l-3rr-i/ce date: 0211112019 bangladesh university

,

L-3rr-I/CE Date: 0211112019BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY, DHAKA

L-3/T -I B. Sc. Engineering Examinations 2018-2019

Sub: CE 301 (Professional Practice and Communication)Full Marks: 210 Time: 3 Hours

The figures in the margin indicate full marks.USE SEPARATE SCRIPTS FOR EACH SECTION

SECTION -A

There are FOUR questions in this section. Question NO.4 is Compulsory. Answer Any

TWO from Question 1 to 3.

I. (a) What are the various methods available for procurement of intellectual services?

Describe the suitability of each method.

(b) A consultant is to be selected usmg QCBS method for design of Airport

Infrastructures. The qualifying marks is 75% for technical proposal. The weightage for

technical and financial proposals are 90% and 10% respectively.

After evaluation of technical proposals received following results are known:

Technical Scores:

Firm A B C D

Technical Score (T) % 80 70 90 76

After opening of financial proposals, the following figures are obtained:

Firm A B C D

Financial Offer Value (Tk) In lakh 100 - 140 110

Prepare the ranking of the finns.

2. (a) For the procurement of twenty storied building construction work with an estimated

cost of 80 crore taka, the project duration is four years. It is planned to have a minimum

billing amount is 10%. Prepare the Tender Data Sheet (TDS) for this project for the

following items:

(i) Tender Security (ii) Specific Experiences in terms of project size and work value (iii) .

Annual Construction Turnover (iv) Liquid Asset.

(b) What contract provision will apply, if the Contractor delays the work without a

reasonable ground? Explain how this provision is applied?

3. (a) Write the specification of brick work considering all aspects of material quality,

workmanship, measurement and payment.

(b) Prepare a BoQ, in a standard fonnat, for brick work of 250 mm thick and 50 m] of

work volume.

Contd P/2

(25)

(10)

(25)

(10)

(20)

(15)

•

•

=2=

CE 301

4. (a) Show diagrammatically the preparation for a career in Civil engineering by ASCE. (17)

(b) Discuss briefly the five basic characteristics ofa profession. (18)

SECTION-B

There are FOUR questions in this section.

Answer either Sea) or S(b). Question 6, 7 and 8 are Compulsory.

5. Explain in your own words what is the Engineer supposed to do in the following cases (as

per ASCE code of ethics lecture-note given to you): (15)

5(a) (i) conflict of interest

(ii) confidential information

(iii) undertaking engineering assignments

(iv) regarding public health, safety and welfare

(v) in all professional relations or professional matters.

OR

5(b) (i) competing with other engineers

(ii) accepting compensation

(iii) interchanging information and experience

(iv) express opinion on engineering subject

(v) avoiding type of conduct or practice.

6. (a) Define communication and briefly describe the process of human communication.

Name (without describing) the basic communication skills and the traits of an effective

communicator. (17)

(b) Name three broad categories of formal communication channels. Describe three types

of internal operational. (18)

7. (a) Show diagrammatically the following: (10)

(i) Scope-schedule-budget relationship

(ii) Various stages ofa project.

(b) Briefly describe the various project characteristics. (10)

8. (a) Define contracts. Briefly describe the terms used to categorize contracts. For a

contract to be binding what are six elements that must be present. (17)

(b) Briefly describe the various types of project delivery systems. (18)

.'

L-3/T-lICE Date: 24/10/2019

BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY, DHAKA

L-3/T-l B. Sc. Engineering Examinations 2018-2019

Sub: CE 311 (Structural Analysis and Design I)

Full Marks: 280 Time: 3 Hours

The figures in the margin indicate full marks.

USE SEPARATE SCRIPTS FOR EACH SECTION

SECTION-A

There are SEVEN questions in this section. Answer Q. NO.7 and any FOUR from the rest.

1. For the beam shown in Fig. I, develop influence line diagram for (i) Reaction at C

(ii) Moment at B (iii) Shear just right of C (iv) Shear just left of E. (28)

2. Compute maximum shear at quarter point of a simply supported beam of 60 ft due to the

wheel loads shown in Fig. 2. (28)

3. Determine maximum moment at one-third point of a simply supported beam of 60 ft due to

the wheel loads shown in Fig. 2. (28)

4. The member AC of the truss shown in Fig. 3 is fabricated I inch longer than specified. After

fabrication of the truss, the temperature drops by 40° F on a cold day of the winter. Calculate

vertical deflection at joint C of the truss due to misfit and drop of temperature. Given: AE is

I ..constant. Thermal expansion coefficient, a = --- per OF.Use virtual work method .. ' (28)

150,000

5. Using method of virtual work, calculate vertical deflection of point C of the beam shown in

Fig. 4. Given: E = 29000 ksi, 1= 500 in4 (28)

6. Calculate rotation of joint A of the frame shown in Fig. 5. Given: E = 29000 ksi and

1= 800 in4. Consider bending strain energy only. (28)

7. (0. NO.7 is COMPULSORY)

Estimate the earthquake base shear and storey shear forces at each level of a 21 m high 7

storied office building located in Sylhet. The structural system is SMRF and each storey

height is 3 m with a plan dimension 30x20 m. The soil type is S4. Given: Dead load =

5 kN/m2, partition wall load = 5 kN/m2, Z = 0.25, 1= 1.0, R = 12, S = 2.0 and C, = 0.083.

Notations convey usual meaning. Use equivalent static force method ofBNBC.

Conld P/2

(28)

. ,

=2=

CE 311SECTION -B

There are SEVEN questions in this Section. Answer any FIVE.

8. Draw shear force diagram, bending moment diagram and axial force diagram of the beams

and columns of the muiti"storied building frames loaded as shown in Figure 6 (Use Portal

Method). (28)

9. (a) Using Cantilever Method draw shear force diagram and bending moment diagrams for the

column ADGJ and girder GHI of the multi-storied building frame loaded as shown in Figure

7. Relative column cross-sectional areas are given beside the columns. (28)

10. Draw shear force diagram and bending moment diagram of the column ABC of the Mill Bent

shown in Figure 8. Also find the bar force in the knee bracing BG and EK. (28)

II. Analyze the suspension bridge as shown in Figure-9. Draw shear force and bending moment

diagram for the stiffening girder of the suspension bridge. Also compute maximum tension in

the cable. (28)

12. Check the static determinacy of the frame shown in Figure-l0. Draw the shear force and

bending moment diagrams for the frame. Given shear at E = - 16 kips. (28)

13. Check the static determinacy of the arch structure shown in Figure-II. Determine the

reactions at A & D and the tension in the rod AD for the loading condition as shown, Also

calculate the internal moments at point F & G. (28)

14. Check the static determinacy of the frame as shown in Figure-12. Draw the axial force

diagram, shear force diagram and bending moment diagram of the beam BCD of the frame.

