Ain Shams UnivercityFaculty of EngineeringCredit Hourc Programs

Affii266-el

EPHS 110: Eng. tech. l(Statics)Communication Eng. Program

Tirne:3 Hours

PLEASE ANSWER ALL QUESTIONS

Question (1): (10 marks)

lf the force Fr = 4# N is acting along the line BHand the force Fz = 6€ N is acting along thecube's diagonal BG. Calculate the resultant of Frand F2 and its moment about:

a) Point Ac) Line AH

b) Line ACd) Line AG

Question (2): (10 marks)

For the shown frame and truss:Calculate the force in member BDin each structure.Calculate the shear force and thebending moment at the mid-pointsof AB and BC on the memberABCof the frame.

Question (3): (6 marks)

For the shown two cases:lf the coefficient of staticfriction atall contact surfaces is 14 = 0.4,calculate the maximum force Pfor equilibrium in both cases.

Question (4): (8 marks)

Calculate the polar second momentof area of the shaded area withrespect to its centroid.

,12,(Dimensions in millimetes)

QHestion (51: (6 marks)

Using the method of virtual work, calculate themoment M which is required for equilibrium of theshown rod in terms of P, L and the position 0.Check your answer using the ordinary equilibriummethod.

G (0,1

D(0,{,1)

1m

A(0,0,{z

(1,1,0)

1m1m B (1.0.1)

20 kg

30 kgP

D

T4mt

D

T4mI

[*Sr*l*smI Fs ,*l* g, *l

30 kg

Pulleys arc smooth

v

Th Tr'I12

y,, A=bh/2

ThI =ht-Is

OF-E4-., x OF!E{t-b- I r-b-l

tr= bh" 13,lr= hb3/3 I lr= bh" l12,lv=hb3 112

Jo = bh (b2 + h2) /3 I .to = bh (b2 + hl t12tr = bh 3

112 ,lr, = hb " tlzl lr = bh 3 /36 , Ir, = hb 3/36

Jc = bh (b2 + h\t12 I t" = bh (b2 + h2)/36

Solution of Final Exam (Communication Engineering Program

uestion

" G (0,1,0) l7(1,1,0)

D(0,1,1)

c(1,0,0)

/ (0,0,1) 1m

zB(1,0,1)

1m

Fr = Fr iaa = r,(#)= 4J' (#)= a i _ q n

Fz = Fz inc = r,(*)- 6^tit-#) = -u i + o i - a i

I=Fr +Fz--6i+roj-rofMoment of the resultant about:

a) Pointr{:

l' j klfrt=E*I=lt o o l=toj+tof

l- o ro -,rlb) Line lC:i! .ac = (M nd in, = ( frr e . itc ) ( incr =

{o o i + r 0 r.(#l

i i+) = -s; * s f

c) Line AH:Since the line AH intersects fr at H and intersects F, atthe diagonal's midpointso the moment of both forces about AHiszero and sequentially the moment oftheir resultant

^E = Fr + F, about AH is also zero. fr), =o

d) Line,4G:Since the line AG is parallel to 4 and intersects F, at G so the moment of bothforces about AG is zero and sequentially the moment of their resultantf = Fr + F, about AG is also zero. fr no = o

1,,t!:-

- Part (a): Frame

800 N

800 N

T4m

Il*r--fsmJ

Fno

RnA 3m B 3m

Free m of frame and all members

lclc

Equilibrium of Part .48C:

EMc =0 :+ (Faosind) (3) = 0

Xf, - g :+ xc * Faa cos 9:.R7

ZFr: 0 + !c: F6p sin 0

The member ABC is subjectedto axial load only xc= ReSo the shear force and bendingmoment are zero everywherealong this member,

Foo =o

xc = R,l

!c:0

Fro= o

D

4m

Xg

Ra

T4m

I

uestion Ol - Part (b): Truss

Or usine the method of section:

Studying the right part:

ZF, -0 * Fao( sin 9):800

) Fao(41 5 ):800

t Fsn: 1000 N

A6(I 3m I 3m I

I XFree of truss and all ioints

il

E 800N

[" H ll l,mmTruUIA B II C

'. " {. '- 'l

Joint Equilibrium equations Result

C If,/: 0 + Fco:0 Fcp:0

D U, - 0 = FBD $15) * Fco: 800 Fno:1000

estion

Case (1): Cable AB is attached.

f--x

T

--.+x

Upper block:

ZFr: a +

xf,:g +Lower block:

ZFr: o :+

=)

