Civil Engineering Reviewer.docx

8/17/2019 Civil Engineering Reviewer.docx

1/97

MATHEMATICS, SURVEYING & TRANSPORTATION ENGINIRING

(NOVEMBER 2011)

Situation 1 – The probability of event A happening is 3/5 and the probability of

event B happening is 2/3

1. What is the probability of both A and B happening?

A. 3/5

B. 1/5

C. 2/5

. !/5

2. What is the probability of only event A happening i.e. event A happening and

event B not happening?

A. !/5

B. 1/5

C. 3/5

. 2/5

3. What is the probability of either A" or B" or A and B happening?

A. 11/15B. 1!/15

C. 3/5. 13/15

#.

Situation 2 – Ans$er the follo$ing proble%s&

!. 'i( )ongr*ent )ir)les are arranged in a )ir)le $ay that ea)h )ir)le is tangent

to at least t$o other )ir)les. +f the radi*s of ea)h )ir)le is 2 )%" find the

peri%eter of the polygon for%ed by )onne)ting the )enters of ea)h )ir)les.

A. 12 )%

B. 2! )%

C. 3, )%

. 32 )%

5. Whi)h of the follo$ing is/are )orre)t?

+. sin-A 0 sin-A

++. )os-A 0 )os-A+++. tan-A 0 tan-A

A. + only

B. ++ only

C. + +++ only

. + ++ only

. A solid re)tang*lar blo) has a vol*%e of 3, )%3

. +f all side %eas*re areintegers" $hi)h of the follo$ing is the least possible s*rfa)e area?

A. 42

B. 2

C.

. 2

#.

Situation – Ans$er the follo$ing proble%s&

6. What is the distan)e bet$een the inter)epts of the line ( 7 2y – 0 ,?


2/97

A. .231B. .6,

C. .34. 5.!!4

. +f 8(3 – 8 9 5" find the range of val*es of (.

A. ∛3 9 ( 9 ∛13B. ∛3 : ( 9 ∛13

C. ∛3 : ( : ∛13. ∛3 9 ( : ∛13

#.

;.

G! STRUCTURA" ENGINIRING & CONSTRUCTION

(NOVEMBER 2011)

$ith theverti)al. =egle)t the $eight

of the boo% and for this

proble%" 1 0 2 0 2%. The

p*lley at is fri)tionless.

+.

1. eter%ine the angle α.

A. !,>

B. 35>

C. !5>

. 3,>

2. What is the tension in )able AC in =?

A. 51.4

B. 25.3

C. 3!.4

. !3.21

3. What is the total rea)tion at B in =?

A. 5!.66

B. !3.21

C. 16.32

. 51.4

#.Situation 2 – The str*t sho$n in the fig*re )arries an a(ial load of @ 0

1! =.

;.


3/97

!.

!.

!.

!.

!.

!.

eter%ine the bearing stress bet$een the

pin and the str*t&

A. !3 @a

B. 3!5 @a

C. 25 @a

. 53 @a

5. eter%ine the shearing stress in the pin.

A. 2 @aB. 3 @a

C. 321 @a. 3!1 @a

. eter%ine the shearing stress in the bolts

A. 154.! @a

B. 14.! @a

C. 123.4 @a

. 16.3 @a

#.

Situation – The )ol*%n sho$n in

the fig*re is loaded $ith a verti)al

load @ 0 3 = and a lateral load < 0,.!5 =. The )ol*%n is 3 % high

and is %ade of steel $ith 3,, %%

o*ter dia%eter" %% thi) and

$eighs 15, =/%.

;.

6. What is the %a(i%*% stress at the base

d*e to the load @?

A. 1.6 @aB. 1.36 @a

C. 2.5! @a

. ,.6 @a

. What is the %a(i%*% stress at the base

d*e to the lateral load?

A. !.6 @aB. 5.2 @a

C. 3.! @a. 2.4 @a


4/97

4. +f the )ol*%n is a solid ti%ber $ith a dia%eter of 25, %%" $hat is the

%a(i%*% shearing stress at the base?A. ,.,4 @a

B. ,.1! @a

C. ,.,!5 @a

. ,.,12 @a

#.

Situation # – The fra%e sho$n in the fig*re is a)ted *pon by $ind

load press*re of 1.!! @a. These fra%es are spa)ed % apart nor%al

to paper. Consider the roller s*pport at B and the oint at as pin.

1,. eter%ine the horiontal )o%ponent of the rea)tion at A.

A. 35.6 =

B. 2.5 =

C. 1.3 =

. 12.6 =

11. eter%ine the verti)al )o%ponent of the rea)tion at A.

A. 23.4 =

B. 2,.2 =

C. 1.5 =

. 1.3 =

12. eter%ine the horiontal )o%ponent of the rea)tion at B.

A. 2.5 =

B. 1.3 =

C. 12.6 =

. 35.6 =

E!


5/97

;. Situation $ – The

sheet pile sho$n in the

fig*re is provided $ith

tension rods spa)ed 3

%eters apart. The

$ooden stringers has d

0 3,, %% and )an be

)onsidered si%plys*pported at ea)h

)onne)tion to the

tension rod. Allo$able

bending and shearing

stresses of the stringer

are 1!.6 @a and 1.!

@a" respe)tively.

D.


6/97

A. 75.! @aB. 7.3 @a

C. 7.1 @a. 73. @a

1. What is the reH*ired e))entri)ity e s*)h that the stress in the top fiber

of the bea% at the fi(ed end is ero?

A. 23, %%

B. 1, %%

C. 2,, %%

. 2, %%

E!

;. Situation – Ieinfor)ed )on)rete bea%s having $idths of !,, %%

and overall depths of ,, %% are spa)ed 3 %eters on the )enters as

sho$n in the fig*re. These bea%s s*pport a 1,, %% thi) slab. The

s*peri%posed loads on these bea%s are as follo$s&

D. ead load -in)l. floor finish" )eiling" et).JJJJJJJJJ.3.2 @a

ive load JJJJJJJJJJJJJJJJJJJJJJJJJ.JJJ.3. @a


7/97

E!

;. Situation ' – Channel se)tions are *sed as p*rlin. The top )hords of

the tr*ss are sloped !< to 1L. The tr*sses are spa)ed % on )enter

and the p*rlins are spa)ed 1.2 % on )enters.

D. oads&

ead load 0 62, @a

ive load 0 1,,, @a

Wind load 0 1!,, @a

Wind Coeffi)ients&

Wind$ard 0 7 ,.2ee$ard 0 ,.


8/97

2!. eter%ine the val*e of the intera)tion eH*ation *sing the load

)o%bination of ,.65 - 7 7W at the $ind$ard side.A. ,.6

B. 1.54

C. 1.25

. 1.6

#.

Situation The )ol*%n sho$n in the fig*re is s*be)ted to shear

for)e parallel to the ,, %% side. Allo$able )on)rete shear stress for

shear parallel to the ,, %% side is ,.1 @a. Con)rete strength fM )

0 21 @a and steel strength for both longit*dinal and )onfining

reinfor)e%ents is !15 @a. The ties are all 12 %% in dia%eter $ith

)lear )over of !,%%.

25. eter%ine the fa)tored shear for)e L* that the )ol*%n )an resist if the

no%inal shear strength provided by the ties is 365 =.

A. 36

B. !2

C. !6

. 532

2. +f the ties are spa)ed at 225 %% on )enters" $hat is the %a(i%*%

val*e of L* in =?

A. !62

B. !21

C. 335

. 34

26. +f the fa)tored shear for)e parallel to the ,, %% side is !,, ="

deter%ine the reH*ired spa)ing of transverse reinfor)e%ent in a))ordan)e

$ith the provision for seis%i) design.

A. 12 %%

B. 1! %%

C. 2!1 %%

. 1,, %%

E!

*! $!21!# S+-ia. P/oiion o/ Si3i- 4i5n

D. $!21!#!# T/an/ Rino/-3nt

$!21!#!#!1 Transverse reinfor)e%ent as spe)ified belo$ shall be

provided *nless a large a%o*nt is reH*ired by 'e). 5.21.6


9/97


10/97

W. f yh 0 spe)ified yield strength of transverse reinfor)e%ent" @a

R. h) 0 )rossse)tional di%ension of )ol*%n )ore %eas*red )enterto

)enter of o*ter legs of the transverse reinfor)e%ent )o%prising area Ash"

%%

S. h( 0 %a(i%*% horiontal spa)ing of hoop of )rosstie legs on all fa)es of

)ol*%n" %%

. s 0 spa)ing of transverse reinfor)e%ent %eas*red along the longit*dinal

a(is of the str*)t*ral %e%ber" %%

AA.

Situation 10 – The girder AB sho$n in the fig*re is s*be)ted to

torsional %o%ent fro% the loads on the )antilever fra%e. The follo$ing

fa)tored for)es are )o%p*ted fro% this bea%&

;a)tored %o%ent" * 0 !!, =%

;a)tored shear" L* 0 2, =

;a)tored torH*e" T* 0 1, =%

AB. The girder has a $idth of !,, %% and an overall depth of 5,,

%%. Con)rete )over is !, %%. The )entroid of longit*dinal bars of the

girder are pla)ed 5 %% fro% the e(tre%e )on)rete fibers. Con)rete

strength fM ) 0 2,.6 @a and steel yield strength for longit*dinal bars is

f y 0 !15 @a. Gse 12 %% Gstirr*ps $ith f yt 0 265 @a. Allo$able

shear stress in )on)rete is ,.6 @a.