C is an internal hinge. (28)

(

C£2>\\L-3/-r-1• - -~--- ~~"--

--- -- --_.- ~.-

IOrt- \O~ 6oK- lL \L IL bb'?-~ lL. .\C- r;1<j¥ .6C> ~b to 50 s-o :;-0

-t- -t- t~ ~ ~ ~

.~ ~"~" +luCD Ii) @)5' 'I to! )' 5'

s' , I 10) r 51 I

Y It ySy5)' '( 5 ~l Y 6" v7 j l'

I\2.-

"

. ~.

~I

I"'/-{:rA ¥~ld:• )( ::>

t:-\0 ~

r:o-sI-

,1J; o,SI j)

t 4--yr-11:1-- -----_._-_."-"- ------ "--"--------- --------------

"::l-- -------. ------- -.--.------..-- ----20£-it

. 161

~b~-===::=~=-~=====-====:==_=_~__J .___ _. .______. _-----ttt I",-A f 2. l;{- I' ( )f1 __ -. ._. __ _ __ • _ _ __

h'~.6

--- ----- --

-------- -- ------ - ----------"---- ---- -_ .._-_ ..'-_._---_ ....•

CE311L-3tT-1

_. /1_-,-..

//

III

K /0/I¥J\~ . 20

F( ')/6 E.

IS I

I I~@IO=~O

._------ -~~-----

A

Sti ffl nit\.~C1irct..r .

{(~-- g@ lat I;: gal) /-_._----

F{~.q------~-

A c

;f I K10' g 30'"

. r•. 12. I

oX,

c.E3\1L-,3fr-l

=5:::;

\( 41<i'ft

~ (,K~(

'D.

--~- ~---------~--------~ --- .~ -~._-

1"C.

Jnkrno.lHing.e.

2,0'

\O\( .

/

,

I,

,.

L-3/T-liCE Date: 06/11/2019

BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY, DHAKAL-3!T-I B. Sc. Engineering Examinations 2018-2019Sub: CE 315 (Design of Concrete Structures - I)



SECTION -A

There are FOUR questions in this section. Answer any THREE.

Use USD method of Design if not mentioned otherwise.

Assume reasonable value for any missing data.

I. (a) Explain the differences between USD and WSD methods of design with particular

emphasis on how safety is ensured in both the design methods. (7)(b) Discuss the behavior of RC rectangular beam in flexure under increasing load by

drawing neat sketches for strain and stress distribution of uncracked, cracked and

ultimate conditions. (8)(c) A rectangular beam made using concrete with fe' = 6.0 ksi and steel with fy = 60 ksi

has a width b = 20 in, an effective depth of d = 17.5 in., and a total depth of h = 20 in.

The concrete modulus of rupture fr= 530 psi. The elastic moduli of the concrete and steel

are respectively, Ee = 4030 ksi and E, = 29000 ksi. The tensile steel consists of four No.

10 bars.

(i) Detennine the cracking moment Mer' (10)(ii) Find the maximum service load moment that can be resisted without stressing the

concrete above 0.45 fe' or the steel above O.4fy. (10)

2. (a) What are the sources of uncertainties in analysis, design and construction of RC

structure? (5)(b) Derive equation for finding Pmax and discuss how a minimum tensile strain (0f =0.004) at failure is ensured by not exceeding maximum reinforcement ratio. Also discuss

the variation of <pwith 0' as given in ACI/BNBC code. (10)(c) A beam section is limited to width b = 14 in. and total depth h = 28 in. Calculate the

required reinforcement if the beam has to resist a factored moment Mu = 725 kip-ft.

Assume two layer tensile reinforcement with d = 24 in. and d, = 25.5 in. Also, assume

d' = 2.5 in. if compression steel is required.

Given: fe'= 3 ksi, fy = 60 ksi. (20)

-,,IIIII

3. (a) A floor system consists of a 6 in. slab supported by continuous T-beams with 30 ft

span and centre to centre spacing of 12ft. 6 in. as shown in Fig. I. Web dimensions as

determined by negative moment requirements at support are bw = 12, And d = 2 I in.

What tensile reinforcement is required at midspan to resist a factored moment Mu = 600

kip-ft? Also, check minimum reinforcement and 0,.

Given: fe'= 3 ksi, fy = 60 ksi.

Contd P/2

(15)

"l

.1

=2=

CE 315Contd ..... O. NO.3

(b) A rectangular RC beam as shown in Fig 2 measures 14 in. wide and has an effective

depth of 23 in. Tension steel consists of eight NO.8 bars in two layers (d = 23 in., dt =

24.5 in.) and compression steel consists of four NO.8 bars is located 2.5 in. from the

compression face. What is the design moment capacity of the beam according to

ACI/BNBC code? Check for yielding of compression steel and Ct.

Given: fe' = 3.5 ksi, fy = 60 ksi.

4. (a) A rectangular beam is 16 in. wide and has an effective depth of 26.5 in. For fe' = 4.0

ksi, fy = 60 ksi, determine the required spacing of NO.3 stirrups for factored shear of

(20)

(i) Vu = 30 kip (14)

(ii) Vu = 100 kip

(b) A rectangular beam has width 14 in. and effective depth 24.5 in. as shown in Fig. 3. It

is reinforced with eight No 9 bars in two rows (d = 24.5", dt = 26.0"). What is the

nominal flexural strength Mn and what is the maximum moment <pMnthat can be utilized

in the design? (15).Given: fe' = 5.0 ksi, fy = 60 ksi.

(c) Why is a minimum amount of reinforcement provided in RC beams? Write the

ACI/BNBC provisions for minimum reinforcement.

SECTION - B


Use USD method of Design. Assume reasonable value for any missing data.

(6)

5. (a) Figure 4 shows the one-way reinforced concrete floor system of a commercial

Building. Design the one-way slab supported by bcams cast monolithically with the slab.

The slab is to carry a service live load (LL) of 60 psf in addition to its self-weight.

Assume floor finish (FF) load of 40 psf and partition wall (pw) load of 60 psf.

Given: fe' =3.5 ksi and fy = 60 ksi. Usc supplied ACI moment co-efficients. Show the

reinforcements in plan and section. (20)

(b) Why are temperature and shrinkage reinforcements required in one-way slab? (9)

(c) What is the minimum length oflap for column splices as per ACI/BNBC code? (6)

6. (a) Design the slab as two-way floor slab system, when there are no secondary beams B2

in the floor system of Fig. 4. The two-way slab panels are 25'x30' in plan, supported by

column line beams only. Given: LL = 60 psf in addition to self-weight of slab; FF = 40

psf; PW = 60 psf; fe' =3.5 ksi and fy = 60 ksi. Use supplied moment-co-efficient tables.

Show the reinforcements in plan with neat sketches.

Contd P/3

(20)

=3=

CE 315Conld ..... Q. NO.6

(b) Calculate the development length of 16 mm and 25 mm uncoated top bars. Repeat the

calculation for bottom bars also. Use fe' =3.5 ksi and fy = 60 ksi. (8)

(c) What do you understand by the tenn "serviceability" in RC design? (7)

7. (a) Detennine both positive and negative reinforcements for the beam B I supported on

three columns as shown in Fig. 4. The beam B 1 is to carry load from one-way slab given

in Question 5(a) above. Consider self-weight of the beam. The size of the beam is 12"x

30". Consider material strength fe' =3.5 ksi and fy = 60 ksi. Show the reinforcements in

long and cross-sections. Use supplied moment coefficient table. (20)(b) What are the BNBC/ACI code provisions for beam stirrups (hoops) for moderate

seismic risk region e.g. Dhaka? (8)

(c) Discuss in short about equilibrium and compatibility torsion. (7)

8. (a) Design the stirrups for the beam shown in Fig. 5. All loads are factored. Calculate the

stirrups with 3 sets of spacings. Width of beam b = 12" and effective depth d = 21".

