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NEWTON'S LAWS

I. Every body continues in a state of rest or uniform motion unless it is acted upon by a force.

2. The rate of change of momentum is proportional to the applied force and takes place in the direction of the force.

3. To every action there is an equal and opposite reaction.

Momentum is Mass x Velocity i.e. mu

P oc d1dt(mu) where P is the applied force

P oc mdu/dt when mass is constant

p= Km a where a is acceleration.

P=ma (Kg mls')

SCALAR AND VECTOR QUANTITIES

Speed is a scalar quantity: e.g.: 50Kmlh but 50Kmlh in the N.E. direction is a vector quantity.

A

TELOCITY REPRESENT A TION

o

Mass The mass ofa body is a measure of the quantity it contains and it is measured in Kg.

Force It is that which causes or tends to cause a change in the motion of a body. It is measured in N N= Kgmls'

If a change of motion is prevented the force will cause a deformation or change in the shape of a body.

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To describe a force completely we need to know the fOllowing:

I. Magnitude 2. Point of application 3. Line of action

If all three are known we have a vector quantity.

TRIANGLE OF FORCES

In order that a body may be in equilibrium under the action of three forces they must either be concurrent or parallel.

20KN 30KN

10m

x

R

Conditions of equilibrium- Co-planar forces

a. Algebraic sum of forces in any two mutually perpendicular direction must be zero.

b. Algebraic sum of moments of forces about any point in the plain must be zero.

LFx=O LFy=O LMo=o+J

R-20-30=0 i.e. R=50KN

MOMENT OF FORCE~ MAGNITUDE OF FORCE X PERPENDICULAR DISTANCE from the point to line of action of force 20 KN

(30 X 10)- Rx=O 30xlO-50x=0 x=3015=6 (force which has the same effect as original forces.

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Resultant and equilibrant are equal and opposite.

Resultant 50KN

Equilibrant 50KN

Concurrent: If three concurrent coplanar forces are in eqilibrium they can be represented by the three sides of a triangle.

Bow's notation

A C

B

c A forces AB, BC, CA ~ scale lcm= ..... KN

b a

Resultant of force systems, components of forces.

~FY=Fsine

Fx_Fcose

Fx+Fy=F FcosB+FsinB=F

4 + 3=5 (Fcose)'+(Fsine)'=(F)'

\j (F)'(cos'e+ sin'e)~

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Determine the Resultant of the given force system with respect to A.

600N

for equilibrium:

LFx=O LFy=O LMO=O

Resultant: Rx=LFx

".

'. " .

1m ~

300N

'. ". 300N

4 ON 4l~00N 300

500

Rx=300+300-600=0

t Ry=-400+400=0 R= [(Rx)' +(Ry )2]112

400

M, = -(300x2)-(300x4) + (400x3) + (600x1) = 0

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The resultant of the three forces and an unknown force P acting at E is a vertical force through A. Determine the force E and the resultant force at A.

ltv. (~ntl'.1n.:.J1+5 <''7 ~ ... "I! tI;)'~~Ii=~ ,-.:.

, l "1 $)

R is the resultant of the remaining forces.

Rx=Ex+75-30=0 => Ex=-45 ....... (2)

Ry = Ey-60 + l00 = R => R= Ey+40 .... (2)

ICC' .. 12~

7S

The best point to t.ke moments is E (bec.use Ey WILL BE THE UNKNOWN)

~ R(5 + 5cos60)=(-60x5)+(I00x1O)-(30x5), where 5 is the radius .nd 5 cos 60 is the side of the triangle. 7.5R=-300+ 1000-150=550

From (2) ...... R=Ey+40=73.3 => Ey= 33.3N

33

a (

45 Force through E = [(45)'+(33.3)'J I12 =55.9N=P .nd. = t.n·133.3 /45=36.5°

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- , - ,

II I (0 10D

,8..

Do.term;~ IoNL P M¢ he ~J/c" ~N.R..N-1i oo

.. ...

H re",u"clIU" W, flOV v

y (0,,'1' c-ruw..tJ

~,,= p~ + 1'£ - 30 = 0 .

+h~ % '~t ~_o

\',,+";-30= 0

1',,-= -75 .30

K~ -<)5 - d:::::=-

R(,.\({)sw) =.. r-GOI..S) + ( /00" d - ( ]O;<''')

/oo "R =- - '300 ..L 1000- I ,"0 :: '5"5 0

Q:: 73 33 ~=----=-:...:-~

73 3J "'40 - _P..LI_-_-_~_oo _33_ ;z:::,

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Determine the resultant of the coplanar system shown and locate with respect to A.

3

.ON 70N

lOON

arc tan3/4=36,86° triangles ABC and DEZ have their sides vertical to each other: i.e. ACB=DZE

Rx= 160-40= 120N Ry= 120-70-100=-50N R=(l20'+50')' 12 R=130 N

.=t.n·'501120, .=22,6·

_ >O~" P

C Take moments aboul A: (40X5)-(looXIO)+(120X8)+(160X4)=800N

M= f x d : 800=130 x d: d=8ool 130=6.15m

................... .. .. . ~.

R

A

T Il.<. r ~s v I ~ "'nt- 's +roM +1..<. hor-'2onta.L

- 6 . " '" .fro,," A

o - 22 . 6

<1"ct

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IOmm xmm

B A

c 20mm

5Kg E

o

480mm

The mechanism shown is part of a load measuring device. Link CD is vertical, and links AB, DF are horizontal. Pin joints at C, D, and F. Top link rest on a device edge at B. The vertical load E is measured by the distance X from B of a counter weight of mass 5Kg. If the load is 4 lannes calculate the distance X.