2. eter%ine the reH*ired area of tension reinfor)e%ent of the girder" in

%%2.

A. !"15!B. 2"632

C. 3"63. 3"313

24. eter%ine the spa)ing of transverse reinfor)e%ent d*e to L*.

A. 136 %%

B. 16 %%

C. 4 %%

. 15 %%

3,. eter%ine the additional area of longit*dinal reinfor)e%ent to resist

torsion" in %%2.

A. 3"5,B. 3"!2,

C. 2"5,. !"12,


11/97

#.

;.

D. Co67

1. T8/8o.6 to/ion7 ;or =onprestressed %e%bers" it shall be per%itted to

negle)t torsion effe)ts if the fa)tored torsional %o%ent T* is less than&H!

I! T* U112

∅√ f '

c ( Acp

2

Pcp )2. To/iona. 3o3nt t/n5t87 The adeH*a)y of solid se)tions *nder )o%bined

shear and torsion shall be s*)h that&

O. √( V u

bw d)

2

+( T u P h

1.7 Aoh)

2

9∅( V cbw d +

2

3 √ f

'

c)

3. Where T* e()eeds the threshold torsion" design of )rossse)tion shall be basedon&K.

. ∅ Tn : T*

. T* 0 2 Ao A t f yt

s )ot V

=.P. Where Ao shall be deter%ined by analysis e()ept that is shall be

per%itted to tae Ao eH*al to ,.5Aoh V shall not be taen s%aller than 3,

degrees nor larger than , degrees. +t shall be per%itted to tae V eH*al to&

@.-a !5 degrees for nonprestressed %e%bers or %e%bers $ith less prestress than

in -b or

Q.

-b 36.5 degrees for prestressed %e%bers $ith an effe)tive prestress for)e not

less than !, per)ent of the tensile strength of the longit*dinal reinfor)e%ent.

I.

!. The additional area of longit*dinal reinfor)e%ent to resist torsion" Al" shall not

be less than&


12/97

'.

Al 0

A t

s ph( f yt f y ) )ot2 V

5. Mini3u3 to/ion /ino/-3nt7 Where torsional reinfor)e%ent is reH*ired"

the %ini%*% area of transverse )losed stirr*ps shall be )o%p*ted by&

T.

G. Av 7 2At 0 ,.,2 √ f ' c b

w sf yt

L. b*t shall not be less than -,.35b$s/f yt. Where torsional reinfor)e%ent is reH*ired" the %ini%*% total area of

longit*dinal torsional reinfor)e%ent" Al %in" shall be )o%p*ted by&

W.

Al %in 0

5√ f ' c Acp12 f y

− A t

s ph( f yt f y )

6. S+a-in5 o to/ion /ino/-3nt7 The spa)ing or transverse torsion

reinfor)e%ent shall not e()eed the s%aller of ph / or 3,, %%.

R. The longit*dinal reinfor)e%ent reH*ired for torsion shall be distrib*ted

aro*nd the peri%eter of the )losed stirr*ps $ith a %a(i%*% spa)ing of 3,,

%%. The longit*dinal bars or tendons shall be inside the stirr*ps. There shall

be at least one longit*dinal bar or tendon in ea)h )orner of the stirr*ps.

ongit*dinal bars shall have a dia%eter at least ,.,!2 ti%es the stirr*p

spa)ing" b*t not less than a =o. 1,.

S. Where&

. A)p area en)losed by o*tside peri%eter of )on)rete )ross se)tion" %%2

AA. Al total area of longit*dinal reinfor)e%ent to resist torsion" %%2

AB. Ao gross area en)losed by shear flo$ path" %%2

AC. Aoh area en)losed by )enterline of the o*ter%ost )losed transverse

torsionalA. reinfor)e%ent" %%2

A#. At area of one leg of a )losed stirr*p resisting torsion $ithin spa)ing s"

%%2

A;. f yt spe)ified yield strength fy of transverse reinfor)e%ent" @aAD. @)p o*tside peri%eter of )on)rete )ross se)tion" %%

A


13/97

AP. Ans$er Key&

[email protected]

AQ.

2 B

AI.3 A

A'.

! A

AT.5

B

AG.

A

AL.6

B

AW.

C

AR.4

AS.1

,

A

A.

11 B

BA.

12 C

BB.

13

BC.

1! C

B.

15 C

B#.

1 C

B;.1

6

BD.

1

B


14/97

CK.

C"! So.ution

CM!

Situation 1

CN!

CP.

C@.

CQ.

CI.

C'.

CT.

CG.

CL.

CW. 'in)e p*lley is fri)tionless" the tensions at sla) and tight sides are

eH*al.

CR. T 0 W 0 3, =

CS. By inspe)tion" α 0 09

C. a 0 2 se) 3,> 0 2.3,4 % d 0 ! tan 3,> 0 2.3,4 %

A. Considering the ;B of the boo%&

YB 0 , T) sin,> ( a 7 T ( d 0 T ( !

T) 0 2$!$ :N

B. Y;


15/97

4H! Situation 2

+. @ 0 1! =

O. Pa/t 17 Bearing stress bet$een the pin and str*t&

Bearing area Ap 0 2 ( -pin-tstr*t

Ap 0 2 ( -1-1,

Ap 0 32, %%2

4;! f p 0 P

A p f p 0148,000

320

f p 0 #%2!$ MPa

. Pa/t 27 'hearing stress in pin& -do*ble shear

'hearing area" AL 0 2 (π

4 -12 0 ,!.2! %%2

'hear for)e" @L 0 @ 0 1! =

. f L 0

PV

A V f p 0148,000

402.124

f p 0 %'!0$ MPa

=. Pa/t 7 'hearing stress in bolts&

'hearing area" AL 0 2 (π

4 -12 0 ,!.2! %%2

'hear for)e" @L 0 @ )os 3,> @v 0 1! )os 3,>

@v 012.162 =

P. f L 0

PV

A V f p 0128,172

804.248

f p 01$!# MPa

4P! Situation

Q. Pa/t I an6 II7

P*ter dia%eter" o 0 3,, %%

+nner dia%eter" i 0 2 %%

I. Area" A 0 π

4 -3,,2 – 22 0 5"5!1.66 %%2


16/97

'. o%ent of inertia" + 0 π

64 -3,,! – 2! 0 54.4,1 ( 1,

%%!

o%ent d*e to @" p 0 @ ( e 0 3 ( ,.1 0 ,.3 =%

o%ent at base d*e to


17/97

E4!

##. Bay" s 0 %

$ 0 ) ( p ( s

#;. $1 0 ,.,-1.!!-

$1 0 .412 =/%

$2 0 ,.1-1.!!-

$2 0 ,.! =/%

#D. $3 0

,.5-1.!!-

$3 0 !.32 =/%

$! 0 ,.!-1.!!-

$! 0 3.!5 =/%

#

#+. ;1 0 $1 ( ! 026.! =

#O. ;2 0 $2 ( .325 0

5.!! =

;2( 0 ;2 sin V

0 1.62 =

;2y 0 ;2 )os

V 0 5.1! =

;3 0 $3 ( .325 0

26.322 =

;3( 0 ;3 sin V

0 .! =

;3y 0 ;3 )os

V 0 25.42 =

#K. ;! 0 $! ( ! 0 13.2! =

#. YA 0 , ;1-2 7 ;!-2 7 ;3(-5 0 BL-12 7 ;2(-5 7 ;2y-37;3y-4

26.!-2 7 13.2!-2 7 .!-5 0 12BL 7 1.62-5

7 5.1!-37 25.42-4

BL 0 1,.4!! = -do$n$ard

#. Y;L 0 , AL 7 BL 7 ;2y 7 ;3y 0 ,

AL 0 < 20!1% :N -do$n$ard

#=. Y right 0 , -'ee fig*re belo$

;3-3.12 7 ;!-! 7 B


18/97

26.322-3.12 7 13.2!-! 7 B


19/97

ER!

ER!

ES!

#T. ;1 0

1

2 K a γ soil


20/97

*A!

f b 0 6 M

b d2 9 ;b

6 (63.853 x106)

b(300)2 0 1!.6

b 0 2'!% 3

*B! f L 0

3 V

2 bd 9 ;L

3 (85,137)

2b (300) 0 1.!

b 0 2'!% 3

*C! Situation %

;. $ 0γ

) ( bh $ 0 2, ( -,.!-,.

$ 0 !. =/%

;#. o%ent at fi(ed end 0 1- 7 !.--3

0 14!.! =%

;;. A(ial stress d*e to prestressing for)e" f pa 0

− Psbh

f pa 0

−540,000400(600)

;D. ;pa 0 2.25 @a

;


21/97

;=. f top 0 2.25 ,.,225e 7

6 M

b h2

;P. f top 0 2.25 – ,.,225-1,, 76 (194.4 x 106)

400(600)2

;@. f top 0 > !% MPa

;Q. Pa/t 7 Lal*e of EeF s*)h that the stress in the top fiber

at fi(ed end is ero&

;I. f top 0 2.25 – ,.,225e 7

6 M

b h2

;'. , 0 2.25 – ,.,225e 76 (194.4 x 106)

400(600)2

;T. e 0 2%0 33

*U! Situation

;L. Gnit $eight of )on)rete"γ

) 0 2! =/%3

ead load press*re 0 3.2 @a

ive load @ress*re 0 3. @a

;W. Weight of bea%&

$b 0γ

) A) $b 0 2!-,.!-,.