Given: fe' =3.0 ksi and fy = 60 ksi. (20)(b) Why does ACI/BNBC code recommend increased development lengths for bars in a

bundled group of reinforcements? Mention the recommended increased lengths for such

b=. ~)

(c) What are the factors that influence development length ofrebar? (7)

:' ."

Given:fc'= 3 ksify= 60 ksi

_Ll----1-

d= 2!"

... -

12'-6"

Fig. 1

...0000 ~ ¥'= 2.5"

" FourNO.8 bars

d= 23"dt= 24.5'

000

t-

I Span=30 ft I,

III,,

-t

EightNO.8 bars

;/o 0 000"0

14"

oO' ~'Given:f,'= 3.5 ksify= 60 ksi

..........0~~ 0

Fig. 2

d= 24.5"

EightNO.9 bars600o 0 0

Given:fe'= 5.0 ksi I

fy= 60 ksi . I. ,. ,

i 14" t-'Fig. 3

i,I,,,

Beam I83(12'x3O")"i

I,,-----------------------r-----------------------+ ,,,I.,,,,,,,,I1

,,,,,,I 82(12'x30') ( f I!=::::::::::::::::::::::::::::::::::::~~::::::::::::=L:::::::::::::::::::::::!, , 'I 'I r 1 I

II i I ONE WAY SLAB iI f I I: CdtJmn C2 I I Column C1 Column C2 :.f "".>.2¥~'" - "(24'J<24') (24'J<24') ,, I ~ 1 • • B1(12'x3()') , . , , ,

'" f CJ:==================================u=========== =====================i ; iI , II I ,, I ,, , I

.) I (

: Beam I: Beam:------B3(12'x3O") : VB4(15'x3O"), I ,

I 82(12'x30') : r~-------------------------------------~,------------+-------------------------------------~~------------I . I I, , ,I , II I ,, I ,, , I, , ,, , ,, , ,

'I Column C2 : I Column C1 Column C2I ,,, •••• ,,J,, , I '''A •.....••• , '''AO.. '' •• '

i \~~;;~====.=====~~~~=~~=============~'=;;~======Il-================;;;;;1. I I., •

; 1 I .

!" i! 11 B2(12'x30'} I : :T-------------------------------------r,---_________ I -----------------------T

~-------------------------------------~;-------------------------------------+I I I II r I JI I t 1I I 1 1I 1 I 1I "7_ .1 1 II I I I

./.' , I II I 1 II 1'1- 11 I I '!: 1 :: t :

3Q'.()"3Q'.()"

Hg-4

40 kip 20K' Ift(- .I r-. lD mcludmq self-wUfoo:;;oo=cv::.60~ . .ALL LOADS ARE FACTORED

••.A c:rossIlIIthcdedit incftCilCS (!l;u the shb (llrtinI,tCS ~m". ~ u tiled:d. ~ 'lIpp'o": ~n onmart~C1edge indicates~ stippof1 Ii whkh croional f~-qa~ iJ: tlt,lil~lc.

"v

\I0)\\

r"

."~ ,. .,: .,i

."'.',..,t...~.."

• A""'SlI),,<hedcdgc _,h ••• he slab """inUcs It ••••• oris fixed at. the SllPl"":" uninlrl<cd edg<:i_.a SIppOIt as. which lOD:ioml ral.uncc is p::gli"iblc. . '. . .

TABU; .1.1.4 . !. .

CoellicierttS ror dead. load positive moments in ~Iabs"Mal'" dl = Cu1wlt . .'. . '. ~ where W::S tot;).' unlromi dead loadM"-,-.j' -= C.,jlwl.. .

casc I..

Ratio' Co.", 2 Case 3 Ca,e4. caseS Case 6 Case 7 .ease 8 Case 9..,

r. 0 0 CJ Q 0 d b '0 Qm~ 7;' ..

.'1.00 C~~I Om6 'O.OIB 0.01.8 'Om7 '.0.027 0.033 0.027 , 0.020 0.023,

C•..•II 0.036 0.018 0.027 .0.027 '0.018 0.027 '0.033 0.023 0.020i

O.95'C ••.",,

0.022 .0.040 O.O~ 0.021 0.030 . 0.028 '0.036 . 0.031 . 0.024 .

C'41 0.033 0.0 6 a.ou 0.024 0.015 0.024 0.031 '0.021 : 0.017'

0.90. C •.", . 0.045 0.022 . .0.025 0.03~ '0.029 . 0.039 0.035 . 0.925 0.026

C.:;", Om9. 0.014 .0.024 0.022 ..0.01.3 0.021 0.028 . 0.019 0.015

0.85 C•.,,' . O.OSO 0.024 . 0.029 0.036 0.031 0.042 0.040 0.029 .0.028

C"", 0.026 0.012 . 0:022 0.p19 0.011 0.017 0.025:: 0.017 . .. 0.013

0.80 Cui' . .0.056 0.026 O.OJ4 '0.039 O.(132 0.045. 0.045. 0.032 0.029

C•.J, 0.023 O.QII 0.020 . 0.016 0.009 0.015 0.022 0.015 o.oio

0.15: C •.J, .0.061 0.028 .0.040 . 0.043 0.633 0.048 0.051 6.036 . 0.031

C""f 0.019 0.009 a.ills '0.013 0.007 0.012 .0.020 0.013 o.pm

0.10 C•.I!I .0.068 0.030 0.046 0.046 0.Q35 0.051 .0.058 0.040 0.033 .

.C~.,,' 0.016 0.po7 0.016 0.011 0.005 0.009 0.017 0.011 0.006

0.65 'c••, 0.074 '0.032 0.054 ! 0,050 :0.036 0,054 0.065 '0.044 0.034

C"".' '.0.01) 0.006 .. 0.014\ 0.009 . "0.004 0.007 '0'.014 0.009 0.005

0.60 Cu' 0.081 0.034 0.062 . 0.053 O.ro7 0.056. 0.073 0.048 0.036

C)"dl. .0.010 0.004 0.011 0.007 . 0.003 0.006 0.012 O.(~P 0.004. . . . . ..

, '0.55' Cui' 0.088 0.035 . :.0.071 0.056" "0.038 : . 0.OS8, 0.0Sl- .0.OS2 ..0.037

,. C'.ill 0.008 0.003 0.009 O.OO~ 0.002 0.004 0.009 0.005 0.003

0.50C.d1 0.095 0.037 G.ORO 0.059 0.039 . 0.061 0.089 . 0.056 . 0.038

C'.-II. 0.006 0.002 0.007 0.004 Q.OOI 0.003 O.OQ7 '0.004 0.002

tS/n;clvres' ~ ./9. H. Nf'ls~",. . .' ... . ..... _ .... - "

-va8- 'tIl.