Solution:

Take moments about F: Tg(500)-(4xIO')g(20) =O

T=(80XIO')/500= 160Kg

Take moments about B: 5g x X- TgxIO =O X=160015=320mm

A charge furnace consists of a trolley and a counterweight A , mountaid on rails B. C and carrying a scoop D. The trolley counterweight, scoop and handle together weigh 117.6 KN and their centre of gravity is on a vertical line between Band C. Calculate the greater safer load in the scoop in KG

I, 5m J

117.6KG~=r;~ _____ ..J;D:...J

J;m,;t ( Take moments about C:

W x (5-1)-(117.6x1)=0 W= 117,6/4=29.4/9.81 =3XlO' KG

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Equilibrium of coplanar DOD concurrent forces

Graphical construction

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To find the equilibrant of the three [orces shown

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J)Ool5r' ¢tc<t-;c.,

'bCo.l~ : len", ?.G\l

/ / "

/ E?ju, 1, brant: .~o f't> s u l!.J.o If'" .:. Gt b

1.) draw force polygon. Closing letter da is the equilibrant. ad is the resultant of the three forces. 2.) choose a pole 0 inside the force polygon and draw the rays OA, OB, OC, OD. 3.) Draw line in the space B parallel to ray ab. Complete link polygon by drawing lines parallel to ac, ad, oa The intersection of the links parallel to Od and On will be a point in the line of action of the equilibrant da. Thus the equilibrant of the three forces is force da whose magnitude and direction is found from the force polygon which passes through intersection of the links [II and IIII of the link polygon.

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l1he f'eJUlt0f7t hcu +0 ho 1ft -fAt- rome inSmel'ff- a 1 rn€ loru.J) -Ksi n60 (d) = -7 (iJsinGO') -+10+ (6WSGO) - 10(6) -:;sli160Ciz

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F5'i n 6L l'b . :<3 (l) -fWJ<9 =-Ij- 3f, - CD

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'2 \

DYN. D.2

1) A bullet 18 fired with a muzzle velocity of 700 m/s. If the length of the barrel 1s 1 a. what 1s the average acceler~tlon1

(Answer: 245,000 mls )

2) An automobile 2 m/s2. How long travelled.

1s decelerRted from a speed of 20 m/s w111 it take to come to rest and how far

(Answer: 10 8. 100 m)

at the r ate of will 1 t have

3) Boy A throws a b:'1ll vertically up wi th a speed.,r 10 m/s fro II! the t op o f a w:J.ll 2.5 m high. Boy B on the ground at the I'J -me instant throw!:; a ball vertically up with a s'Peed 0 r 12 _/s. Determine the time at which the two b~lls 1'1111 be at the 8all8 height ahove the ground. What 1s thi s height?

(Answer: 1.25 s, 7.34 m)

4) A particle of dirt falls from an elevator that 1s moving up with a velocity of 3 m/s. If the particle reaches the bottom of thn ~levator shaft in 2 6. how high above the bottom of the elevator abaft "as th e elevator when the particle Gtarted fal ling.

(Answer: 13.62 m)

5) A r a dar equinped police car notes n car travelling at 110 km/h. The police CDr s tart R pursuit 30 s after tho observation and accel~r"t ~

to 160 ltm/h in ?O s. A:~eum1n g the r;nep.ds nre maintained 'In Cl s traigh t road, ho w far from the observation p~st will the chase end.

(Anewer: 3.91 km)

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UNIFORM MOTION

CONSTANT ACCELERATION OR DECELERATION

Velocity: A vector quantity !mown in magnitude and direction.

Initial velocity u (m/s) Final velocity V (m/s)

Acceleration: vector quantity-rate of change of velocity (m/s f ) or (m/s2)

Distance: travelled S metres (m) I l n l €,. : ..5" cond [ s \

VELOCITY TIME GRAPH

VELU

(m/s) ~( .,' ___________ YL.l "-"body has initial velocity u m/s at t=0 '" and accelerates uniformly to velocity

u (m/s) at time t seconds

o r(TIME) (s) Acceleration = rate of change ofvelocity= dv/dt

a9f-tL -t-

1f=U+at .... (I)

slope of'\f~t

Distance traveled is velocity x time = area under the graph \/-t

S=(u+u)/2 x t or II2(u+v)t.. .... (2)

su\:~)) into(2) gives: S= I 12[ u+u+atjt

S= ut+ 112at' ...... . (3)

(I) squared: J'=u2+a't2+2uat = u2+2a(ut+ I l2at')

.if=u2+2as ....... ( 4)

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ANGULAR UNIFORM MOTION

Initial velocity w1 rod/s Final velocity w2 rodl s a acceleration rod/s' 8 distance rod t time s N rpm

Mechanical sheet 1

!.A=D

w=2nNI 60 rod/s

w2=w1+ot _ 10

8=1I2(w1+w2)t _20 8=w It+ 1/20t' _.30 w,' = w,'+208 _ 40

ex l )C. ....r=7oe:::. MIS T,~:,----------~~= 4 - • I ~ - .

7cO

acceleration 0=700/t m/s - (1)

Distance S=1I2X 700 X t 350t = L (2)

tL-_---'~'___'_-,-'--'--_'Fram •.. (2) time s

(1) becomes: 700111350=0 :. 0=245,000m/s'

or u=O u=700m/s u'=tt+2os s=lm 700'=0+20 x 1 o=? m/s 0=700x700=245,OOOm/s'

t=1I350s

C'

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SIGN CONVENTION

Velocity and acceleration are regarded as positive if they are positive in the direction in which the distance is regarded as positive .

2/ . Velocity

20 acceleration =2mfsl

, a = dv/dl = (0-20)/(1- 0) = -2m/s' => 1= 105

Distance travelled is area under the u-t graph

s= 1/2x20xlO= 100m

MOTION UNDER GRAVITY

Any object in motion above the earth surface is subjected to acceleration due to gravity- which may be taken as constant in problems in applied mechanics and equal to 9.8m/s2 and is directed towards the centre of the earth.

mass mKg

a=g '--+---'

F=ma=mg (the weigt of the body)

e.g. if m= lOKg weight =mg= lOx9.8=98N

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From mechanical sheet I ,exercise 4

A :::U:="'II)I~Z~O 10 ·················

............. .. . ... ......... 12\ 'CJ

Upward motion CD - ®

u=3m/s u=O a=-9,8m/s2

t=? S=h\ m

U2 =u2 +2as

0=2(-9.8)hl 9/19.6=hl =0,459m

S= II2(u+u)l=hl 112(3 +O)tl =0.459m

II =(2x0.459)/3 =0.3065

Downward motion

u2=0 a=g=9.81 S=h1+h2 t=2-0.306s

S=ut+I/2at' S=0+0.5X9.81X(I.694)'

Height of elevator above the bottom of shaft when particle start falling : h,= 14.075-0.459=llJiI.