Xf,- g =)

R' : 196 (cos 15o) = 189.32 Nf : F, R* = (0.4) (189.3) = 75.728 N

T : f + 196 (sin 15') = 126.46 N

R+f. (sin 15o) : 294 +R. (cos 15o)

R = 457.27 Nf : p,X = (0.4) (457,27) = 182.9 N

P : T* /'(cos 15o) +R. (sin 15o) +/= 431.5 N

Or after solving the upper block to calculate the tension 7, then use the systemof two blocks together as:

490 N

f'= rr, R'

f*= lr, R*

System of two blocks: (After calculating 7)

X.&:0 + R+T(sin15'):490 :e:+ f = t+; = (0.4) (457.27) = 182.9 N

Xf,-g +" P:T* Z(cosl5")+f =Case (2): Cable AB is removed.

R: 457.27 N

P : 431.5 N

/ \,.'196 N490 N

System of t'wo blocks: (Directly)I .:\'i'

XFy: 0 :) R : 490 N+ f : tt"A = (0.4) (490) = 196 N

Xf,:g :+ P:f + P:196N

In this case the inclination angle (15") will give no effect on results of P.

Note:This case can be also analyzed using the same procedure in case (1) bystudying the upper block firstly to calculate (R- and"f.) then stddying the lowerblock to calculate (R +/and P)

uestion

II-r1

3I3I

I1ixttl

,i

From symmetry about x-aris: F = 0

:a!n4ax:z/e:7,.!/.::/!:a2ailola:7:r/:!.:a,r/i4 ta:.1:,i/t127:/a::1.7/q. !/:rrf/.4-/_.t4t:/:r/7u4

t,!,-?*:*9.#),.i, !,.,!..*:,L,1*:),-l

Part (1): (positive)

& = -1'5 mmAr:30x3:90mm2

(r,), = (:Xro)' n2 = 67samm4 , Q )r= (:o[rf t3 = 270*m4

Parts (2) & (3): (positive) Azat: 2 x (9 x 3) - 54 mm27za3 = +4'5 mm

(1,)ru, -- {zRectangler }* [ (9X3F t rz +(e x 3) (13.s)2 ] = 9882 *m4

Q )ru, = {zRectangles }, [ (3XeF /3]=14s8 mm4

Parts (4) & (5): (positive) Aqas = 2 x [(9 * 3) / 21: 27 mmzi+as = *3 mm

(/, )ou, -- {zrriangtes }'t (qX3I / 36 +(e x 3 / z) (r1)2 I = 3zso.s m*4

Q )ou, : {zrriangles }. t (:) 0Y t Lzl -364.s*-4Total shaded area [(1) + (2) + (3) + (a) + (5)l:m-Aqas:i71;;2."

\"X =(4\+ 4a,$z*z* A+*sI,4a)l(At+ 4*s* A+*s) = l.lmm , 7 =0(/, )roo, = (r, ), + (I *)zu+ (/, )+es = 19912.5 mm4

(r )r", = krl * Q )ru, * Q r)ou, = zoe2smm4

(J o)roor= (/, )ro,rr * k, )r", = 22aa5*m4 :

Polar moment of inertia at the centroid C: l

(./" )roo, = (,ro)ro, - 4,^(x' + Y2)= 22005 - (17 l)t(l. 1)2 + (0)21 = 21798.1mm4

uestion

Principle of virtual work: d Uau ror.es & couptes in equilibrium = 0

6Ucorpt"u +dUForce Prisht+ *df/por"6 Pd*=0

- M6e *Frisht .6 it +Fun . 6 7a = 0

Where:

\ = (Lsin d) i + (Lcos d) j5 VA = [(Z cosd) 6e]i + (-I sin 0) 6ej i

fu = (*Lsin d) i + (-Lcos d) j6 78 = [(-I cos 0) \qi +f(Lsin g) 6e] iPrinciple of virtual work:

- M 6e+(ri;. [ (Zcosd) 60 ? +(-Isin q 50 j 1

+(-ri). [ (-Zcosd) 60 i +(Isin q 60j ] = o

- M 6e + (PI cos 0) 60 + (PLcos d) 60 = 0

,:.t 2. "t' : € F,A/E/"ttui fr//& t?&2 E/!;

(-M + 2PLcos? ) 60 = 0 iM .=?!I c9101

Ordinary method of equilibrium:

ZMo =0 =+ M *(P)(Icosd)-(P)(Icosd) = Q

= ir)-1L{[-ffii.e^i

M\)A

IJ