$b 0 5.6 =/%

;R. Weight of slab&

ps 0γ

) t ps 0 2!-,.1

ps 0 2.! @a

;S. ;a)tored floor press*re&

p* 0 1.!-3.272.! 7 1.6-3.

p* 0 13.4 @a

;. #H*ivalent load on bea% d*e to fa)tored press*re&


22/97

DA. $*1 0

pu s

6 3−( s ! )

2

x 2$*1 0

(13.96 ) (3 )6 [3−( 37.5 )

2

] x 2$*1 0 34.! =/%

DB. Total fa)tored *nifor% load -in)l*ding bea% $eight

$* 0 1.!-5.6 7 34.!

$* 0 #!1 :N?3 → Pa/t 1

DC.

D. o%ent at D" D 0−wu !

2

12 D 0

−47.71(7.5)2

12

D 0 223.!3 =%

D#. Iea)tion at D" ID 01

2 $* ID 01

2 -!6.61-6.5

ID 0 16.41 =

G*! a(i%*% fa)tored shear in bea% D


23/97

DK. ead load press*re 0 62, @a

ive load press*re 0 1,,, @a

Wind 0

1!,, @a

Bea%

$eight 0 64 =/%

;b( 0 2,6

@a;by 0 2,6

@a

V 0 ar)tan

-1/!

V 0 1!.,3>

D.

D. Wind )oeffi)ient&

Wind$ard )oeffi)ient 0 ,.2

ee$ard )oeffi)ient 0 ,.

D=. ead load $ 0 62,-1.2 764

$ 0 4!3 =/%

DP. ive load $ 0 1,,,-1.2

$ 0 12,, =/%

D@. Wing& $$$ 0 1!,,-1.2-,.2

$$$ 0 33 =/%

DQ. $l$ 0 1!,,-1.2-,.$l$ 0 1,, =/%

DI. Pa/t 17 *e to dead and live load only

D'. $= 0 -$ 7 $ )os V $= 0 -4!3 712,, )os 1!.,3>

$= 0 2,64.,15 =/%

DT. $T 0 -$ 7 $ sin V $T 0 -4!3 712,, sin 1!.,3>

$T 0 514.65! =/%

DG. ( 0 w " !

2

8 ( 02079.015 (6)

2

8

( 0 4.35 =%

DL. f b( 0

M x

# x f b( 09.356 x 10

6

6.19 x104

f b( 0 151.1!.@a


24/97

DW. y 0wT !

2

8 y 0519.754 (6)2

8

y 0 2.334 =%

DR. f by 0

M y

# y f by 02.339 x 10

6

1.38 x 104

f by 0 1%!#'$ MPa

DS. Pa/t 27 ead 7 ive 7 Wind on $ind$ard side

$=2 0 ,.65-$= 7 $$$ $= 0 ,.65-2,64.,15 7

33

$= 0 111.22 =/%

D. $T2 0 ,.65-$T $T 0 ,.65-514.65!

$T 0 34.15 =/%


25/97

,.1 @a

fM ) 0 21 @a


26/97


27/97

) 1,, 7

350−h x3

h( 0 [ -,, – 2 ( !, – [-12 7 [-25 7[-12

h( 0 262.5 %%

1,, 7

350−h x3 0 12 %%

+C. Therefore" *ses s 0 1,, %%

I4! Situation 10

+#. * 0 !!, =% )over 0 !, %%

L* 0 2, = fM ) 0 2,.6 @a

T* 0 1, =% f y 0 !15 @a

b 0 !,, %% f yt 0 265 @a

h 0 5,, %% Bar dia%eter" d 0 12 %%

Allo$able shear stress in )on)rete" ;v) 0 ,.6 @a

+;. #ffe)tive depth" d 0 5,, – 5 0 !35 %%

+D. Pa/t 17

* 0 !!, =%

+


28/97

+P. *1 0 * %a( 0 !,6 =%

+@. * 0 *1 7 *2 !!, 0 3.32 7 *2

*2 0 63. =%

+Q. Tension steel area" As 0 As1 7 As2

+I. As1 0 N%a( bd As1 0,.,154-!,,-!35

As1 0 2"6, %%2

+'. *2 0 T2 -d – dM 63.6 ( 1, 0 As2 -!15-!35

5

As2 0 533 %%2

+T. As 0 2"6, 7 533 0 , 1 332

+G. Pa/t 27

L* 0 2, =

+L. Av 0 2 (π

4 -122 0 22.2 %%2

+W. Ln 0

V u

ϕ Ln 0280

0.85

Ln 0 324.!12 =

+R. L) 0 ;v) b$ d L) 0 ,.6-!,,-!35

L) 0 132.2! =

+S. Ls 0 Ln – L) Ls 0 324.2! – 132.2!

Ls 0 146.16 = U 1/3 √ f c'

bw d

+. ' 0

A % f yh d

V s s 0226.2 (275 ) (435 )

197,170

s 0 136.2 %%

OA. a(i%*% spa)ing -d/2 0 216.5 %% or ,, %%

OB. Therefore" s 0 136 %%

OC. Pa/t 7

Al 0

A t

s ph( f yt f y )cot

2)

V 0 !5>

( 0 !,, – ! ( 2 0 3, %%

y 0 5,, – ! ( 2 0 !, %%

Aoh 0 ( y 0 3,-!, 0 125"! %%2


29/97

Ao 0 ,.5Aoh 0 1,"1!

%%2

ph 0 2 -( 7y 0 1!32

%%

O. Tn 0 T u

ϕ 0

180

0.85 0

211.66 =%

O#. Tn 0

2 Ao A t f yt

s cot )

211.66 ( 1, 0

2 (106,814 ) At (275)

s cot 45*

O;.

A t

s 0 3.,5 %%

OD. Al 0

A t

s ph( f yt f y )cot

2)

Al 0 3.,5-1"!32 (

275

415)

)ot2

!5>

Al 0 3" !2, %%2

@H!

@I!

@@!

@;!

@"!

O. MATHEMATICS, SURVEYING & TRANSPORTATION

ENGINIRING

(MAY 2012)

1. R and S are inversely proportional $ith ea)h other. Diven that R 0 15",,, $hen

S 0 12"5,,. ;ind R $hen S 0 32"!,,.

A. 6"!22.35

B. "56.!5

C. 6"!4.5

. "45.32

2. The s*% of seven )onse)*tive integers is ero. What is the s%allest integer?

A. !

B. 1

C. 3

. 2


30/97

3. The s*% and prod*)t of three distin)t positive integers are 15 and !5"

respe)tively. What is the largest integer?

A. 5

B. 4

C. 15

. 6

!. What is the )*rved s*rfa)e area of a spheri)al seg%ent -$ith t$o bases if the

dia%eters of the bases" $hi)h are 25 )% apart" are 1,, )% and 1!, )%"

respe)tively.

A. 11"63.!3 )%2

B. 1,"56.43 )%2C. 13"63.3! )%2

. 12"32.65 )%2

5. The area of a par on a %ap is 5,, %%2. +f the s)ale of the %ap is 1 to !,",,,"

deter%ine the tr*e area of the par in he)tares -1 he)tare 0 1,!%2

A. !,

B. ,

C. 1,

. 12

. #val*ate the interal& ∫π

3

2 π

3

cscx cot x dx

A. 1

B. ,

C. [

. 1

6. ;ind the general sol*tion of the follo$ing differential eH*ation&

. yF 7 3yM !y 0 ,

A. y 0 C1 e!( 7 C2 ( e(

B. y 0 C1 e!( 7 C2 ( e

(C. y 0 C1 e!( 7 C2 e(

. y 0 C1 e!( 7 C2 e

(

#.

;.

D.


31/97

A. 3!.!=B. 3 =

C. 32.12 =

. !2 =

2. 'i( - steel )ables are *sed

to s*pport a )ir)*lar

%o*lding having a dia%eter

of 2 % and $eighing 3.=/%. The )ables are eH*ally

spa)ed aro*nd the %o*lding

and atta)hed to a single hoo

3 % above the %o*lding. +f

the allo$able stress in the

)able is 1,5 @a" $hat is the

reH*ired dia%eter?

A. %%

B. 6 %%C. 4 %% . 1, %%

3. A verti)al steel rod is fi(ed at the top and s*pports an = load at the lo$er

end. The rod is 1,%% n dia%eter and 25 %% long. Gnit $eight of steel is 66

=/%3. What is the total elongation of the rod?

A. 12.632 %%

B. 12.53 %%

C. 12.463 %%

. 12.12 %%

!. A hallo$ )ir)*lar t*be has an o*tside dia%eter of 5 %% and is 5 %% thi). The

t*be is fi(ed -)antilever at one end and s*be)ted to a torH*e of ! =% at its

free end. What is the %a(i%*% shearing stress in the t*be?