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D2/5 No3 HMIV

Vel

<\4<4 H •

.. .........•.. / ,:,.~./' ....

110 Km/hr= llOxtOl /3,6xlOl =30.56m/s

160Kmfhr= 160,. lQl m/33.6x 103 =44.44m/s

slope of the graph = area

Area under the graph = distance travelled

t

Distances travelled by two vehicles are the same if pc catches c after 2 seconds

:.30.561 ~ 112(44.44)20+44.44(1-50) :.30.561 = 444.4+44.41-2222 : .13.861 = 129.945

: .Distance from observation post = 30.561 = 30.56:d29.94 = 3.911 Km

Acceleration of police car=44.4 /20=2.22 ("" ,f)

SHEAR FORCE AND BENDING MOMENT

Consider simply supported beam carying single concentrated load W as shown. Divide the beam into two parts at section xx distance x from L.H. end.

y

R1 R2

Summation of forces in y direction

LFy=O => R,+R2=W : . R,=W-R2 .... ... (J)

ConSidering left hand position of the beam:

R,-SF=O .·. SF=R,···········································G

Consider the R.H. position of the beam:

SF+R2=W or SF=W-R2 ... .. ... .. ... ..... .. @

The internal force is called the shearing force and is the algebraic sum of the forces to either sides of the section, were the shear force is required . Considering the equilibrium of the section there would be an unbalanced moment of R,x. This is the external bending moment at xx, and is balanced by the internal moment resistance of the section.

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THE BENDING MOMENT (B.M.I:M

..... at any section of the beam is the algebraic sum of the moments of forces to either side of the section.

L

M",,=R1X-W1(x-a) or R2(L-x)-W2{(L-x)-b}

SIGN CONVENTION

X

I

X X

X

I I

X

(NEGATIVE)

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Simple supported beam with concentrated load W at mid-span w

~2 ,~ " ~ X>O<U2 X>U2<L

" '. " , SFxx=W/2 SFxx=W/2-W= . , = -W/2 " .....

W/2, X X !W/2

;.. X

• L ,.:

+ KN !----1'-----1 SHEAR FORCE DIAGRAM

MXX=WXJ2

X=O/ X=U2 M=o 5 M=WU4

W' U , 4

Mxx=W/2 X-W(X-U2) = -W/2X+WU2 (X=L

W/2(L-X), X=U2, M=WU4 M=O

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Simply supported beam with uniformly distribyted load (UDL). along the entire length.

x W/unit L

L ~ :

WU2 ~ .

. ~ ; :

Totalload=WL KN/m m

Ifl. w SFxx=- -WX= - (L-2X)

2 2

X=o ]

SF= Ifl. 2

_L X--2

SF=O ]

: -WU2

X=L

_Ifl. SF---2

HOOKE'S LAW STRESS cr B

' 0

plastic zone

o,L---~----------~

STRAIN e

HOOK'S LAW: cr/e=E (Young's modulus of elasticity)

C in N/m2, KN/m2, MN/m2

c

A: limit of proportionality-elastic limit B: yield point C: maximum load 0 : fracture

specimen :gauge length 50 mm

C.SA=1td2~ (original csa is used)

Yield stress cry= LOAD AT '8' original C.S-.A.

Ultimate tensile stress (U.T.S.)= LOAD AT 'C' original C.S.A.

Nominal stress at fracture D= LOAD AT FRACTURE 'D' original C.S.A.

Stress at fracture= LOAD AT FRACTURE 'D' final C.SA

ego A tie bar on a vertical pressing machine is 2 m long and 4 cm in diameter. What is the stress and extension bar under a load of 100 KN?

10 KN

1=2m •••••••••••• · •••••••• ! ........... dl

C cr/E=E :.

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p

.A. =E dl

I

:.dl=E.. =cr .!.. AE E

(1 00X2)/(7[/4(4/1 02)X205X1 06)=0, 78X1 0·3m

stress=load/csa=P/A =100/(7[/4)(4/1 02)2=79.6X1 03KN/m2

11 - 1- 3':2.

6 = P A

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1 d.t.. loolNV ""d._-{====:~I+!d-: ~ 'OO"lrY

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F ~ ::20<; >< I 09

AJ/rr)2.

0' = .E.. _ A

.-

e- -'. ell --ei -=. p.e r-

-:. 100" 2. _ _ 0 776 "'" ~ ( t"~ 1(2.05" >< lob - fcf

Of"''':U CR'Jd..e..f' ?)

38 "'''' Id 101

PI::J.l .... ,----'-----Lf ___ J----=-j--~ pJ:fV ./._.J .... _-.,..,d -""2.1""\ _~_37 ....... -i>

L, Lz L3 63 ~ 81./ /tiN /h7(.

-/om L J)r:kC-cl; -0'" ! =- £'1 1- /2+ h _. !!...:: r;; G

f(ll. E 1J1

T p1.J _ Ibe

-- qt; . ?. 7 2101<IO b

fiN 11t)"t.

")-7 t _

q.5 )7""IQ,1", ( ~16 ·732. + (;01 ·717-t

~ 0 - 002

III ClL)'2 ~ 103

?, / '2b2. l/{ b)

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SHEET 1.

Stress and Strain

1. A 38 mm diameter Mild Steel bar 6 m long lengthens 1.6 mm under a

load of 70 KN. Determine (1) the stress in MN/ m2., and ( 2) tile modulus

of Elasticity.

2. A wrought iron bar 7.5 m long is SO mm diameter for 1.B m of its

length, 65 mID diameter for 2 m of its length and 38 mID diameter for

the remaining 3.7 m of its length. The bar is subject to a tension

which produces a stress of 84 MN/mz on the 38 mID diameter portion.

Take E - 210 X 109 N/ M2 and determine the elongation of the bar.

3. The following observations were made during a stress - strain experiment

on a wire of 0.5 mID mean diameter and having an unstressed length of

3 m.

Load (N)

Elongation (mm)

o o

8 16 24 32 40 48 56

0.55 1.07 1.51 2 . 06 2.65 3.11 3.68

calculate the value of Young's modulus and deduce the probable material

of the wire.

