2008 Midterm

Report 13 Downloads 183 Views
McGill University Department of Civil Engineering and Applied Mechanics CIVE-207A SOLID MECHANICS Midterm Exam Date: Oct. 27,2008, Time: 3:30-5:30PM, room: MC304

Notes:

1) This is a closed book examination. No crib sheet is allowed.

2) Calculators that can store text in memory are not permitted.

3) This questionnaire has 3 pages in total.

4) The examination will be marked out of 30.

5) You can keep the questionnaire.

Problem 1 (10 marks): Steel rods (A=0.05 in2 , E=30x10 3 ksi, a=6.5x10-6fF) are attached at points A and C to the "rigid" right-angle bracket in Fig. 1. The bracket is supported by a smooth pin at B, and the rods at A and C are initially stress-free. (a) Determine the axial stresses induced in rods (1) and (2) if the temperature of rod (1) is decreased by 50°F. Clearly indicate if they are tensile or compressive stresses. (b) Determine the transverse shear force VD and bending moment M D on the cross section at D that result from this temperature change. (c) Determine the angle of rotation of rigid bracket due to the temperature decrease in rod (l). Indicate if it is clockwise or counter clockwise.

1---5 in. -----1-5 in.-!

Fig. 1:

I

i

i

.

Ai

T'~~~ !

I 8 in.

I

I

5 in. (1)

1

(2)

c ..........""""""'""""~g I-

1

CIVE-207A

Oct. 27, 2008

Midterm Examination

Problem 2 (14 marks): The uniform aluminum-alloy shaft in Fig. 2 is attached to a rigid wall at the end A and is welded to a rigid flange at end C. (a) The holes in the flange were supposed to align with holes tapped in the wall plate, but actually an initial external torque (TS)i must be applied at B in given direction in order to rotate end C through an angle ~o = 5 degrees (180 degrees= n radians) to perfectly align the holes in the flange with the holes in the wall plate. What initial torque (TS)i (in N-m) is required? (b) Once the holes are aligned, the bolts are inserted and the flange is tightened to the wall plate, the applied torque (TS)i is removed. What is the maximum shear stress created in the shaft after the applied torque is removed? Sketch the internal torque diagram. (c) Subsequently the final torque (Ts)fis applied again at B in the same direction to its maximum. Determine the maximum torque (TS)f that may be applied if the resulting maximum allowable shear stress in the shaft does not exceed its allowable shear stress 'tan=120 MPa. Sketch the internal torque diagram when applied torque TS reaches maximum. (Givens: G=26 GPa, d=50mm, L[=2m, L2=lm. Using n=3.1416.) Fig. 2:

-reI

_1.

B ___- Lo'--- ,. JT-- . B

.

~---------- L 1-----.-----­

2

CIVE-207A

Midterm Examination

Oct. 27, 2008

Problem 3 (6 marks): A cantilever beam is subjected to two concentrated loads, and its section has the dimensions shown in Fig. 3. (a) Sketch shear force and moment diagrams for the beam, expressing the shear values in terms ofP and the bending moment values in terms ofPL, where L=4m. (b) If the allowable flexural stress (tension or compression) is O"au=180 MPa, determine the maximum value of load P that can be applied to the beam. The moment of inertia about the neutral axis is I=122x10 6 mm4.

Fig. 3:

P/4

68.6mm

P

NATI~

c

B

A

248.9 mm

t

~--2m~2m--+ll1!il Formula sheet: Axial deformation

Basic definitions Average nonnal stress in an axial member

.

F

(Javg

L AjE Fi L;

e=

or

i

j

Force-temperature-deformation relationship

Average direct shear stress

FL

e= -

V

+a!J.TL

AE

Av

Torsion

Average bearing stress F (Jh

e = FL AE

=A

'a\'g =

Elongation in axial members

Maximum torsion shear stress in a circular shaft

TR



'ma, =

T

Ah Average nonnal strain in an axial member e cavg = L

where the polar moment of inertia I p is defined as:

A verage normal strain caused by temperature change

Angle of twist in a circular shaft

eT

=11. (j,T

p

I = ~ rlf- r 4 ] = .2:. [D4 -,f1 p 2' 32 '

n



~r

01= A

FlO:: q 1',5" .elo ( \~?16t))

, =(\'lrt::sl (T)

I

Flo':: -2~ ="2(q=t;sj= - Iq~ £b ((.aUApt1eSS(O'1)

0;.: *=-3,qt:~i(CJ L~=(). Vo;' -~, ;: -q 1i S'" £.10

L~~, ~:: -

C9

uao- v'b\0.h;t\:

~

r:;crJ :: - 4-H, r ti(,-Itl,

e FL 2.

rJ..:: Cf

.' 2

It

(lqr ~ to ) (t' )

:: -eA (t)' ::~'="'o-X-( fS-,)-(0-,o-s---:i",t;:-)-(S"~/f-)

=- 2..c)%JC(D~ ~.

()b-

:=..

0, l) llq C>

(cc rAJ)

Ai

'to::

§"

0

S'=

0 3 \1\.l~ s-)C 't8'\)o ::; ~,o~=t2

~ = tlf'(4= i (3.\~lbJ(W'M~+" b\~S'l3 \A.\W\'t.

® ~ = Til. ITs\. o

q.~,

.:>

.::(o.Ol(}2)(ZblCl~;;1.)(6\?~3~4)

L(

2cn>o lMW\

(6, ~t2)(2.(,lC\6 ~)C (,\ ~ 3 1MJ-)

\ - (L\+LI.) .,.

Tb'f'jf ~

cPoii-Ip

'1(

(!) __ ~o C:r1r

Q~ (J,lA

30ooIM-1o\.\

- ::.

,Th .

_ \ \ ®\ l

pt.

o

1\ I\!IM,

Tor~d~r~'.

Fio'"~tz.) ,

Ti::: ~ [ll,X I~~) (('I~~3~)(O.~l)+~'lq.SXI~

T

~ ~ [ 131\' N';' t21\q.S'.v~J =2J~~ Nw.< ~)

x t

" ' - - - - ­ _ _- - J

lts

X€

f()~J We.. : ~\,~V\ T'2.= TA\' I T. ='2.1 b&NIAt .

cr.) ~ tiT,,:: 2J~ ~ Z.~IIS ::: ~1.( ~ N~ =: 0;) . f

btl b """"

l\(ZM4j- T2.(\"")~~!