..Ralia C", I Ca;sc2 Cas<) Case 4 6.••5 Case 6 Case.7 Case 8 .e..t9

. . . .

. '. 0 0,9=1 0 W U 'c:=f P0mQ-. .', .'. .i . ,

I.do C•.••.; 0.04~:0.016 .

.0:050 oms .0.Q11 0.033 '0.061. ' ••••.1 (1.045 0.050 0.071 0.061 . 0.033

0:95 c...,; O.osa (l.OSS 0.079 . .0.07.5 I. om8 0.065c•.••, .0.041. .0.072 . 0.04.5 0.067.'. 0.056 0.029. '.

'0.90 C.:.. 0.055, .'0.060 0.080 0.079 0.043 . 0.068. Cl-e. 0.037 0.010' 0.040 0.062 0.051 0.025

0.85 Caw, 0.060 .0066 0.082 0.083 0.049 O.on.. C~J' O.OJI 0.&5' 0.034 0.057. 0.046. 0.021

0.80 c......• '0.065 0.071 0.083 0.086 0.055 .omsc•...., 0.027 0.061 0.029 0.051 0.041 .0.017

0.75 c.....1 '0.069 0.fl76 0.085 0.088 0.061 0.078, CllI(I .. 0.022 0.056 0024 . . O..O.H OW6 0.014

0.70 ' •.•••• 0.074 0.081. 0.086 0.091 0.068 0.081. c•.•., 0.017 O.OSO 0.019 0.038 0.029 0.0f!

0.65 c.." O.op. .0.085 . 0.087 0.093 . 0.074 O.08j.c•.••• 0.014 .0.043 0.015 0.03\ ..0.024 0.008

0.60 c•..,. 0.081 0.089. 0.088 0.095 0.080 .0.085. ,C•.••., 0.010 0.035 0.011 0.024' 0.018 0.006

.. ' .0.55 .c~•• 0.084. 0.092 0.089 0.096 0.085 .0.086

C•."", 0.007 0.028 0.00)l 0.019 0.014 0.005

0.50 c•...,. 0.086 .,. - 0.094' 0.0Q0 0.097 0.089 0.088C""'r M06 O.OZZ 0.006 0 ..014 0.010 . 0.003

TAB.LE III .Coemcitnls lor ~egat.ive moments in slabs". M""", ~.Cur.wI1. . '. . '. .M. c' ~ where. w at !obl'Gnlform.dead plus live load,b,Mt Q . j., ••• "Ij:

@ .:z:;e~j/).of ConC/'eI.f2

12M E<i;'Hd/"i / '. Pif-2S

~:J<r.

~-V

"

11~l'

~

\~~'V'l7~-

ROltiO Ca.~; C"l~2 C.I~CJ Ca."C" . CaseS en,. 6 Co,.. 7 C...,3 Case 9I.

0,

m'" 1;,- 0 0 c:J D D 0 0 CJT ..

100 W•• 0.51l 0.5(l 0:.13 0..\0 O.R3 0.71 0.2~ 0.33 0.67. W. O.lO O:lli n.R~ OSn . 11.17 0.2'} 0.71 0.67 0.33

0.55'", ~

095 ~:0.55 !UIl 0.55 I).H6. 0.75 '0.33 . 0.33 .0.71CI,45 O.4S' ().811 0.4.\ 0.14. 0.25 0.67 0;62 0.29

w. O.r;o. 'C.(~) . 0.2.1 0.60. 0.8H , 0..79' .0.38 . 0.43- 0.750.90 w. OAO C.41l 0.77. 0,40 .1l.12 0.21 0.62 . 0.57 0.25 .

085 w. (I.M :.()J,(, 0.211 (\.66 0.9(1 (1.~3 0:43 0.,49 0.79. W. 0.14 0034 0.72 0.34 lUll 0.17 0.57 0..51 0.21..

0.30 ~. 0.71 q.71 0.33 0.71 0.92 0.36 0.49 . 0.55 ,0.83. . 0.2Y n.2Y .- 0.67 0.29 0.0.8. 0,14 0.51' 0.45 .0.17

0.75 W" 0.76 0.7(,' . . O.:'l9 .... 1l.~6 0.94 0.88 . . 0.56 0.61 .0.86. W•••. 0.24 . 0.21 (I.M 0.24 1l.06 0.12 0.44 0.39 0.!4

0.70 ~:O.XI . 0.8' 0.45 O.X1 0.95 1l.91 0.62' 0.68 0.8911.19 .0.19 0.55 0.19 0.05. .11.09 0.38 0.32 0.11

0.65 ~:n.RS 11.85 n.51 n.x; 0.96 0.93 0.69 0.74 0.920."5 0.15 . . 1),47 0.15 1l.04 0.07 0.31 , 0.26 0.00

0.60 ~:0.8Y I .o.R9' 0.61 '0.89 . '.0.97 ' .0.95 .' Q76 . . 0.30 0.940.\1 . 0.\1 .. 0.39. . 1l.1I. O,OJ 0.05 0.24 0.20 . 0.06

..055 IV. 0.92 . 0.92 0.69 0.92 0.98 . 0.96 .0.31 '. 0.35 0.95

'0.31. "

0.04 0.19 0.15 0.05.. "Y,. 0.1l8 .0.08' 0.1ll!. 0.02 <

0.50 ~:0.94 0.94 . 0.76 0.94 .0.99 0.97. 0.36. 0.39 . 0.97'.0.06 . p.06. n.24 0.06 0.01 .1l.03 0.14 0.\1 : 0.03

..• A crosshOllChcdct!xe indialo th~1 the :dah cntltinuc: ••:lCm5:>.or ~ fixed :ll.,the support; linunrnMted edit bKlicatua 5tJP(XlM nl1l'hicht~ion:d ~h;lurx:e i.••hq;fj~iblc. . . .'

TABLE 11.6

. Ratio of load W i~I. and I", directions for she~r hi slat} nnd load on supportf\.Ratio Ca::;e J Casei Case 3 Ca..,., 4 C~", 5 ~",6 : C •.«:7 Case 3 C,se9.I. 0 0 0 p.0 d ~ 0 Cm= --,.

1.00 C•.1l .0.036 0.027 0.021 0.032 0.032 0'.0)3 0.1)32 0.028 0:030Gb•tI 0.036 0.027 .' 0.032 0.032 0.027 0:032 .0.035. ' 0:030 0.023

0.95 C..,.lI .. 0.040, 0.030 . 0.031 0.035 0:034' OA~8 . '0.036 0.031 '0.0,32. : ~b-.11 0,033 0.025 0.029 0,029 0'.l!24 O~Q2., ,0.032: 0.027 0.0.25.

0.90 Cd•1I 0.04; . 0.034 0.035 O.oj~ omi. 0.042 . 0.040 0.0)5 . 0.036C".11 0.029. 0.022 0.027 0.026 0:021 0.025. ,().029 .. 0.024 0.022..'