4. A tensile test was made on a copper specimen 19 mm diameter and

100 mm gauge length. The maximum load was 62.5 KN and the load at

fracture was 47.8 KN . The diameter at the fracture was 10 . 4 mm and

the distance between the gauge points at the moment of rupture was

152.5 mm. Determine:- (a) the ultimate strength, (b) the nominal

stress at fracture, (c) the actual stress at fracture, (d) the

percentage elongation, and (e) the percentage reduction in area.

5. A specimen of material 0.3 m long has three parallel parts of

diameters 25, 50 and 75 mm and o f lengths 75, 100 and 125 mm respectively .

When the specimen is compressed by a load of 40 KN, the total length

is reduced by 0.0457 mm. Find the value of Young's modulus and the

strain in each part of the specimen.

cont'd ...

-2-

6. The compressive load on a holloW' cast iron column 1s 400 KN; outside

diameter of column 150 mm; compressive breaking stress of cast iron

600 MN/m~. Find the thickness of metal required if the safety factor

is 5.

7. A tie bar is 1 m broad and 25 mm thick and a longitudinal load of

2.5 MN reduces the breadth by 0.16 mm. It is also found that a stress

of 250 HN/m2 produces an elongation of 0.4 mm on a length of 200 mIll.

What is the value of Poisson's ratio for the bar?

8. A steel bar of rectangular cross section 50 mm x 12 mm 15 subjected to

a pull of lOa KN in the direction of its length. Taldng E as 210 x

l09N/m2 and Poisson's ratio as 0.3, find .the decrease in length of the

sides of the C:::OS9 section and the percentage decrease in the area of C the cross section.

9. A bar of steel 75 mIll x 25 mm cross section is subject to an axial pull of

180 KN. calculate the decrease in length of the sides if Young's

modulus is 210 x 109N/ m2 and Poisson's ratio is 0.286.

10. A bar 3 m long and of rectangular cross section 50 mm x 75 mm carries

11.

a tensile load of 300 I<N." Find the lateral, longitudinal and volumetric

strains of the specimen . Take E = 210 x 109N / m2 and'J =-0.25.

A p1ece of steel 450 mm long and 25 mm x 25 mm cross section is

to a tensi1e load of 150 MN/m2 in the direction of its length.

0.3 and E is 210 x 109N/ m2 calculate the change in volume.

subjected

If \) is

12. A block of steel 250 x 75 x 100' mm 1s subjected to a compressive load

of 600 I<N in the longitudinal direction. Ftild the longitudinal and

lateral strains and the total change in volume when 'J is 0.28 and

E is 210 x l09N/ m2.

13. If within the elastic limit a bar of steel stretches 0.001 of its length

under Simple tension, find the proportional change in volume, Poisson's

ratio being 0.25.

cont' d •.•

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(

r SUNDERLAND POLYTECHNIC -'§P-

MECHANICAL ENGINEERING DEPARTMENT

Temperatur e Stresses SHEET 4

1 . A steel bar 38 mm diameter and 4 am long is raised in temperature

thr ough 60°C after whi ch its ends are firmly secured. After cooling

to normal temperature again the length of the bar is found to be

2.

1 . 4 mm less than when it was at its highest t emperature . Determine the

total pull in KN exerted by the cold bar and the stress in it.

o Os = 0 . 000011 per C

o If a bar of steel 25 mm diameter and 3 m long is heated to 55 C above

the temper ature of the atmosphere and then firmly gripped at its ends,

find the tension in the bar when cooled to the temperature of the

atmosphere if during cooling it pulls the end fastenings 0.65 mm

nearer together.

o = 0.000011 per C.

3. Two parallel walls 7 . 5 m apart are stayed together by a steel bar

2S mm diameter passing through metal plates and nuts at eac h end.

4.

The nuts are screwed up to the plates whi le the bar is at a temperature

of lSOoC . Find the pull exerted by the bar after it has cooled to ISoC .

(a) if the ends do not yield, (b) if the total yielding at the two ends

is 6 mal.

o a s = 0.000011 per C

o A steel metre rule measures .100.02 em at 20 C; determine the temperature

at which it gives correct values .

a = 0 . 000011 per °c

- 1-

5. A flywheel boss :5 strengthened by a steel hoop shrunk on. The boss

i s 0.5 m diameter and the ring is bor ed 0.406 mm s maller i n diameter

than the boss. To what temperature must the ring be raised in order

that it may just slip over the boss? What is the tensile stress

in the hoop after cooling assuming that the boss is unyielding .

6. A brass r ectangular bridle is to be shrunk onto a steel rectangular

bar 75 rom by 100 rom . The bridle is requi red to just slide over the

steel bar when its temperature is rai sed by 2500c . Calculate the

internal dimensions of the bridle and the tensile stress in the sides

after it is shrunk on and cooled. Assume that the steel bar is

unyielding.

o : 0.000018 per c.

7. A heat exchanger tube of hard drawn brass is 0 . 6 m ~ong , and has an

internal diameter of 10 rom with a wall thickness of 2 . 5 mm. The tube

is r igidl y built i n at both ends during manufacture of the pl ant at o atmospheric temperature of 20 C. The heat exchanger is heated to a

temperature Which is 40 deg C greater than the normal working temperature o of 100 C. Determine whether the tube will have failed, and if so , how?

You may assume for brass that the modulus of elasticity is unaffected by

temperature changes and the yield stress (Oyt) may be assumed to vary as

(

where t is temperature in deg C and Oyt the yield stress at temperature t .

(1971)

For brass a = 20 x 10- 6;oC ,

(Fails as strut)

8. A clock pendulum is made in the form shown in Fig. 2/2 . Calculate the

length A such that L does not change with temperature. An amateur

repairs the pendulum by fixing the inner steel par t to both the Aluminium

parts throughout their lengths. Calculate the change in length L of the o pendul um after repair for a temperature change of 20 C, and also calculate

the change in stress for the newly fixed parts, if each member of the

pendulum has a cross sectional area of )0 mm2 .

, ~

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~q; 8. (cont'd . )

Steel a = 11 x 10- 6 0 per C

E = 200 x 10 9 N/ m2

Aluminium 2S x 10-6 0 a = per C

E = 6S X 10 9 N/ m

2

(1. 76 ro , 22.1 MN/m2 , 11. 05 MN/m 2)

9, A steel component has the dimensions shown in Fig . 2/3 when it is at room

temperature.

The temperature of the component is increased by 60 deg C. Calculate the

axia l load i n the component .