A. 6.5 @a

B. 4.6 @a

C. 42.3 @a

. !.2 @a

5. A de)orative )on)rete bea% is si%ply s*pported over a span of %. The bea%

$eighs ! =/% and the )ra)ing %o%ent is 3 =%. What is the safe *nifor%

load of the bea%?

A. !.!! =/%

B. !.! =/%

C. 5.2! =/%

. 3.! =/%

. A 2. % )antilever bea% )arries a *nifor%ly distrib*ted load of 2, =/%

thro*gho*t its length and a )on)entrated load of 3, = at a point 2 %eters fro%

the fi(ed end. What is the bending %o%ent at the fi(ed end?

A. 41.3 =%

B. 6.6 =%

C. 123.4 =%

. 44.2 =%

6. A 12 % long bea% is si%ply s*pported at the left end and at 3 % fro% the right

end. The bea% $ill be s*be)ted to a *nifor%ly distrib*ted %oving load. What


32/97

total length of the bea% %*st be s*be)ted to this load to prod*)e %a(i%*%

negative %o%ent at %idspan?A. 4 %

B. 3 %

C. 6.5 %

. !.5 %

E!

;. Situation 1 – The

hoo is s*be)ted to

three for)es @" Qand ' as sho$n. @ 0

35 = and Q 0 !5

=.

. eter%ine the angle α

s*)h that the res*ltant of

the three for)es is , =

a)ting horiontally to the

right.

A 22.5>B 21.6>

C 2!.4> 23.12>

1. +f angle α 0 ,>" find the %agnit*de of the for)e ' s*)h that the res*ltant

for)e is horiontal to the right.

A ! =

B 51 =

C !2 =

!5 =

2. ;ind the %agnit*de of the for)e ' s*)h that the three for)es are in eH*ilibri*%.

A. !3.6 =

B. !,.43 =

C. !5.4 =

. 3.5 =

E!

;. Situation 2 –

The horiontal

distan)e fro%

A at one end of

the river to

fra%e C at theother end is 2,

%. The )able

)arries a load

of W 0 5, =. The sag EdF of the )able is 1 %.

D.

3. ;ind the distan)e (1 s*)h that the tension in seg%ent AB of the )able is eH*al

to that seg%ent BC.


33/97

A. 4 %B. 1, %

C. 12 %. 11 %

!. Cal)*late the tension in seg%ent BC $hen (1 05 %.

A. 2,.5 =B. 16!.4 =

C. 15.!3 =. 16.42 =

5. What is the total length of the )able $hen (1 0 5 %?

A. 2,.13 %

B. 2,.6 %

C. 21.12 %

. 14.6 %

E!

;. Situation – The 1.%

dia%eter )ir)*lar plate

sho$n is s*pported by

eH*ally spa)ed posts along

its )ir)*%feren)e. A load @

0 115, = is pla)ed at

distan)e ( 0 ,.!5 % fro%

post A.

. =egle)ting the $eight of the

plate" $hat is the rea)tion at post

A?

A. 3!.2 =B. 6.6 =

C. 141.6 =

. 14!. =

6. =egle)ting the $eight of the plate" $hat is the rea)tion at post B?

A. 6.6 =

B. 3!.2 =

C. 14!. =

. 141.6 =

. Considering the $eight of the plate" $hat is the rea)tion at C? the plate is !5

%% thi) and the *nit $eight of steel is 66 =/%3.


34/97

A. 14!. =B. 141.6 =

C. 3!.2 =

. 6.6 =

#. Situation # – The

billboard" 3 % high by ! %

$ide" is s*pported as sho$

in the fig*re. The total$eight of the billboard is 3,

=. < 0 1.5 %" V 0 ,>.

Wind press*re" H 0

1. @a

Wind press*re

)oeffi)ient" ) 0 1.,

4. The horiontal )o%ponent of the rea)tion at A is nearest to&

A. 14.5! =

B. 21.4 =

C. 1.3 =

. 12.!5 =

1,. What is the a(ial stress str*t BC $hose )ross se)tional di%ension is %% (

6 %%?

A. 4!.1 @a

B. 6.3 @a

C. 6.5 @a

. 1,2. @a

11. +f the str*t AB $ere repla)ed by a 1 %% ∅ steel )able" deter%ine thenor%al stress -in @a in the )able.

A. .5 @aB. 4,.1 @a

C. 4.3 @a. 44.1 @a

#. Situation $ –

A girder

$eighing 1

=/% is

s*spended on

a paraboli)

)able by a

series of

verti)al

hanger. The length of the bea% is 2! % and the sag of the )able is 3

%.

12. What is the verti)al )o%ponent of the rea)tion at A?

A. 2!, =B. 25, =

C. 21 =. 265 =

13. What is the tension in the )able at the )enter?


35/97

A. !6 =B. !12 =

C. !32 =. 521 =

1!. +f the allo$able )able tension is 3, =" $hat is the %ini%*% sag?

A. !.5 %B. 3.5 %

C. 5 %. 5.5 %

#. Situation % – 'teel tan $ith an o*tside dia%eter of ,, %% has a

$all thi)ness of %%. The tan is *sed as storage of gas *nder a

press*re of 2.2 @a.

15. eter%ine the val*e of the tangential stress in the tan $all.

A. 3.2 @a

B. ,.3 @a

C. 4.! @a

. 4,.2 @a

1. eter%ine the val*e of the longit*dinal stress in the tan $all.

A. 3.5 @a

B. !3.1 @a

C. 3!.6 @a

. !,.2 @a

16. +f the allo$able tensile stress in the $all is 12! @a" to $hat val*e %ay the

gas press*re be in)reased?

A. 3.65 @aB. 2.63 @a

C. !.123 @a. 3.346 @a

#. Situation – The solid pole

sho$n in the fig*re is loaded $ith

verti)al load @ 0 3= and lateral

load < 0 ,.!5 =. The pole is 3 %high 2, %% dia%eter and

$eighs 22 =/%3.

1. What is the %a(i%*% )o%pressive

stress at the base?

A. ,.65 @a

B. ,. @aC. ,.5 @a

. ,.52 @a

14. What is the %a(i%*% tensile stress at

the base?

A. ,. @a

B. ,.52 @aC. ,.65 @a

. ,.5 @a

2,. What is the %a(i%*% shearing stress in the pole?

A. ,.,,46 @a B. ,.,,5! @a


36/97

C. ,.,132 @a . ,.,115 @a

#. Situation ' –

The barge

sho$n in the

fig*re s*pports

the load $1 and

$2. ;or this

proble%" $1 0

1!5 =/%" $2

0 24, =/%" 1

0 3 %" 2 0

%" 3 0 3 %.

21. What is the length of barge EF so that the *p$ard press*re is *nifor%?

A. 15 %

B. 12 %

C. 2, %

. 1 %

22. What is the shear at 3 % fro% the left end?

A. 12 =

B. 151 =

C. 14! =

. 16! =

23. At $hat distan)e fro% the left end $ill the shear in the barge be ero?

A. ! %

B. 5.5 %

C. 5 %

. !.5 %

#. Situation – A )on)rete pad s*pports t$o distrib*ted loads of 112

=/%" as sho$n in the fig*re. +t reH*ired to deter%ine the %a(i%*%shear ad %o%ent in the pad d*e to these loads.

2!. What *nifor% base press*re EHF is ind*)ed by these loads?

A. 2! =/%

B. 32 =/%

C. ! =/%

. !2 =/%

25. What is the %a(i%*% shear a)ting on the )on)rete pad?

A. 2! =B. !2 =

C. 32 =. ! =

2. What is the %a(i%*% %o%ent on the pad?

A. !2 =%B. 2! =%

C. 32 =%. ! =%


37/97

E!

;. Situation 10 – A 1,% long bea% is si%ply s*pported at the left end

and at 2 % fro% the right end. The bea% $ill be analyed for

%a(i%*% shear at the %idspan that )an be ind*)ed by a %oving load.

26. What is the ordinate of the infl*en)e diagra% at the %idspan?

A. ,.3

B. ,.!5

C. ,.25

. ,.5

2. What is the ordinate of the infl*en)e diagra% at the free end?

A. ,.3

B. ,.!5

C. ,.25

. ,.5

24. The bea% $ill be s*be)ted to a *nifor%ly distrib*ted %oving load. What

total length of this bea% %*st be s*be)ted to this load to prod*)e %a(i%*%

shear at the %idspan?

A. ! %

B. %

C. 3 %

. 5 %

#. Situation 11 –

The tr*ssed

bea% sho$n is

5.! % long. A

%an of $eight

EWF is standing

at the %iddle of the bea%. =egle)t the $eight of the bea%.

3,. The )apa)ity of the rod is 2=" $hat is the safe %a(i%*% $eight of the %an

in g?

A. 132 gB. 124 g

C. 15 g. 16 g

31. +f the %an $eighs 5 g" $hat is the tensile stress in the rod if its dia%eter

is 1,%%?


38/97

A. 12.4 @aB. 1!.35 @a

C. 16.6 @a. 1.6 @a

32. What is the total length of the rod?

A. .12 %B. 5.4 %

C. 5.3! %. 6.32 %

#. Situation 12 – The tr*ss sho$n is %ade fro% D*io 1,, %% ( 15,

%%. The load on the tr*ss is 2, =. =egle)t fri)tion.