0,85' C(l.U 0.050 .0.037 0.040 0.043 "0.041 0.046 'O ..MS . 0.041) 0.039'.. Cb•U 0.026 01119. 0:024 0.023 q.019 .0:022. .0.026 0.022 0.0200.80 C.,.I1 .. . '6.056 0'041: O.I14S O.O4ll 0.044 0.0.51' ,0..051 0.644 0.042Cb•1l 0.023 0.017 0..0.22 0.020 0.016 .0.019 ,'0.023 0.019

..0.017

0.75' C~l/ O.l)~1 0.045 O.IlSI o.on 0.047 o.ms. ' n.OS6 0.049 0.046elf•1I . 0.019 ',0.914 . 0.019' . 0.016 .(1:013 . 0.016 0.020 :0.016 .0.0130,'70 CI{I.t, 0.068 0.049 . 0.1l5) 0.057 '0.051, 0.0':.>0 '0.063 '0.Q54 0.050C"./f 0.016 0.012 0.016 0.014 '11.1111 O.I)IJ 0.017 0.014 .0:011

0.65 C~lI 0.074 0.053 .0.064: 0.062 0.055 0.064 . 0.071l .. 0.059 0.054,.. Ch•U 0.013 .0.010 11.014 o.~Hi 11.O09 O.(JIO 0.014 0.011 0.009

0.60 C~II 0.081 :0.058 0.071 . . 0.067 n.059 0.068 0.077 0.065 0.059C".11 0.010 0.007. 0.011 0.009 . 1l.007 0.ClQ3 0.011 0.009 0.007

0.55 C~lI 0.088 . 0.062 0.030' 0.072 0.063 1(073 'O.OgS": 0.070' 0.063.. Cli.1I 0.008 . 0.006 0.009 0.007 . 0.005 '0.006 0.009 .0.007. 0.006

.__0.076..

0.50 Co.t1 O,O9l .. 0.066 0.08K 0.077 n.067 '0.Q78 '. 0.092 ' 0.067C"-" 0.006 0.004 :0.007 (l.005 < 0.004 0.005 6.007 . 0.005 0.004

TABLE-I2.5Coefficienls for live load positive moments in slabs"M4.~tr =.C~.~wl~:Whc~ HI";' tot.d uni'ronn. liVe"load ..Mb.ptu .•11 ..= C.-.uwl;

tI A. crO$."halchcdcdgc irdiatatNt the',dab [IlntiRUC$ a:(tl.".~.oct.,fl,l,t.'d ;t:.lhe $tIppuR;aa UM1~ke4 edge jndk4tesi." support al whi.h !Union,11tS1""" •• , ll<gfigihk:. . ..

~;:j

s:i

~C1'S

.-- -' - '---.~,..- '\

'~

Discontinuous end 1Summary, of ACI moment unrestrained: 0 11

1 1 1 1 1 1coefficients: (a) beams with Spandrel: 24 14 10 11 16 11more ,than two spans;

Column: 1 1(b) beams with two spans 16 14only; (c).slabs with spans [lotexoeediog iOIt; (d) beams

"in '\vbic.1:!.thesum.of wlumn'_'_,'.~fft:n:~.se:,e~ceedsS ~in1es:,',' .

(aj;'.<:thdiuio'ofbeain stiffnesses"" at ,each end or' !he span.

Discontinuous ena0 1 1 0unrestrained: 11 111 1 1 1 1 1Spandrel: 24 14 9 9 14 241 1 1 1Column: 16 14 14 16

(b)

1 1 1 1 1 1 112 14 12 12 16 12 AC; w

(e)

r li1

121

121

121

12

(d):.. ~~ ~! t~;.(

"',

',' '

1.

Date: 20/0112020BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY, DHAKA

B. Sc. Engineering Examinations

Sub: CE 317 (Design of Co ncr etc Structures-II)



Assume reasonable value for any missing data.


SECTION -A


(a) Design a square "tied column with about 2.5%' reinforc"emen"ttOsuppo;.t ;:';;iking' ..unfactored loads: PDL=900 kip and PLL=600 kip. Given: fc'=5.0 kSl and fy=72.) ksl. Also, 9desi n the ties re uired.(b) A 16 X 28 inch column is reinforccd with Tcn NO.9 bars as shovm in Fig.1. Constructthc nominal strcngth interaction diagram for the column with five points corresponding topure axial load, pure bending, balance condition, 8,= 0.001 (tensilc) and 8,=0.005 (tensilc). 26Also find corresponding <j> for the above points. Assume bending about Y -Y axis. .Given: fc'=4 ksi and f =60 ksi.

2.

4.

(a) A 60"x60" square pier of a bridge is reinforced with seventy-two No-9 bars arrangeduniformly around the perimeter. Material strengths are fc'=4.0 kSl and fy=60 ksl. Checkadequacy of the short column using Load Contour Method for:

Pu=4000 kip, Mux=4000 kip-ft, Muy=5000 kip-ftUse su lied column stren'!th interaction dia ram chart assumin =0.9(b) Why is 'P value for compression lower than thosc for flexure or shear? What does thehorizontal cut-off in the ACI/BNBC desi n stren th interaction dia ram si nif.?

(c) What is a slender colunm? Write ACIIBNBC kl/r limits below which the effects ofslcnderness may be ne lected for swa and non-swa frames.(a) A circular spirally reinforced column carries unfactored working loads: PDL=700 kipand PLL=500 kip. Design the column with about 2.5% reinforcement. Also design the ACIs iraJ. Given: fc'=4.5 ksi and f, =72.5 ksi.(b) A ground floor column of a 12-storied building is to be designed for the following loadcombinations (axial force and uniaxial bending)-Gravity load combination Pu=3000 kip, Mu=IOO kip-ftLatcralload combination Pu=2000 kip, Mu=IOOOkip-ft

Architectural considerations require that a circular tied column of 40 in. diameter is to beused. Material strengths are fc'=4ksi and fy=60ksi.Find the required column reinforcement and show in sketch. Use supplied column strengthinteraction desi n chart assumin reinforcement distributed alon the erimeter and =0.9.(c) Explain the Seismic design philosophy under different levels of earthquakes.

(a) Writc the seismic detailing provisions of a two-way slab ,,~thout beam, which is part ofan IMRF s stem, as er BNBC.(b) Discuss different modes of failure of a high-rise shear wall.Cc)A shear wall of a 18-storey building is subjected to following factored loads:

Pu=650 kipVu=500 kipMu=6000 kip-ft

The wall is 20 ft long, 180 ft high and 12 inch thick. Design the shear wall with f c=4ksiand fy= 60 ksi. Ignore axial force as it is less than balanced load of the section.

20

7

8

II

17

7

10

8'II.,.