For steel the coefficient of linear expansion is 12 x 10-6 per deg C and

the modulus of elasticity is 2 x 10 5 MN/m2• (1969)

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SHEET 1.

Stress and Strain

A 38 mm diameter Mild Steel bar 6 m long lengthens 1.6 mm under a

load of 70 KN. Deeermine (1) the seress in MN/m', and (2) the modulus

of Elasticity.

A wrought iron bar 7.5 m long is 50 mm diameter for 1.8 m of its

length, 65 mm diameter for 2 m of its length and 38 mm diameeer for

the remaining 3.7 m of its length. The bar is subject to a tension

which produces a stress of 84 MN/m' on the 38 mm _ diameter portion.

Take E = 210 x 10~ N/M2 and determine the elongation of the bar.

3. The following observations were made during a stress - strain experiment

on a wire of 0.5 mm mean diameter and having an unstressed length of

4.

3 m.

Load (N)

Elongation (mm)

o o

8 16 24 32 40 48 56

0.55 1.07 1.51 2.06 2.65 3.11 3.68

Calculate the value of Young's modulus and deduce ~~e probable material

of the wire.

A tensile test was made on a copper specimen 19 mm diameter and

100 mm gauge length. The maximum load was 62.5 KN and the load at

fracture was 47.8 KN. The diameter at the fracture was 10.4 mIll and

the distance between the gauge points at the moment of rupture was

152.5 mm. Determine"- (a) the ultimate strength, (b) the nominal

stress at fracture, ecl the actual stress at fracture, Cd) the

percentage elongation, and (e) the percentage reduction in area.

5, A specimen of material 0.3 m long has three parallel parts of

diameters 25, 50 and 75 mm and of lengths 75, 100 and 125 mm respectively.

When the specimen is compressed by a load of 40 KN, the total lengtii

is reduced by 0.0457 Mm. Find the value of Young's modulus and the

strain in each part of the specimen.

cont'd." "

-2-

6. '!he compressive load on a holloW' cast iron column is 400 KN; outside

diameter of column 150 mm; compressive breaking stress of cast iron

600 MN/m~. Find the thickness of metal required 1f the safety factor

is 5.

7. A tie bar is 1 III broad and 25 mID. t.'lick and a longitudinal load of

2.5 MN reduces the breadth by 0.16 Mm. It is also found that a stress

of 250 HN/m2 produces an elongation of 0.4 mm on a l~~qth of 200 Mm.

What is the value of Poisson I s ratio for the bar ?

8. A steel bar of rectangular cross section 50 mID. x 12 mm is subjected to

a pull of 100 lCN in the direction of its length. Taking E as 210 x

l09N/m2 and Po1sson I S ratio as 0.3, find the decrease in length of the

sides of the cross section and the percentage decrease in the area of .

the cross section.

9. A bar of steel. 75 111m x 25 mm cross section is subject to an axial pull of

180 KN. Calculate the decrease in length of the sides if Young's

modulus is 210 x 109N/ m2 and Poisson's ratio is 0.286.

10. A bar 3 til long and of rectangular cross section 50 mm x 75 mal carries

a tensile load of 300 KN. Find the lateral, longitudinal and volumeeric

strains of the specimen. Take E = 210 x 109N/ m2 and V • 0.25.

c

11. A piece of steel 450 mm long and 25 tam x 25 tam cross section is

to a tensile load of 150 MN/m2 in the direction of its length.

0.3 and E is 210 x 109N/ m2 calculate the change in volume.

subjected (

If 'V is

12. A block of steel 250 x 75 x 100 mm is subjected to a compressive load

of 600 KN in the longituctinal direction. Find the longitudinal and

lateral. strains and the total change in volume when 'V is 0.28 and

E is 210 x 109N/ml.

13. If within the elastic limit a bar of steel stretches 0.001 of its length

under simple tension. find the proportional change in volume. Poisson's

ratio being 0.25.

cont' d •• •

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A spherical cell of thickness 19mm diameter O.9m is subjected to an internal pressure of 5MN/m2

.

Calculate the stress in the material.

.1 " t~ l'1mm .c··· ~ •. ... Z ..... "r · ...... ···.Pc: s Nul",

;4- P ~

\ J .i . c; . ~ .';; ..

cr8=1td/4t=[(5xO.9)/(4x19)]x103=59.2MN/m2

Determine the thickness of a pipe 250mm bore to carry an internal pressure 2.5MN/m2 if the tensile stress in the material is not to exceed 45MN/m2

Pd/2t=45MN/m2

t=(2 .5xO.25)/(2x45)=O.0069m=6.9mm

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FLUIDS

LIQUIDS IN MOTION

Liquids are incompressible i.e. density remains constant.

Types of flow

1: Turbulent flow (particles of fluid flow in disorderly manner) 2: Viscous flow (stream line or laminar flow)

Particles of fluid move in orderly manner and retain the same relative positions in successive crossections

3: Uniform flow (The c.s.a. and velocity of the liquid are the same at each successive area).

Mean velocity

Volumetrc flow Q is the volume of the water flowing per second

m ,

d m csa=a m2

[,

Q=aU ....... (1 )

Q=1l/4d2U density p=m/v :. m=pv

mass flow rate per sec m= pav

Mean velocity U=Q/a from (1)

Continuity of flow: for the continuity of flow in any system the quantity of liquid entering the system must equal a quantity leaving the system. Mass flow/sec ct paU=ct in a liquid :. in a liquid au=ct and Q=ct p=ct

(

Oil nows through a pipeline which contracts from 450 mm diameter at A to 300 mm diameter at B and then forks . One branch is 150 mOl diameter discharging at C the other being 225 mm diameter discharging at D. Irthe velocity A IS 1.8m1s and velocity at D is 3.6 m/s what will be the discharges at C and D and the velocities at Band D?

. Qc=? n : C

. ~

~-~Q":? A

1.8m1s U.

j----------------,---------~ --' ~6m/ s

450mm din 300mm dia

o 225mm din

for continuity of now: QA=QB=QC+Qo

, Qs= QA ·. (lld,-)/4*U s =0.286

=> U B=(O. 286*4 )/( 110.3 ')=4. OSm/s

QA/Qs=as*Us/aA *UA= I :. Us =UA.(a4as)=UA(diam A/diam 13)'=

( =1.8*(0.4SI0.3)'=4.0Sm/s oil leaving the system at C and D= oil entering the system at A.

