2

2nr dr

lc

TP = -2Tr ry r 3dr + 2Trry JR r 2dr c 0 c

4 2TC { 3 3) =-rye +-ry \R -c 1C

2c

3

R3 c3 =2Trryl--3 12

rg. -=-

NATIONAL UNIVERSITY OF SINGAPORE Department of Mechanical Engineering Module Code:

Module Title: _ _ _ _ __ _ _ _ _ _ _ _ _ __

Solution to Question No.

~,-ve11 -M td-

"Zl' -=

I fD M P~

___j_

....._ ~'( :::: zY & -= -:J"-Cl)

-f-aY

-II

J~c{

t

=

-

Dy

I J-o

C"'~ ~> -~

= / f7j

)(/(!

L4- =- {2 7·~3 ?( 10t-

_L

32- (}0

~-v?t y/elcl.

(y

= ~ G fCL

G

"'rt'.t{

t)'Y)Wl

-foy-{ ueP

tyc;_

==

:r

Ly .T

C

ly =-

-

;s-o xJ27-23 x'!O

t-

30

-·b) C.,ve~ -+4tq;t

6.36 X/0~ N

T :::.. I· 2- T r

m ·M

~ I· 2- x 636 xiD ~ _{-

7 6J /(_ /0 lrJe. -/(p,v-e

I

3

-=- 2 IT Zy [--_g:_ 3. -

Ll\ /rD

c3

_-

e~;

12

-

1\JWI~

J 3 C ] /2-

=71/3

XI{ Ct)

SetBy: ___________________________

Date: _ _ _ _ _ _ _ _ _ _ __

rg. ?

NATIONAL UNIVERSITY OF SINGAPORE Department of Mechanical Engineering Module Code:

Module Title: _ __ _ _ _ _ _ _ _ _ _ _ _ __

I

Solution to Question No.

~

c' )

c =- d1·14-VI1 ~ tA.S·,-~j

@-4

Oy L

~

'oyL y

(J=

->

l-~?r>( IO

>(~

11 · 1~

m~

O· J 1 ~-I

C) 1/~

)-t C{)

0

s~ss-

vevj

&==

7 6·3 :>?rv]

f4-o6. :s

~~ tr->L ;_ri·J.

S=f-,