17 :11I

SECTION - B

There are FOUR questions in this Section. Answer any THREE.Assume reasonable values for any missing data

---~--- . - ."----~. IJ CO YvIm&U< 'a1 bw!JI~ -?1 ~ be- d.eJ~da5~-C:Lftd:-j2!dte" ~ZJl' d,>5~. 71ie (n~(~

~.u.~ die ;20,XPo a::».d--dY:Y- Me >f-ficd:t.L#_eyt:-----1~~drLg::-h'mJ..5_~~~~£dJ.r:vefacYf---.-i.oo f22.-f- c-1J AtAf2-&Z~o.sd d~ktYLI-h£J-P-4_t~_dJr'~_$f-we;y~t ~ .-a!.P.b .. MA;;~/al ~lrtPjlfi.s ttV2e tJ.- '" :3-S-lUtc< xvf. 7$:. ~ 6.0 1<:5/ -

laJ~-rte-5~7m-ess-fbr-d~'7- -rEJ-CXfa.<'VisfpOi1c:tt.'3.-;;6-h~--fcd6rrey-~111) ;5ke.zt12 ..~--e:arrc.eh1ed-----:?J V.5Z£ -

!;/I'c.aI --b) Pe..5/rJ1. CL / YIfeJi.l C/Y' ,Demel ~ :ft(e.. -;2Z/, ---~--.--. ~.~- '--_._.I-I --~b~.JI~_~~_a..~-f~

I ./ ---~rdi:M_Y.J.~}t..ef, e.> •.!

-b~W-Li--t6-e- ~_p-{de.. ~ *-:f£.e- blVuf:.j;Y- _@_----,.6W:.e1-{1.t-t?~---.£A5_d:/!.L'ci;J1e.5.5 !z::: Z:.5 -;7-

___ ~d_~5.J~e. 4AeM..- '£.eJ;yforcefrled- ~___ -I-Me.. sUz,.b, Loew& cx.J ~C-<! 5/r8t.zzff?

em.e. ..-Jah1e.. (:/5 /11 tf .. ~ •

-CiJ, TIe. j1~-af- 6L- f2!'le-~~ ~*1~?teO- (jj)I{ p' I_a .L' ,;L. I t..LP----fiO--t::M._!«.l2:Le1:!2'k_~=lJ:J_=.....A.Ly-=-_.p.7. e.5 t.577 "--

-----/:;(.e..-CP..t.'Ov,'J'1_p4 'X-d~_.d--6kl!!)1._t'r. E&:t!?____ 71-e-._~b»< >-t ~'}teS a.. bL.-::::- czpt?kt'p- a~

U -==' 5lJO 1<I,I.2_+J.<Jt?.?'lk.lt:':f1J _ tie-..lJ:J.d:vi,dad---f2./1e- ~a~-4--do/~. 12~t7tL-#-e- p:le---~~-qJ-vJ.---'--d.kw_lAe.9zet~ce.men:& ~---~'b..d_,6!L.~d.0.~~&Lr<'Ul~-tc£'"3-0 .t..5 i C?hvL-_~1;_-=-b.CU(..5.-I__r • ._

-4--:-(g)-f}-P-t'5~.>/OY/~bJ!):J£t.d. O7I1~e-l:- bea.'ht.. _@)_-----4l5- <f~Yj--.lYJJ4~~ P- .pf2~e-5> If------,tffp JyV iM f6e .slee! iWom.dr~ qv~.

pe Sjy....f?d-Jt~_~_,tzekc~~5P0 k/l/

- __ ~du,~=e-__-b.,~_b2.5eJ. [B-a-&:t!z.m. k 0--- ~ef..uJ&'--.

----=Of-d-.:.£.!qyk- 9---:ihef'zf?- .is a..- I!..lreo.1od tlf-(/~~Jft!p .. 6J~f2J{b-_-/:1[..e- g..ke.rne fi we-7.6:...ad~---d Wol~ iJ_t.JJefi. -f(,e.. /'y,/.;,a!

ee;.~cfiA7J1 ~ M_p-J£e.>lr@endnLJlt'vek/.--_ .

- 3-- -

'iJ-&L ~ UJ,,"cl'j'cJn L:7?tL.£J,-~ a:;r-___ eoJ1JJA~~-L05.>~. (}J;)_Co-klyu1e-:-ab~e--~l-l:V-~----=l-f-ciI a.>1-6e ~;d !y_~---hb---»C-_/ffr. ?:&Le~j/e---5k.p--0~~e___ ,b..elioJ:!:!:= 17..6 re ,

,Q:) --ne_iJJ~i.~CtJk:Y1>!b ,d ,:L()~,x..,gp~)Y)~5~__ Gfl 1____ .dee;JL>::J~:f._~ie! --fr:*1 A&lvi~~ '

_____ 121.----'" '-Q./2..JtijJ._~---.e.J-h"" ,30tJ kiP- . /)eJ{j.YI &L.. I____ ,6.~e p-f-~~~b_P:/fczm ;t !i:/F-----h~__4~ .,~'7i:e-A!~.!e-/.ftil;Lf; e.ut7i.::J-_____ ~u?;e -0 ,,4.£i2o--,e;s4 ;:'t/(e-i:.!:- =' 3000:?-Jl_'_

-----G..):Lt:!.~fJ_",,60., dim_f25-I._' ._.5..M:uL.7£e~inftrce -_____ -~)?1~'e~,nfs//1.--" ti.m:J_e~_'_.5_'e_e-/z__@s .

_'Z..(g)-lJ.lit.lj5.>--:I£fLJYi.v--"J.~:f¥J-~tJ~he.J/Y'-eY.f~ -@_'. . . C£r.J.clL.e.-/£.~v..e8--~c.d Cpn~.'

--~C?)-,_b-X(2lf2in,} W7-:-f-JLe.5-keYed CC>1cJ?e.je. !Jea.m'>_-e3-_____ dz=e~.{U~_th..e3__~_~~boX-L1.ee--ho/i 2

___ C£), }1.t?-.k.e a..- P-Jz.dtJr,;Y.!o~ de.511b_# a.. -1ec.;h''o!'1 -(!i)-____ r1f_f-'z.eJlteyeA b.~~..?z.eJ.i.5~_____ ~,.eJ...:-t2..p451J..J-LJ!.-m - ne. N-ePicd( d....qdf._____ '.!Jf-i£.e Aee-L~4f----:J}veJt a.5 tfP{) »1", "

~_~_~T,£~I/.e ;me~~_fOY'c.~ -<sled____ ~-0_._'~ff.~'Si.~lJ.,/'::1~f-p-c<J q:j~k 5;" ei1--Jze.

CPndle/e. uwla [email protected]:f--.-/~ .&d -L;z..tJ M.P"'t '

C£j/r------r- I

2.5" iY 2.5/1

-+-i- I -+-i-Given:fc'= 4 ksi andfy= 60 ksi

16'f

TenNO.9 bars

!Y ,r---} 281r

Fig. 1

- COLUMN (24"X24")

il _!

~I=-=-==1 .1 _

1 -

1 _J _L _

DL=900kip'LL=500kip

20" DIA PILE.18" 5'.0" ._5'.0" . 5'.0" 18".