Qc+Qo=QA=0.286Ol3/s

Qo=ao*Uo=1l14*0.225'.3.6=0.143

Qc=O.286-0. 143=0. I 43Ol3/s=ac*Uc= 1l14*0. IS ' .Uc

Uc=(O. 143*4)/(11*0. I S')=1l14(O. IS ),Uc=8.09m/s

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•

(

Re! No. 90021/30005

SUNDERLAND POLYTECHNIC

POLYTECHNIC INTE&~DIATE CERTIFICATE 1982

ENGINEERING I

!IomiC.1) 7th lUIl'-, 1982 9.30 ~.m. - 12.30 p.m.

G

3.

Cand.Wa.tu aM Ilequ.l:r.ed to o.ttempt ALL qUUWIlJ

.i.J. Section A ~ THREE ollOm secUOI1 B. Both

JeC-UDnD ClWlI} equa.! maJr.!u,.

SECTION A

A horizontal rod of length 1.8 m and ot negligible weight rests on

two 9uppor~s at its two ends. A weig~t of 60 N 1s suspended from

the rod at a distance of 0.75 m from one end. Determine the two

end reactions.

If one support can only withstand a reaction of 10 N. what is the

greatest distance from the ather support at which the weight could

be suspended ?

The vertical side of an oil tank is 4 m wide. The depth of all in , the tank 1s 2.1 m. Determine the total horizontal thrust on the

side due to the all in the tank and the position at the Centre of

Pressure. Specific gravity of oil i9 0.8. For wnter p = 10 3 kg/m 3.

In the crank and connecting rod mechanism

force on the piston is 5 kN and the angle

sh.own in Fig. o e is 25 .

Q3, the

Determine for this position (a) the force in the connecting rod, and

(b) the reaction between the piston and the cylinder wall.

-, /

I 1J%CG SkN - --, \ j

I , J Fig. Q3.

--- -' cont ' d .... .

it •

• P . I.e. ENGINEERING r ...... . ... • .... 2

4. The triangular framework shown in Fig. Q4 is loaded by a vertical

farce of 10 kN as shown. Determine graphically the magnitude and

nature of the forces in the members and the vertical support

reactions,

R ... Fig. Q4.

..

5. Steam enters an engine at a pressure of 9 bar and ls' 10% vet. rt is'

exhausted from the engine at atmospheric pressure 1.013 bar, 18$ wet .

Determine the loss of enthulpy per kg of steam passing through the

engine.

pbar

1.013

9.0

419

743

,,£g

2257

2031

hg

2676

2774

)

) All kJ/ kg.

A copper wire of diameter 1 . 6 mm and length. 3 m extends 1.27 mm

when carrying a mass of 10 kg. Determine (a) the stress, (b) the

strain, Bnd (c) the modulus of elasticity of the copper.

A body of mass 4 kg rests on an inclined plane. WRen the inclination o

of the plane is 24 to the horizontal, the body just slides down

without acceleration . Determine the value of the horizontal force

r equired to push the body up the plane with no acceleration.

cont'd .....

(

(

f

• P. I.e. ENGINEERING I .. • .. •. .. ... ... 3

10.

B motor car travelling at 90 km/b

on a level road can be brought to rest by the brakes is 35.6 m. It

the motor car is travelling ~t the same speed up an incline of 1 in 15,

determine the shortest distance 1n which it can oe brougbt to a halt

assuming the braking force remains the same.

Air is to be compressed in a diesel engine

pressure and temperature are 105 kN/ m2 and

cylinder. The initia­o 16 C respectively snd

the tinal pressure is to be 7 ',Qi/m2. In order to tgn1 te the final

charge at the end of compassion, the te~peratu~e required

1s IIOaoe. Calculate the ratio of initial to final volumes for this

to be possible .

A mercury U-tube manometer is used to measure the pressure above

atmosphere of water 1n a pipe, the water being in contact with the

mercury in the left hand limb as 9hown in Fig. Q10. What is the

gauge pressure at A if the mercury is 30 em belo~ A in the left

band limb and 20 cm above A in the right hand limb? Specific

gravity ot mercury = 13.6. Density ot water = 10 3 kg/ m3 .

Fig. QIO.

J d

o M r

cont'd .. .•.

• • P.I.C. ENGINEERING I .•..•....•. •. .. 4

SECTION B

11. The accompanying Worksheet shows a cantilever frame attached to a

rigid wall at X and Y. Using the Worksheet and a scale of 1 em = 2 kN,

draw the force diagram for the frame when a load of 12 kN Is carried

at Z. Hence determine (n) the loads carried by eacb member of the

frame, and whether tensile or compressive, and (b) tbe reactions

exerted by the wall at X and Y.

12. Prove that in a polytropic process the work done between two points

W __ P l Vt - p!V z

1 and 2 is given by ;-n - 1

0.045 ml of air at a pressure of 210 kN/m2 and a temperature of lSoC

is compressed to a pressure of 1900 kN/m2 according to the law

pyl on = C.

Determine 0 the mass of air

(b) the temperature at the end of compression

(c) the work done on the gns.

For air R = 0 . 287 kJ/kg K. PV = mRT.

13. For the beam show in Fig. Q13, draw the shear force diagram to scale

showing the values at significant points. Hence determine the

_ magnitude and position of the maximum bending moment.

~ k.\ 40 2: 20 k~l ...... IOk~/ ..... c J)

A. <!>

--6-,=-~ 4-_~-E.. F

~Aw- 4~

Fig. Q13

cont'd . • . ..

(

r ,

(

-

P. I.C. ENGINEERING I ............... 5

14 . A miners' cage starts from rest to descend a pit shalt with an , acceleration of 0.5 m/s. After the lift has descended a distance of

15.

14 m a stone is dropped down the shaft from the surface .

Determine (a) the time the stone is in motion before it catches up

with the 11ft,

(b) the velocities of both lift and s~one at impact.

A pipe 300 m long tapers from a diameter of 1 . 2 m at its upper end

to a diameter of 0.6 m at its lower end and has a slope of 1 in 100. , The pressure at the upper end is 69 kN/ m. Neglecting losses

determine the pressure at tbe lower end when the rate of flow is

5.5 m3/ min.