;.

D. Allo$able stresses for D*io&

Co%pression parallel to grain 0 11 @a

Co%pression perpendi)*lar to grain 0 5 @a

'hear parallel to grain 0 1 @a

'hear longit*dinal for oints 0 1.!5 @a

33. eter%ine the %ini%*% val*e of (.

A. 1, %%

B. 15, %%

C. 1, %%

. 1!, %%

#.

3!. eter%ine the %ini%*% val*e of y in %%.

A. 3!.4

B. 2.

C. 13.2

. 14.5


39/97

35. What is the a(ial stress in %e%ber AC in @a?

A. 1.2

B. 1.4

C. ,.6

. 2.6

#. Situation 1 – The lap oint of a tension %e%ber is sho$n in the

fig*re. The plate is 252 %% $ide and 12 %% thi). The bolts are 2,

%% in dia%eter and the holes are 3 %% larger than the bolt dia%eter.

'teel is A3 $ith ;y 0 2! @a and ;* 0 !,, @a. +t is reH*ired todeter%ine the )apa)ity of the oint based on gross area" net area" and

blo) shear.

;.

3. eter%ine the safe val*e of @ based on tension on gross area.

A. !5, =

B. !2, =

C. 5,, =

. !, =

36. eter%ine the safe val*e of @ based on tension on net area.

A. !34 =

B. !21 =

C. !53 =

. ! =

3. eter%ine the safe val*e of @ based on tension on blo) shear.

A. !23 =B. !4 =

C. !45 =

. 521 =


40/97

#. Situation 1#

– A % long

fi(edended

bea% )arries a

*nifor%ly

distrib*ted

load of 2,

=/%. Gse # 02,, D@a and +( 0 6.5 ( 1,

%%!.

;.

34. eter%ine the %o%ent at the fi(ed end.

A. , =%

B. 55 =%

C. 5 =%

. 5, =%

!,. What is the %a(i%*% shear in the bea%?

A. , =B. 55 =

C. 5 =. 5, =

!1. Co%p*te the verti)al defle)tion at the %idspan.

A. ! %%

B. 6 %%

C. 5 %%

. %%

#. Situation 1$ – A fi(ed end bea% has a span of 1, % and s*pports a

s*per i%posed *nifor%ly distrib*ted load of 2, =/%.

;.

D. @roperties of W !5, ( 6,&


41/97

!3. What is the average shearing stress in the bea%?A. 2!.35 @a

B. 2.42 @a

C. 23.15 @a

. 14.32 @a

!!. eter%ine the %a(i%*% shearing stress in the bea%

A. 2.42 @a

B. 14.32 @a

C. 2!35 @a

. 23.15 @a


42/97


43/97


44/97

#. Situation 1% – A b*ilt *p

se)tion )onsisting of W 35,

( 4, $ith t$o 12 %%

plates $elded to for% a

bo( se)tion as sho$n in

the ;ig*re ',1. The se)tion

is *sed as a )ol*%n 1,

%eters long. The )ol*%n isfi(ed at both ends and

bra)ed at %idheight abo*t

the $ea a(is -Sa(is. The

)ode provision is given in

;ig*re ='C@,1. Gse ;y 0 2! @a.

;.

D. @roperties of W35, ( 4,&


45/97

D. ;a 0 [1−( K!

r )2

2 + c2 ] $ y $ , # , ;.'. 0

5

3

+3(

K!

r )

8 + c−

( K!r )3

8 + c3


46/97

!4. What is the %a(i%*% bending stress in the )ol*%n d*e to a %o%ent of 6,

=%" abo*t the (a(is of the se)tion?

A. 11!. @a

B. 123.4 @a

C. 4.5 @a

. 111.1 @a

5,. What is the )riti)al -%a(i%*% effe)tive slenderness ratio of the )ol*%n?

A. !.2B. 6.1 C. 5!.!. 5.2

#. Situation 1' – The de) of a bridge )onsists of a ribbed %etal de)

$ith 1,, %% )on)rete slab on top. The s*perstr*)t*re s*pporting the

de) is %ade of $ide flange steel bea%s strengthened by a )over plate

1 %% ( 2, %% one at the top and one at the botto%" and is spa)ed

1.2 % on )enters. The bea%s are si%ply s*pported over a span of 25

%. The loads on ea)h bea% are as follo$s&

;.

D. ead load 0 12 =/% -in)l*ding bea% $eight and de)

Wheel live loads&

;ront $heel 0 1 =

Iear $heel 0 62 =

Wheel base 0 !.3 %


47/97

A. 123 @aB. 1,6 @a

C. 42 @a. 4 @a

52. Cal)*late the %a(i%*% bending stress in the bea% d*e to live load pl*s

i%pa)t.

A. 64 @a

B. 2 @a

C. @a

. 5 @a

53. Cal)*late the %a(i%*% average $eb shear stress in the bea% d*e to live

load pl*s i%pa)t.

A. 6. @a

B. .5 @a

C. 4.1 @a

. 12.! @a

#. Situation 1 – The W!5,( bea% is s*pported by a )on)rete $all

and a 13,%% $ide bearing plate as sho$n. The bea% rea)tion is

25, =. All steel are A3 steel $ith ;y 0 2! @a. Con)rete strength fM )

0 26.5 @a.

;. @roperties of W!5,( are as follo$s&

D. d 0 !5, %%

bf 0 14, %%


48/97

E!

;. Situation 20 – The floor fra%ing plan of a reinfor)ed )on)rete

str*)t*re is sho$n in the fig*re. The bea%s are 2, %% $ide and 52,

%% deep and the slab is 11, %% thi). Pther than )on)rete $eight"

the floor is s*be)ted to additional -s*peri%posed dead load of 3 @a

and live load of 5.2 @a. Gnit $eight of )on)rete is 23.5 =/%3.

D. *e to spa)e )onsideration" the )ol*%ns # and < are to be re%oved.

This $ill %ae girder B#


49/97

I!

@!

;!

"!

. Situation

21 – The

floor

fra%ing plan

of a

reinfor)ed

)on)rete

str*)t*re is sho$n in the fig*re. Then the )ol*%ns # and < are

deleted" girder B#


50/97

. Situation 22 – The re)tang*lar footing sho$n is s*be)ted to a(ial

load of @ 0 12,, = and a %o%ent of 0 3, =%. it is reH*ired to

deter%ine the safe gross bearing )apa)ity of the soil to s*pport the

given loads. The *nit $eights of )on)rete and soil are 23.5 =/%3 and

1 =/%3" respe)tively.

.

3. What is the %a(i%*% fo*ndation press*re in @a?

A. 25 @a

B. 26! @a

C. 26 @a

. 321 @a

!. What is the %ini%*% fo*ndation press*re in @a?

A. ! @a

B. 4 @a

C. 2 @a

. 5! @a

5. What is the %ini%*% reH*ired gross allo$able soil bearing )apa)ity to )arry

the given loads?

A. 31, @aB. 2, @a

C. 24, @a. 3,, @a

#. Situation 2 – The T

bea% sho$n res*lted

fro% %onolithi)

)onstr*)tion of the

bea% and slab. The

effe)tive flange $idth

is 11,, %% and the

*nifor% slab thi)nessis 12, %%. $idth of

bea% is 3!, %%" total

depth of the Tse)tion

is 54, %%. The

)entroid of steel is 6,

%% fro% e(tre%e

)on)rete fiber. Con)rete strength fM ) 0 21 @a and streel strength f y 0

!15 @a. Gse strength design %ethod.


51/97

;.

. Cal)*late the no%inal strength of the bea% for positive %o%ent negle)ting

the )ontrib*tion of the top reinfor)e%ent.

A. 56.2 =%

B. 5,3.2 =%

C. !5.1 =%

. 52.5 =%

6. Cal)*late the no%inal strength of the bea% for negative %o%ent.

A. 24. =%B. 321.4 =%

C. !32.12 =%. 23.!3 =%

. Cal)*late the reH*ired no%inal shear strength of the bea% if it is s*be)ted

to a fa)tored shear of 22, =.

A. 24.! =

B. 24.5 =

C. 25. =

. 231.4 =

#. Situation 2# – A reinfor)ed )on)rete bea% has a $idth of 3,, %%

and an overall depth of !, %%. The bea% is si%ply s*pported over aspan of 5 %. 'teel strength f y 0 !15 @a and )on)rete strength fM ) 0

2 @a. Con)rete )over is over 6,%% fro% the )entroid of the steel

area. Gnit $eight of )on)rete is 23.5 =/%3. Pther than the $eight of

the bea%" the bea% )arries a s*peri%posed dead load of 1 =/% and

live load of 1! =/%. Gse the strength design %ethod.

4. eter%ine the %a(i%*% fa)tored %o%ent on the bea%.

A. 135 =%

B. 121 =%

C. 1 =%

. 13 =%

6,. +f the design *lti%ate %o%ent )apa)ity of the bea% is 2,=%" deter%ine

the reH*ired n*%ber of 2, %% tension bars.

A.

B.

C. 4

. 6

61. +f the bea% $ill )arry an additional fa)tored load of 2!, = at %idspan"

deter%ine the reH*ired n*%ber of 2, %% tension bars.