~.-~-----$=---~<:, iii ir;, i. . i .. i i

__ ,G'LD..r-b." _ __. __ LJ. .\J.} c:::J-"4J \.1- MI I. I I ~9 . . , I

in i! . ! .! ! !.' L~--,-Eb-----ilJ----~-,...,.ll __------L L__ ::Lj. 1~~ •

- '~,I"'=J=-

" 300mm ,_

MIDSPAN SECTION

EEoo'"

..'r self-wt.=4.5 kN/m

/ _. LL=11.0kN/mf==C.=~~~~l_. 12 m

iIii:IIIft"l~,

!IIi,:

Fig-i~. :3 --- II~--==== ~-----------.---- --------- -=--=--=--=--=--==_.:-.....t

DESIGNAIDS 775

2,0

'"01,9 0

/I

'"1,8 "1,7 0,08

1.6 0,07

1.5 '0,06

1.40,05

r 1.3

~ 0,04~ 1.2,. o.'l~, 1.1 0,03,.,:

/Iff Q.~I~1,0 0,02

~/I 0,9 Pp 0,01

I,,0

r 0,8; I

~ I 0,7, I~

,Il •

0,6I "1 1 0.5i III

I 0.4

I II 0,3

,I.. 0,2I'1 0,1

_. _ .._--~----- -- ----

-

!i1C:i

cIi....•

"

I1,•

III!-1

0.00,00 0.05 0,10 0,15 0,20 0,25 0.30 0,35 0.40 0.45 0,50 0,55 0,60 0,65

GRAPHA.I6Column strength interaction dingram for circular seclion with ')' = 0.90 .

•

www.engbookspdf.com•

http://www.engbookspdf.com

DESIGN AIDS 767

i

4;2 Interaction diagram

0 f' 4 ksi/I

c

~ty 60 ksi II~hlly 0.90 .- o 0 ~ 0 0

R I 01"~'1 b e , ••0.08 ::1:1 l e ~ "- - "0 0 0

0.07

~C 0'0.06 , /J

C',.O,OS ~ , , ~C. I <:>() •

,0.04 C' <:l; .

:0.03 :

0.02 \\-.\t 7-

Pg, 0.Q1 ,

~ .,0°.e\"'~ ..

,~fl

_0.002.,. ~ - - .' - t, ty,

-Ii

<'';0.005

2.0

0.00.00 O.OS 0.10 0.1S 0.20 0.2S 0.30 0.3S DAD OAS 0.50 0.55 0.60 0.65

I".i.,jIj1iI

\ !

GRAPHA.8Column strength interaction diagram for rectangular section with bars on four faces and 'Y;:: 0.90.

Fdi/m 77es/~ of COr-cnz.I..•.. S~/G£ / /f).f4.. Ed(J ~_ ; Po /" 1\ .AJ/ls"-'1

:01; VO"r"'J /' / /

•

www.engbookspdf.com

•

i---~---~.. -.' .~-.••...~...- .. . ~.'.~--_-......~-_._._-_.•.•.~-_••.•.••..----

http://www.engbookspdf.com

L-3ff -liCE Date: 29/1012019

BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY, DHAKA

L-3/T-I B. Sc. Engineering Examinations 2018-2019

Sub .:CE 331 (Environmental Engineering- I)




SECTION-A

There are FOUR questions in this section. Answer ALL.

I f~

1. (a) What are the Sustainable Development Goals? What is hydrologic cycle and why is

it important? Identify the pathway for transmission of diarrhoeal and other faecal-oral

diseases from faeces. (12)

(b) What are the major greenhouse gases? What are the principal impacts of climate

change in Bangladesh? Illustrate the natural services that a healthy biodiversity

commonly provide to minimize overall environmental degradation. (12)

2. (a) Draw a diagram to demonstrate the "Essential Elements of Water Supply System".

Briefly describe "Pond Sand Filter", "Shallow Shrouded Tube Well" and

"Unaccounted-for Water".

ORDraw a flow diagram of public water supply chain. Explain briefly "Tara Pump",

"Rainwater Harvesting" and "Non Revenue Water".

(16)

(b) State the hydraulics of groundwater flow towards a well and deduce the

mathematical expression for the yield of a well in an unconfined aquifer. (10)OR

State the hydraulics of groundwater flow towards a well and deduce the mathematical

expression for the yield of a well in a confined aquifer.

3. (a) What are the purposes and methods of well development? Draw a flow diagram for

the reverse recirculation rotary drilling method for a well boring.

ORWhat are the differences between "Screen Well" and "Gravel Pack Well"? Draw a flow

diagram for the direct rotary drilling for a well boring.

(13)

(b) The population of a town as per the census records are given below (table I) for the

year 1955 to 2015. Estimate the population for the years 2035 and 2050 using Logistic

method. (13)

Contd P/2

,.=2=

CE 331Coutd ... Q. No. 3(b)

Table -IYear Population1955 40,2001965 44,5001975 60,4001985 75,6001995 99,0002005 1,24,3002015 1,58,900

OR

With the populations given above in Table-I, estimate the population for the years

2035 and 2050 using Incremental Increase method.

4. (a) Describe the sources and significance of the following water quality parameters:

(i) Nitrogen and (ii) Phosphorus.

(b) Explain thc process of ion exchange for demineralization with necessary diagram

and equations. How can dual and mixed media filters achieve the gradation necessary

for longer filter runs?

SECTION - BThere are FOUR questions in this section. Answer any THREE.

(10)

(19)

5. (a) What are the typical differenccs between water quality of surface and ground

water? What are the different water treatment applications of gas transfer? What is the

purpose of filter backwashing?

(b) Describe the mechanism of reduction of surface potential by double layer

compression in alum coagulation.

6. (a) What are the properties of an ideal disinfectant? What is the differcnce between

Reverse Osmosis and Electro-dialysis?

(b) What is water hammer? Explain the phenomenon of water hammer by suitable

sketches. How can you reduce the water hammer effect in water works practices?

(c) Describe briefly the steps of laying underground water pipes.

Contd P/3

(8+8+4)

(15)

(8+4)

(2+8+3)

(10)

,t•

=3=CE 331

\

7.

8.

(a) Discuss briefly the steps of the water safety plan processes with flow diagram.

(b) Drinking water is supplied to a pourashava from ground water having both

dissolved iron (10 mg/L) and Arsenic (1.0 mg/L). The water supply is intermittent. It is

expected risk of ingress of contamination when there is low or no pressure in the

system. Calculate the risk score and category for these three hazardous events (i) High

iron concentration, (ii) High As concentration, and (iii) ingress of the contaminations to

the distribution system.

(a) What are the basic assumptions made in the Hardy-Cross method? Deduce the

expression for detennining flow correction in Hardy-Cross Method.

(b) Discuss briefly the layout of water distribution system.

(c) An elevated cylindrical water tank of 80000 gallons capacity has to be designed. If

concrete work per sft in the floor and wall of shell costs Tk. 120.00 and Tk. 180.00

respectively, what would be the most economic dimensions for the tank?

(15)

(~O)

(2+10)

(12)

(11)

{

,

.,....."

.( ..~:'.