Bernoullis equation is v' P

z + -- + -- = constant . 2g pg

Density of water = 10 3 kg/ m3 •

• '/ eJO

C 1'2 kN

(

(

5qle. : I c ...... ~ 2. kr-l

MiI-;..D • •

M/),-:. - 40)\.<\ -:: - ,GO Me. - j 4(') :<'1 0 h(11i . (, "7 )q;) - C;O X3) 0::::. \ '2b. 01

tJ.p-:: -(4DfC-\4)~\1I·67)(lo) - (IO()X5)-=- ~ 6';

ME.~ - (LiO",1 ~ -d I 1!{7J\ 12) - ( loox7) - (4D1'-1) _IJ Mf=-:- - t IAClA'20)-tJII I.67)\lbH l'DO~II )t (1683J!(Lj) - (l'lOX 3)

r

(

4

60kN 40kN 20kN

, -.F 1(;, Q. I I

"

P . I.C.. 1981. E"GINEERIrlG r. ... . .•• 3

13.

14.

15.

A mass of 5 ~r ,en~s vnrtically ano 1s attached to a cord wound

round the horizont~l axis of a flywheel. The flywheel has a mess of 2S k~

and a rAdius of 7yr3t1on about its axis of 15 em. and the axle 1s 5 em d1e~eter.

If the mass 1s releosed. ~nd in fa1l1nf!; corrrnences to turn the fl y~ ... h2el. calculate

a) the acceleration of tr.2 mASS .

b) the time taken for thR mess to f~ll vertically from rest throu~h

a distence of ~ m.

c) the tension 1n the cord.

Nsp,lect friction at the axle bearings.

State garnculll's theorem and describe. with the aid of a sketch. the

!J!"'inci::lle 0';: ..::: lJ!: ntur1. meter.

A lJ~ntur1 netsr h9;S e throat diameter of 7.S en end 1s fitted to :3

15 em diameter "Iate r mi:lin. ,rJ., nercury manometer fitted to the main and to

the throat indicate9 3 ~i~~er9nCQ in head of 40 cm of ~~rcury. !f the meter

has a coefficient of O,QS . calculate by workin~ fron first principles tha

dlschar[9 in l1tr9!/~in,

,4, ",/;ISS of ~ir "nclosed in a cyl1ncer f1tt9 d t"i th a aiston underr'oes

the followin~ cycla of r9vRrsl~1~ orocesses:

1 2. ~~1acatic cc~~r9s5inn,

2 - 3, h~at transiqr at const~nt volumA to th~ air.

3 - 4, ~dl abatic exoansior, to the in! tisl volume.

4 1, heat tr~nsfp.r at const ant volume frOM t~e air to the initial

st~'Ca,

If the lr.1tial conditions are pressure 0.965 band tomoerature 37°C.

the volume compression ratio 1s 8 , ~nd the maximum pressure during the cycle

is 58,5 b. determine:

a) the oressure anc temperature at the or1nc1o~1 nolnt~ of tha cyclAI

b) the Magnitude and direction of the he~t tr~nsf~r 09r k~ of air

during the process:

cl the net work tranSTqT during the cycle csr k~ of .'3ir:

dl the thermal efficiency of the cycle.

y • 1,4 Co a 1.005 kJ/k3

(

(

13. Fig. 3 shows II triangular plate immersed in oil with one edge parallel to the free surface. Deter· mine the total force acting on the plato, and the distance of the centre of pressure from the free surface of the oil. The specific gravity of the oil is 0·88 and the density of water 1000 kg{m3. The second moment of area. of a triangle about an (\ris parallel to ita base and passing through the centroid is given by

FREE SURF ACE

E N

Fig. 3

14. 0·1 m3 of a gas is compressed from a. pressuro of l:!O l\lNjm2 aod temperature 25°0 to a. pressure of 1-2 1t1N{m2 according to the la.w PV1.2 = constant.

Determine (a.) the work done on the ga.s. (b) the change of internal energy of the gas,

and (c) tho quantity of heat received or rejected by the gas.

0 0 = 0'72 kJ/kg K

R = 0-286 kJ {kg K

w PV=mRT

AU = mCto(T2 - TIl

15. .A. bea.m 12 m long is simply supported over a. span of 6 m and overhanss bot~supports by_the same amount. Tho right hand overhanging portion carries II. uniformly distributed load of 15 !tN/m aDd II. concentrated load of 30 k.i.'l" at the IUtreme end. The left hand overhanging portion carries a uniformly distributed load of 30 k..1{/m and n. concentrated load of:!O ki'i at the extreme end. In addition a load of 150 k..L'f is concentrated at midspan. Dnnv to scale the shear force and bending moment diagrams indicating clearly tho principal values on each diagram.

b J v

c ..

£, d

~ ,

•

/

"3'ol<-~ ~ \0 k~ C t-

20 kj

G- I:=..

•

6 c. .. le : - \ c......... ~ \ 0 \<.r\

t.1 J

(

. (

Ref. No. 90021/30006

SUNDERLAND POLYTECHNIC

POLYTECHNIC INT ERMEDIATE CERTIFICATE 1980

ENGINEERING I

Wwnuday , 218t J.[ay, 1080 9.30 a.m. to 12.30 p.m.

1.

"

3.

Candidatu ar~ requirtd to att~mpt all quuticm.! in S~ction .::f. and thr ee from S~ct ion B.

Both Stcti01U carry equal marks.

SEartON A

Determine how long B trnin will tako to travel 1200 m if it b.o.a an initial speed of 108 kmfh and it haa a constant acceleration of 0·1 m/s~.

A gae is expanded according to the lo.w PV = c from 0. pressure PI and volume VI to pressure P~ and volumo V~. Determine the work done by the go.s in terms of the given quantities.

A crane lifts a mBSll of 10 X 103 kg. Determine the power required to give the load an accelera.­tion 1'0 m/8~ when it is moving at 0·1 m/s.

4. Fig. 1 shows a system of (.'Opiano.r forces. Determine the magnitude and direction of the resultant and its position relative to the point A.

A

600~ " , , 1 OON

, E -

, 1m

400 N SOON

Fig. 1

5. An aluminium rod is 36 cm iong and 15 mm diameter. It is subject to a tensile Ioo.d of 40 kN. Determine (a) the stress, (b) the strain and (c) the extension. if E = iO X 109 N/m~.