A. 1!

B. 4

C. 1,

. 12


52/97

#. Situation 2$ – The

se)tion of a )ol*%n

is sho$n in the

fig*re. ;or this

proble%" b1 0 3,,

%%" b2 0 1, %%"

d1 0 25, %%" d2 0

35, %%. fM ) 0 2@a" f y 0 !1! @a.

62. eter%ine the lo)ation of

the gross )on)rete area

%eas*red fro% ya(is.

A. 21 %%

B. 22 %%

C. 26! %%

. 253 %%

63. eter%ine the lo)ation of the plasti) ne*tral a(is of the )ol*%n %eas*red

fro% the ya(is. =egle)t the area of )on)rete o))*pied by the steel.

A. 262 %%

B. 3,2 %%

C. 22 %%

. 242 %%

6!. eter%ine the fa)tored %o%ent * d*e to fa)tored load @* 0 32,, applied

!,, %% fro% the ya(is. Ass*%e that the )ol*%n is reinfor)ed s*)h that

plasti) ne*tral a(is is 24, %% fro% the ya(is.

A. 352 =%B. 36 =%

C. 32 =%. 36 =%

#. Situation 2% – The )ol*%n

sho$n in the fig*re is

s*be)ted to shear for)e

parallel to the ,, %% side.

Allo$able )on)rete shear

stress fir shear parallel to the

,, %% side is ,.1 @a.

Con)rete strength fM ) 0 21 @a

and steel strength for both

longit*dinal and

reinfor)e%ents is !15 @a.

The ties are all 12 %% in dia%eter $ith )lear )over of !, %%.

65. eter%ine the fa)tored shear for)e L* that the )ol*%n )an resist if the

no%inal shear strength provided by the ties is 365 =.

A. !21

B. 51!

C. !

. !52


53/97

6. +f the ties are spa)ed at 23, %% o )enters" $hat is the %a(i%*% val*e of

L*" in =?

A. !!

B. 521

C. 34

. !1

66. +f the fa)tored shear for)e parallel to the ,, %% side is !,, =" deter%ine

the reH*ired spa)ing of transverse reinfor)e%ent in a))ordan)e $ith the

provisions for seis%i) design.

A. 15!. %%

B. 112.5 %%

C. 125. %%

. 2,.1 %%

#. Situation 2 – A prestressed )on)rete bea% ha a $idth of 3,, %%

and an overall depth of ,, %%. the prestressing tendons are pla)ed

at a distan)e EeF belo$ ne*tral a(is of the bea% and the applied

prestressing for)e is @ 0 15,, =. There is 15` loss of prestress.

6. eter%ine the )o%pressive stress in )on)rete $hen @ is applies at the

)entroid of the bea%.

A. .!3 @aB. .21 @a

C. 6., @a. 6.5! @a

64. What is the %a(i%*% )o%pressive stress in the bea% $hen e 0 12, %%?

A. 1!.32 @aB. 1.62 @a

C. 15.5 @a. 1.42 @a

,. eter%ine the val*e of e))entri)ity EeF s*)h that the res*lting stress at the

top fiber of the bea% is ero.

A. 1,, %%

B. 12, %%

C. 2,, %%

. 15, %%

#. Situation 2' – The se)tion of a prestressed do*bletee )on)rete floor

oist is sho$n in the fig*re. The prestressing for)e in ea)h tee is 65,

=. Gnit $eight of )on)rete is 23.5 =/%3.

;.


54/97

D. The properties of the do*ble tee se)tion are&

Area 0 22,",,, %%2

+ 0 14, ( 1, %%!

y1 0 4, %%

y2 0 26, %%

y3 0 65 %%

'i%ple span" 0 %


55/97

E!

*!

G!

H!

+. Situation 0 –

Ans$er the

follo$ing H*estions&

6. Whi)h of the follo$ing deals $ith for)es at rest?

A. +%pa)tB. Kineti)

C. 'tati). yna%i)

. Whi)h of the follo$ing for)es deter%ines $hether a body $ill be at rest or in

%otion?

A. Ies*ltantB. #H*ilibrant

C. Wor. o%ent*%

4. #nergy by virt*e of velo)ity

A. @otential

B. Kineti)

C. Wor

. %o%ent*%

#.

;. Situation 1 – Ans$er the follo$ing H*estions on a(ial defor%ation of

rigid bodies&

4,. Within proportional li%it" the stress is dire)tly proportional to strain.

A. #lasti) li%it

B. So*ngMs od*l*s

C. @oissonMs Iatio

.


56/97


57/97

CS.

C.

1

A.

2

B.

3

C.

!

.

5

#.

;.

D.

6


58/97

E;! 1

#. A 0

√ 1.82+2.42+0.92 ππ

A 03.1321 %

AB 0 A 0

3.1321 %

AC 0 √ 2.42+1.82

0 3%

#. ;ACy 0

2.4

3 ;AC 0 ,. ;AC

#=. ;ABy 0

2.4

3.1321 ;AB 0 ,.62 ;AB

#P. By sy%%etry" ;AB 0 ;A

#@. YB 0 ,

;ACy-2.6 0 W-,.4 ,.;AC-2.6 0 -,.4

;AC 0 ,.!16 W

#Q. 'et ;AC 0 15 = W 0 3 =

#I. YC# 0 ,

2-;ABy-2.6 0 W-1. 5.!-,.62;AB0 W-1.

;AB 0 ,.!35,1 W 0 ;A

#'. 'et ;AB 0 15 Kn

#!#'2 :N (5o/n)

ET! 2


59/97

EU!

#L. V 0 ar)tan-3/1 0 61.55>

#W. Total $eight" W 0 3. (π

-2 0 22.14 =

#R. Y;L 0 , ( T sin V 0 22.14

T 0 3.46! =

#S. ;t 0 T ( A) 1,5 0 3"46! (π

4 -d) 2

d) 0 .4 say 33

E!

;A.

0 1, %%

0

25 %

@ 0

=

γ

s

0 66

=/%3

# 0 2,,

D@a -for

steel

;B.

Area" A 0π

4 -1,2 0 6.5! %%2


60/97

;C. Weight of rod" W 0γ

s Ls 0

66",,,]6.5!/1,,,2^-25

W 0 15!.14 =

;. #longation d*e to )on)entrated load @&

;#.

. 1 0

P!

A-

. 1 0

8,000(25,000)78.54 (200,000)

. 1 0 12.632 %%

;;. #longation d*e to o$n $eight&

;D. . 2 0

1

2

W!

A- . 2 0

1

2(151,189)(25,000)

78.54(200,000)

. 2 0 ,.12,3 %%

;


61/97

;P. 0w !

2

8 3 0w(6)2

8

$ 0 .!!! =/%

;@. 'afe *nifor% load 0 .!!! – ! 0 #!### :N?3

*=! %

*R! A 0 3,-27

1,-2.-1.!

A 0 !2 :N<

3

*S!

*T!

;G.

*V!

*!

*D!

;S. o%ent 0 $ ( Area *nder the infl*en)e diagra%

;. a(i%*% negative %o%ent at B $ill o))*r $hen the *nifor% load is

$ithin C only. Total length 0 3

GA! Situation 1


62/97

DB. Diven&

@ 0 35i

Q 0 -!5 )os ,>i –

-!5 sin ,>

Q 0 22.5i – 3.46

GC! Pa/t 17

D. Ies*ltant" I 0

,i

D#. I 0 @ 7 Q 7 '

,i 0 35i 7 -22.5i3.4 7 '

' 0 42.5i 7 3.46

'( 0 42.5 = " 'y 0 3.46 =

D;. 1

0 ar)tan

# y

# x 1

0 ar)tan38.91

92.5

1

0 22!'$9

Pa/t 27

DD. Ies*ltant is

horiontal

to the right $ith1

0

,>

D i 7 sin ,>

I(i 7 , 0 -12.5 ,.5'+ 7

-3.467,.'

D+. , 0 3.46 7 ,.'

' 0 #$ :N

G@! Pa/t 7


63/97

DK. @ 7 Q 7 ' 0

,

35i 7 -22.5i –

3.46 7 ' 0 ,

' 0 12.5i 7 3.46

'( 0 12.5

=

'y 0 3.46=

G"! ' 0

√ 12.52+38.972

' 0 #0!2 :N

D.

GN! Situation 2

DP. Pa/t 17

The tensions in the )ables are eH*al $hen their angles of

in)lination

are eH*al. 'in)e A and C are on the sa%e elevation" therefore (1

0 (2 0 1,.

GP! Pa/t 2 & 7

G=!

DI. )

0 ar)tan -5/1 0 6.4>

1 0 ar)tan -15/1 0 .14>

& 0 1, >

1 –

& 0 15.12>


64/97

D'. ;ro% the for)e

polygon&

DT.

T 1

sin 1 0T 2

sin) 0

W

sin &

DG. T1 0

50

sin15.12 * sin86.19 *

0 11!21

:N Pa/t 2→

DL. T2 0

50

sin15.12 *

sin78.69 * 0 16.42

=

DW. ength of )able&

0 (1 se) V 7 (2 se)

1

0 5 se) 6.4> 7 15 se) .14>

0 20!1 3 Pa/t →

GD! Situation

DS. Pa/t 1 & 27

=egle)ting the $eight of the plate&

D. YA 0 ,

2IB -1.35 0 115, -,.!5

IB 0 11!%% :N 0 IC


65/97


66/97

HO! Situation $


67/97

+D. 2

l 0 p0

4 t 2

l 02.2(584)

4 (8)

2 l 0 #0!1$ MPa

+


68/97

+'. 'hear" L 0 < 0 !5, =

ia%eter" 02, %%" r 0 1!,%%

'hear stress" f v 04 V

3 πr f v 0

4(450)

3 π (140 )2

f v 0 0!00 MPa

IT! Situation '

IU!