Hl9H (P.'I'.O.

6. A. jet of 'Water emerges horizontally from an orifice in the side of a. water tank. It is seen to pass through a point 150 mm below the ceDtre of the orifice and 450 mm borizontnUy from the t..:l.nk. By considering each drop of \'i'ater to be Q. projectile determine the \'"clocity of the jet when it leaves the tank.

7. A gas.o.ir fuel hils a ciw.rncteriatic gBS constant of 0·7 kJ/kg K, and is supplied. to an engine. Ita pressure is measured as 80 mm water gauge when atmospheric pressure is measured as ;05 mm of mercury. The temperature of the fuel is 30 0 e. Determine the density of the fuel. For water, p = 1000 kg/m3. Specific gmvity of mercury = 13·6. P-V = mRT.

8. Determine the values of the hoop and longitudinal stresses sei up in the shell plate of a torpedo, the diameter of which is 0'5 m and the thickness of the plate !)·38 lllnl. The internal air pressure is 12'4 MN/m2.

O. A steel bllr of diameter 18·75 mm Ilnd of length 3·6 m has its temperllturo raised from 15°e to 55°C when the ends are then secured to prevent contr.l.ction. If the bllr then cools to its originnl temperature Ilnd there is no yield in tho end fastenings, determine the magnitude and nature of the "'" stress in the bar. E = 204- X 109 N/m~. Coefficient of linear espansion = 0·000012 per °e. (

10. Draw the complete force diagram for the Crome structure shown in Fig. 2. Hence determine the magnitude of the forces in the members CG, BF and SF only, stating whether they are in tension or compression.

30kN 20kN

Fig. 2

SECTlON B

11. A particle is acted upon by forces P, 2P, 3v'3P !lod -!P, the angles between the first Bnd second. the second aDd third. and the third and fourth forces being sag, goo and 150Q respecti.ely. Determine the magnitude of the resultant force Bnd its direction relative to the first force .

12. A qunntity of steam nt a pre!lsure of :!·1 lI,rn/m2 and dryness froction O·g occupies a volumo of

2

0·2562 m3. It is now e:qlo.nded until the pressure becomes 0·7 }IN/m2 when the dryness fraction is 0'754-. Determine the change of internal energy.

( ,

( 1 •

2.

3 .

ReT. No. 90021/30005

SU~!OERLArJO POL YTECI-!N!C

FOLYT!::C~r-JIC INTERMEDIATE CERTIF!CATE 1981

ENGIrlEERWG I

------------

Candid~tes are reQuired to ~ttn~ot ALL questions

in Section A and THREE from SRction 9.

equ~l marks.

50th sections cerry

=:ECTION A

D~termine the force required to null a block of nass 100 kv uo a plane

inclined at 300 to tho horizontal when the force is: (al Parallel to the olano.

(b 1 Horizontal.

The cosTTicient of fr1c~ion can bo talt.cn as 0. 2. ."

A flyw'leel is rot~tin~ ~t unifor . • soeec of 150 r~vhnin. If it is broul!ht

un~fornly to r est !n 11) ;:'Iins . cetemine the an"ular retardation I, r2!c/ s z • Hew

Many revolutions cr,~s the flywh~Rl make curin~ the r~tard3tion oeriod ?

Ths cranknin of a st3~m en~1ne is 0.5 n from the crankshcft ~xls. which

is rotating unifomly dt 300 rev/min. I,oJhat 19 the rravnitude ~nd directicn of

the acceloration of t he crankpin ?

4. A car movin e at 72 k~/hr elan? a hori=cntal r~ad 15 ~rou~ht to ~st in

50 m. If the mess of the car is 1000 It.C!:. what 15 t'":c C!ve r aqa rnf.3rdii11" Torce ?

,. A mass oT 100 k~ falls freely from rest throu~h a vertical heipht of G M

o "

when its vslocity is dacreas9d to 5 m/s by impartin!!; enerp,y to a r,ech1nF~.

Calculate the enerr,y ~iven to the machina.

A bar of steel 50 em long has a cross-sectional eraa oT 400 mm2 and is

subjected to a ciraet pull of 2Q kN . Oatennine the stress in the bar and thFl

olongation aT E • ?OO X 10 9 N/r::.

Cont / ovp.r/

P.LC •• 1981. ENGINEERHlG I. ... • ••• 2

7.

8.

9.

10 • .

11.

12.

Celculate the discharge 1n I1trss/m1n from ~n orifice 40 mm diameter

in the side of a tank contalnln~ water at ~ constant head of 2.5 m above the

centre of the orifice. The coefficient of dischar~e can 'be taken as 0 . 6.

Show that. for a thin cylindrical 5hell . t~e hoop st~~9 1s double the

lonc1tudlnel stress.

In a non-flow orocess 1000 kJ erg rejected and the internal ener~y

increases by 200 kJ. P.ow much work 1s done and 1s the orocess an expansion

or compression ?

A quantity o~ ~as whose originel ores sure and volume are 1.5 band

O.S ml, resoectively. 1s comoressed to a volume of 0 .1 m3 • the lew of

compression being PV I' 2 =- C. Determine the Tinal pressure and the work don i

SECTION B

Fig. Q 11 nmresents a simnly supportec pirc!:!r the Joints of \·.h1ch j'T'tay

be regarded as cinned.

Determine the forcos 1n the IT1AmOers 1. 2. 3 ~ nd 4. 9tatin~ wh~thp.r the

nembers are 3trut~ or ties.

(

A beam AECD. ,9 m lone. 1s simnly supr.ort-:=c ~t .fJ • .:!n1 C. 7 cm .:mart. The

beam carries a un1 fonnly d1str1butad loee of " 0 krJ/ m l'"l!n bet • ."een :'he s uooorts,

e verticel concentrated load of 1S kN at fi. '",hich is 1 m --rom .'\. end ~ '/8rtice1

concentrated load of 20 kN at O. Draw the ShE ~ .1np Forca and cendinr. r10ITl13nt

diagrems to sca19 anc mark thereon the Drinc1o~1 values end ~os1t1on of

contraf1exure on the b8a~. Oatermi~8 the oos1tion and value of the maxiMum

bending moment.

Cont. oV'3rl