+L. W 0 1!5-3 7 24,-3 0 13,5 =

+W. o)ation of W&

W( 0 1!5-3-1.5 7 24,-3-1,.5

( 0 6.5 %

+R. ;or the *nifor% press*re at the botto% of the barge" ( 0 /2.

0 2-6.5

0 1$ 3 Pa/t 1→

+S. Gp$ard press*re" H 0

W

! 0

1305

15 0 6 =/%

+. 'hear at a point 3 % fro% the left end -B&

LB 0 H-3 – $1-3 LB 0 6-3 1!5-3

LB 0


69/97

@B! Situation

@C!

@4!

@E!

@*! Pa/t 17

OD. H 0

$orc

! H 02 x 112(1.5)

7

H 0 #' :N?3

O


70/97

@"! Situation 10

O.

O=. 'hear 0 $ ( Area *nder the infl*en)e diagra%.

OP. ;ro% the infl*en)e diagra%" the *nifor% load %*st be $ithin AB and

C to prod*)e %a(i%*% area. The total length is ! 7 2 0 % 3

@P! Situation 11

OQ. The bea% is ass*%ed hinged at B.

The for)e in the str*t is W.


71/97

OI. V 0 ar)tan-,.4/2.6

V 0 1.!35>

O'.

OT. Y;L 0 ,

2T sin V 0 W

OG. Pa/t 17 T 0 2= W 0 2-2 sin 1.!35>

W 0 1.25 = 0 125 =

OL. ass" 0

W

g 01265

9.81

@! M 12'! :5

@D! Pa/t 27

OS. 0 5 g

By ratio and proportion fro% the previo*s H*estion&

O.

T

85 4g 024"

128.99 4g

T 0 1.31 =

KA. 'tress" f t 0

T

A r f t 0

1,318

π

4 (10 )2

f t 0 1%!' MPa

KB. Pa/t 7

ength of rod 0 2 √ 2.72+0.92

ength of rod 0 $!%2 3

;C! Situation 12

;4!


72/97

K#.

K;.

KD.

K

K=. At oint C& 2; sin 1 0 2,

; 0 1. =

KP. A(ial stress on %e%ber AC 0

$

100 (150) 0 1!2$'

MPa

K@. I1 0 ; sin 1

0 1,= I2 0 ; )os 1 0 1 =

KQ. Considering I2&

KI. Pn s*rfa)e ab& V 01 0 32>

K'. ;ab 0

p x 5

p sin2

)+5 cos2) 0 .226 @a

KT. I2 0 f ab ( Aab 1",,, 0 .226 ( y ( 1,,

1!#$ 33

KG. 'hear& ;v 0 1 @a

I2 0 ;v 7 Av 1",,, 0 1 ( -1,,(

F 1%0 33

;V! Situation 1


73/97

KW. Pa/t 17 Tension on gross area&

Ag 0 252-12 0 3,2! %%2

Allo$able tensile stress" ;t 0 ,.;y 0 ,.-2! 0 1!.

@a

KR. @ 0 ;t ( A @ 0 1!.-3,2!

@ 0 #$0 :N

KS. Pa/t 27 Tension on net area.

Allo$able tensile stress" ;t 0 ,.5;* 0 2,, @a

=et area" An 0 -252 – 3 ( 23-12 0 214 %%2

K. @ 0 ;t ( An @ 0 2,,-214

@ 0#!2 :N

A. Pa/t & Blo) shear&

B. @ath 1&

Tension&

At 0

-3(2 2(23-12

At 0 4,

%%2

C. AL 0

2]3(2 73

–

2.5(23^-12

AL 0255

%%2

. ;t 0 ,.5;* 0 2,, @a

;L 0 ,.3;* 0 12, @a

#. @ 0 ;t ( At 7 ;L ( AL @ 0 2,,-4, 7 12,-255

@ 0 !4.62 =

;. @ath 2&

Tension&

At 0 -3(3 2.5(23-12

At 0 156 %%2

D. AL 0 ]3(2 7 3

– 2.5(23^-12

AL 0 126 %%2


74/97


75/97

T. a(i%*% %o%ent" %a( 0 A 0

B 0 w !

2

12

G. %a( 0

20.834 (10)2

12

L. %a( 0 163.13 =%

W. a(i%*% shear" L%a( 0 IA

0 IB

R. L%a( 0w!

2

S. L%a( 0 1,!.1 =

. Pa/t 1& a(i%*% bending stress&

f b %a( 0

M +

I x f b

%a( 0

173.613 x 106(

450

2 )

274.7 x 106

f b %a( 01#2!2 MPa

A. Pa/t 27 Average shearing

stress&

f v ave 0

V

d t w f v ave 0104.168 x 10

3

450(10)

f v ave 0 2!1$ MPa

B. Pa/t 7 a(i%*% shearing stress&

f v %a( 0

V6

I x t

C. Q 0 YAy

Q 0 15,-15-21,76.5 7 21,-1,-1,5

Q 0 6,4.65 ( 1,3 %%3

. t 0 1, %


76/97

ME! f v %a( 0104.168 x 10

3(709.875 x 103)

274.7 x 106(10)

f v %a( 0 2%!2 MPa

M*! Situation 1%

D. C) 0

√2 π

2 -

$ y 0

√ 2π 2(200,000)

248

C) 0 12.16


77/97

. 1

0

K!

r

+ c

1 0

37.66

126.17

1 0 ,.24!

=. ;' 0

5

3+

3

8 1 −

1 3

8 ;' 0 1.665

P. ;a 0 (1−1 2

2 ) $ y

$# ;a 0 133.!6 @a

@. @ 0 ;a ( A @ 0 133.!6-14"45,

@ 0 2%%2!' :N Pa/t →

M=! Situation 1

I. A 0 2A1

A 0 2-!5,

A 0 4"12, %%2

'. +( 0 2+(1

+( 0 2-3.1 (

1,

+( 0 6.2 ( 1,

T. +y 0 2-+y1 7 A (12

+y 0 2 ]2.41(1,

7 !5,-42^

+y 0 !4.2! ( 1,

%%!

G. r( 0 √ I x A r( 0 √ 76.2 x 106

9,120 0 41.!1 %%

L. ry 0

√

I y A ry 0

√

49.24 x 106

9,120 0 63.! %%

W. Pa/t 17

A(ial load 0 4,, =

A(ial )o%pressive stress&

R. f a 0 P

A f a 0900,000

9,120

f a 0 '!%' MPa


78/97

S. Pa/t 27

o%ent abo*t (asis" ( 0 6, =%

. Bending stress&

=A. f b 0

M x c

I x f b 0

70 x106(

250

2 )

76.2 x 106

f b 0 11#!' MPa

=B. Pa/t 7 Criti)al slenderness ratio

=C. ( K!r ) x 01(4000)91.407 0 !3.6

( K!r ) y 01(4000)

74.48 0 $#!## C/iti-a.←

=. Situation 1'

=#.

=;.

=D.

o%ent of inertia

of the bea% $ith )over

plate&

=


79/97

=K. f b 0

Mc

I x f b 0937.5 x 10

6(441)

4,222 x 106

) 0 !!1 %% f b 0 !2$ MPa

=. Pa/t 27 Bending stress d*e to live load pl*s i%pa)t

a(i%*% %o%ent in the bea% d*e to t$o %oving loads&

=. %a( 0( P!− Pd )2

4 P!

==. @ 0 total load 0 4, = d 0 $heel

base 0 !.3 %

@s 0 s%aller load 0 1 = 0 bea% length 0

25 %

=P. %a( 0

(90 (25 )−18 ( 4.3 ))2

4 (90)(25) 0 52!.! =%

=@. +%pa)t fa)tor 015

!+37 015

25+37 0 ,.2!14 U ,.3

-o

=Q. a(i%*% %o%ent $ith i%pa)t&

0 %a(-1 7 +%pa)t fa)tor

0 52!.!-1 7 ,.2!14 0 51.33 =%

=I. f b %a( 0

Mc

I x f b %a( 0

651.33 x106(441)

4,222 x 106

) 0 !!1 %% f b %a( 0 %' MPa

='.

=T. Pa/t 7 a(i%*% average shearing stress.

=G. a(i%*% shear o))*rs at the rea)tion $here the heaviest

load is nearest.


80/97

=L.

=W. YI1 0 , 25 I2 0 1-2,.6 7 62-25

I2 0 .4,! =

=R. a(i%*% shear in)l*ding i%pa)t&

L%a( 0 .4,! ( -1 7 +%pa)t fa)tor

L%a( 0 4.4,!-17,.2!14

L%a( 01,6.43 =

NY! f v ave 0

V

d t w f v ave 0

107.93 x103

850 (15)

f v ave 0 '!#%$ MPa

N! Situation 1

PA. Pa/t 17 oad" @ 0 25, =

Allo$able bearing stress of )on)rete" ;p 0 ,.35fM ) 0 4.25

@a

PB. @ 0 ;p A 25,",,, 0 ,.25 ( W (

13,

W 0 1!' 33 a 200 33


81/97

PC. Pa/t 27

A)t*al

bearing press*re&

P.

f p 0

P

130(200) 0

4.15 @a

P#. ( 0

1,, – 0 2 @a

P;. t 0 √3 f p x

2

$ b

t 0 √3 (9.61 ) (62 )

2

0.75(248) t 2#!# 33

Pa/t 7 Web yielding stress at toe of fillet -end rea)tion&

PD. f a 0

P

( " +2.5 4 ) t w f a 0250,000

[130+2.5 (38 ) ] 10

P


82/97

OM! ive load&

$1 0pl ( ' $1 0 5.2-2.5

$1 0 1 :N?3 Pa/t 2→

P=. Pa/t 7 ;a)tored )on)entrated load at #&

;a)tored load&

$* 0 1.! $d 7 1.6 $l $* 0 1.!-16.3! 7 1.6-13

$* 0 !.!3 =/%

OO! ;a)tored )on)entrated load at #&

I# 0 $*-.2 I# 0 !.!3-.2

I# 0 2'! :N

OP!

O=!

OR!

P'.

OT! Situation 21

OU!

OV!

O!

PR. Pa/t 17 'hear at B d*e to )on)entrated and *nifor% loads&

PS.

P. LB 0 IB1 7 IB2 LB 0 [-5-6.5 7 [-26, 7 26,PA! LB 0 2''!$ :N


83/97

PB!PC!

@. Pa/t 27 a(i%*% shear at # d*e to )on)entrated load

@#.

@;. +n ;ig*re -2&@D. L#1 0 26, =

@


84/97

Q. #ffe)tive epth" d 0 54, 6, 0 52, %%

Q. Balan)e" )balan)e 0

600d

600 + fy 0 3,6 %%

Q=. c1 0 ,.5 sin)e fM ) U 3, @a

QP. Pa/t 17 'trengh of bea% for positive %o%en" negle)tingtop bar.

Q@. As 0 2!5! %%2

QQ.

=R! Au3in5 7

Q'.QT. Asf y 0 ,.5 fM ) A) 2!5!-!15 0 ,.-21A)QG. A) 0 56",2 %%

2

U Af QL.QW. A) 0 bf ( a 56",2 0 11,, ( a

QR. a0 51.4 %%

QS.

Q. ) 0 a / cf ) 0 1 %% U)balan)e -f s 0 f y

IA.

IB. n 0 T-d a/2 n 0 Asf y-d a/2

IC. n 0 2!5!-!15-52, – 51.4/2

R4! n 0 $0!2 :N3

RE!I;. Pa/t 27 =egative %o%entID.

I


85/97

RP!

IQ. T 0 C) 7 CM sII.

I'.

IT.IG. Asf y 0 ,.5 fM ) a b 7 AM s fM sIL.

IW. f s 0 ,,

c -d

c a 0 c1)

IR.

IS. 1!63-!15 0 ,.5-21-,.5)-3!, 7 2!5! ( ,,c - 70

c

I. ) 0 ,. %% U )balan)e -f s 0 f y

'A.

'B. fM s 0 ,,80.68 - 70

80.68 0 64.!25 @a U f y -PK

'C.

'. a0 c1) 0 .%%SE!

';. n 0 C)-d – a/2 7 CM s-d – dM

'D. n 0 ,.5 fM ) a b -d – a/2 7 AM s fM s -d – dM

'


86/97

T#. * 0$" L

2

8 * 053.738(5)

2

8

T*! * 0 1%! :N3TD.

T


87/97

G@. * 0

P" L

4 7 16.43 0 !6.43 =% _ * %a(

-do*bly

GQ. *1 0 * %a( 0 33,.1! =%GI. As1 0 As %a( 0 2"5 %%

2

G'.

GT. *2 0 * – *1 0 136.64 =%

GG.GL. *2 0 T2-d – dM 136.64 ( 1,

0 ,.4, As2-!15

-!1,6,

GW. As2 0 1",5 %%2

GR.

GS. As 0 As1 7 As2 As 0 2"5 7 1",5G. As 0 3"6!3 %%

2

LA.LB.

LC. As 0

π

4 db2 = 3"6!3 0π

4 -2,2 =

L. = 0 11.4 say 12 Ja/

VE!

V*! Situation 2$

LD. Pa/t 17

L


88/97

VY. The plastic centroid of a column cross section is the point through

which the resultant column load must pass to produce uniform strain in

failure. It represents he location of h resultant force produced by the steel

and concrete.

VZ.

WA.

WB. C)1 0 ,.5 fM ) A1 C)1 0 ,.5-2-65",,,WC. ()1 0 125 %% C)1 0 165 =

W.

W#. C)2 0 ,.5 fM ) A2 C)1 0 ,.5-2-3",,,

W;. ()2 0 !25 %% C)1 0 1!44.! =WD.

W


89/97

W'. ́

0 21! 33

WT.WG. Pa/t 7

WV. The eccentricity of a column

load is the distance from the load to

the plastic centroid of the column

WW.

WR. * 0 @* ( eWS.

W. * 0 32,, ( ,.11

DA! * 0 $2 :N3

RB.DC! Situation 2%

R. b$ 0 !5, %% f y 0 !15 @a

R#. h 0 ,, %% Allo$able shear stress of )on)rete" ;v) 0

,.1 @a

R;. fM ) 0 21 @a

RD.

R


90/97

RW.RR. L* 0 Ln L* 0 ,.5-561.!

DY! L* 0 #'$!' :N

R. Pa/t 27

SA. s 0 23, %%

SB. Av 0 3 (π

4 -122 0 334.24 %%2

SC. Ls 0 f y d

s Ls 0339.29(415)(535.5)230

S. Ls 0 326.3 =

S#. Ln 0 L) 7 Ls Ln 0 14.! 7 326.3

S;. Ln 0 52!.!6 =SD.

S


91/97

C. Ash 0 ,.,4sc fc

f y 334.24 0 ,.,4

s(358)(21)

415

. s 0 2, %%

#. ini%*% reH*ire%ent a))ording to 'e)tion 5.21.!.!.2&

a b/! 0 112.5 %%b -25 0 15, %%

) 1,, 7

350 -

3

;. h( 0 [-,, – 2 ( !, – [-12 7 [-25 7 [-12

D. h( 0 262.5 %%


92/97

Q.I. Pa/t 17

'. When e 0 ,

T. f ) 0P

f ) 01275 10

3

300(600)

G. f ) 0


93/97

AA=. f top 0 P

A +

Pc

I f top 0

2(750,000)220,000

+(220,00 x 195)(90)

1890 x 106

f top 0 6.11 @a

AAO! f bot 0 P A − Pc

I f bot 0

2(750,000)220,000

+(220,00 x 195)(270)

1890 x 106

f bot 0 < #'!%0# MPa Pa/t 1→

AA@. 'tress at %idspan d*e to loads&

f top 0

Mc

I f top 0

211.36 x 106(90)

1890 x 106

f top 0 1,.,5 @a

AAQ. f bot 0 Mc

I f bot 0211.36 x 10

6(270)

1890 x 106

f bot 0 3,.14! @a

AAI. Pa/t 27 'tress at botto%" fibers d*e to servi)e loads and

prestress&=ote& There is a loss of prestress of 2,` at servi)e loads.

AAS! f bot 0 3,.14! – !.,4-1 – ,.2,

f bot 0


94/97

AA! pa 0

w

b pa 07.603

2.5

pa 0 !0# :Pa

AAR. Situation 2ead load" @ 0 6!, =

ive load" @ 0 !, =

AAS. ;a)tored load" @G 0 1.! @ 7 1.6 @ 0 1"1 =

AA. ;a)tored base press*re" HG 0

Pu

A ftg 01,818

2.4 (2.4) 0 315.25 @a

ABA. #ffe)tive depth" d 0 !5, 4, 0 3, %%

ABB. Pa/t 1 & 27 ;a)tored shear on footing" L*&

ABC.


95/97

AB. d 0 ,.3 %

AB#. Wide bea% shear&

( 0 [ -2.! – ,.35 – d 0 ,.5 %

AB;. L* 0 H* ( Area L* 0 315.25 ( -2.!-,.5

L* 0 $0!# :N Pa/t 1→

ABD. @*n)hing shear&(1 0 ,.! 7 d 0 ,.6 %

(2 0 ,.35 7 d 0 ,.61 %

AB


96/97

ABQ. = 0

A ¿̄ A s¿

= 0

4663

π

4 (20 )2

= 0 1#!' a 1$ Ja/

ABR.

ABS! Situation

0

ABT. Ans$ers&

ABU! @art 1& Stati-

ABV! @art 2& Ru.tantAB! @art 3& ;inti-

ABR.

ABY! Situation

17

AB! Ans$ers&

@art 1& Hoo:

"aK

ACA! @art 2& Poion Ratio

ACB! @art 3& Youn5

Mo6u.u


97/97

ACC.

AC.

AC#.

AC;.

ACG!

AC

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Civil Engineering Reviewer.docx