L-3fT-IICE Date: 19/10/2019BANGLADESH UNIVERSITY OF ENGINEERING AND TECHNOLOGY, DHAKA

L-3fT-I B. Sc. Engineering Examinations 2018-2019

Sub: CE 341 (Principles of Soil Mechanics)



SECTION-A

There are FOUR questions in this section. Answer Q. No.1 and any TWO from the rest.Question NO.1 is COMPULSORY

Assume reasonable value of missing data, if any. Use attached chart where necessary.

1. (a) Define the following soil types:(i) Verved day (ii) Boulder clay (iii) Loess

(b) Derive an expression for shrinkage limit of soil having known specific gravity.

(c) Classify different types of soils based on particle size according to ASTM D422.(d) Define principal typcs of secondary structures in soils.(e) A clay sample (liquid limit = 65%, plastic limit = 32% and natural moisture content =49%) was collected from a depth of 6 m below the existing ground level. Water table is atthe ground level and saturated unit weight of the sample is 18 kN/m3. If the sample isnormally consolidated, estimate the undrained shear strength (Su) of the sample at thatdepth.

(f) Deduce an expression for Rankine's passive earth pressure due to a cohesive frictional

(c-~) backfill.

(g) Draw neatly Casagrande's Plasticity Chart according to Unified Soil ClassificationSystem (USCS) showing the classifications of different soil deposits.(h) Draw typical results for consolidated drained (CD) direct shear tests conducted onloose and dense sands.

2. (a) Liquid limit test was carried out on a soil sample using Casagrande's apparatus andthe following data were obtained:

No. of blows (N) 16 20 24 30 35Water content (%) 57.1 55.2 53.8 51.9 50.6

Plot the flow curve and determine the liquid limit of the soil. Also calculate the value oftoughness index if the plastic limit of the soil is 24%.(b) Classify the following two inorganic soils according to USCS:

Soil A: Percent finer No. 200 sieve (0.075 mm) = 93Liquid limit = 52%Plastic limit = 31%

Soil B: Percent finer NO.4 sieve (4.75 mm) = 92Percent finer No. 200 sieve (0.075 mm) = 8D60 = 1.7 mm, D30 = 0.43 mm, DIO= 0.08 mmConsistency limit of fraction passing No. 200 sieveLiquid limit = 43%Plastic limit = 21%

Contd P/2

(6)

(~)

(5)(6)

(6)

(6)

(7)

(5)

(13})

(11)

=2=CE 341Contd ..... Q. NO.2

(c) What are the merits and demerits of direct shear test? Briefly describe the behaviour

of saturated samples of clay in consolidated drained (CD) triaxial compression test and

also show with neat sketches two examples of consolidated undrained (CU) analyses in

clays. (11)

(d) A smooth vertical wall of height 10 m retains a soft clay backfill of unit weight 16.5

kN/mJ• Undrained shear strength of the clay backfill is 35 kN/m2 For undrained

condition of the backfill, calculate the following. (II)

(i) Depth of tension crack and unsupported height of wall

(ii) Active earth force before tension crack forms.

(iii) Active earth force after the formation of tension cracks.

3. (a) The following results were obtained at failure in consolidated undrained (CU) triaxial

compression tests conductcd on three specimens ofa clay sample: (14)

Specimen Cell prcssure Deviator stress Pore water pressure

No. (kN/m2) kN/m2 kN/m2

I 75 150 300

2 199 272 418

3 25 50 100

Draw the Modified Failure Envelope in a plain graph paper and hence estimate the values

of effective shear strength parameters C' and r/f. Also write down the Mohr-Coulomb

equation for the effective stress failure envelope.

(b) Explain the concepts of Rankine's active earth pressure, passive earth pressure and at-

rest earth pressure. (IO~)

(c) For a soil, the following results were obtained from grain Size distribution and

Alterberg limit tests: (8)

% finer No. 200 sieve (0.075mm) = 91Liquid limit = 56Plastic limit = 30

Classify the soil based on AASHTO Classification system.

(d) The following results were obtained in a consolidated drained (CD) direct shear test

conducted on a clay sample. (14)

p

(~;). Also comment on the stress history of the sample.

Specimen Effective Normal Stress Peak Shear Stress Residual Shear StressNo. (kN/m2) (kN/m2) (kN/m2)

1 75 80 14

2 150 105 27

3 250 145 44

Determine the effective peak friction angle (~'») and effective residual friction angle

Contd P/3

=3=

CE 341

4. (a) A specimen of a saturated normally consolidated clay sample was fully consolidatedin the triaxial cell under a cell pressure of 200 kN/m2 Pore pressure within the specimenat the end of consolidation was zero. Deviator stress was then applied under undrainedcondition and increased until failure took place. The values of deviator stress and porepressure at failure were found to be 150 kN/m2 and 50 kN/m2, respectively. A secondspecimen of the same elay was fully consolidated in the triaxial cell undcr a cell pressureof 300 kN/m2• Pore pressure within this specimen at the end of consolidation was zero.Deviator stress was then applied under undrained condition until failure took place.Calculate the following analytically: (13)

(i) The values of effective angle of internal friction (~') and undrained angle ofinternal friction (~u)of the sample.(ii) The values of pore pressure at failure (ur) and pore pressure parameter A at failure(Ar) of the second specimen.

(b) Liquid limit of a soil in air-dry statc and over-dry state were found to be 85% and56%, respectively. Plastic limit of the soil in air-dry state was found to be 31%. What isthe group symbol of the soil? Also mention the group name of the soil. (8)(c) For the retaining wall shown in Fig. I, the backfill supports a uniformly distributedload of 30 kN/m2 Draw Rankine's active earth pressure distribution on the wall anddetermine the total active force (thrust) per unit length of the wall. Also state thcassumptions in Coulomb's theory of earth pressure.

(d) Draw the following qualitative curves: (1l~)

(i) Pore water pressure versus axial strain for saturated samples of normallyconsolidated and overconsolidatcd clays in consolidated undrained (CU) triaxialcompression tests.(ii) Skempton's pore pressure parameter B versus degree of saturation.(iii) Skempton's pore pressure parameter A at failure (Ar) versus overconsolidation(OCR) ratio.

Also mention the major factors affecting shear strength of soils.

SECTION -B

There are FOUR questions in this section. Answer all FOUR questions.Assume reasonable value (values) for missing data only

5. (a) Why compaction test is done in the laboratory? Write down the salient features of theStandard Proctor and Modified Proctor test method. Discuss the factors affectingcompaction of soils. (10)(b) Two sites are being considered as "borrow" sites. The in-place unit weight of the soilof the first site is 101 Ib/ft3; the water content was found to be 10%. On the 2nd site, theunit weight of the soil was found to be 96 Ib/ft3, at a water content of 14%. Theconstruction site required 315,00 ft3 of soil in a compacted state, at a unit weight of 126Ib/ft3, at a water content of 14%. The unit (per ft3) price from each "borrow" site wasTk. 9.00 for the material and Tk. 10.00 for transportation. In addition, for the materialthat required the additional water, Tk 1.50 was estimated to be the additional cost for unitvolume (ft3) of soil. Calculate the cost of material from each site. Assume Gs is equal to2.65 for both the sites. (15)

Contd P/4

..J

\

=4=

CE 341Conld ..... O. NO.5

(c) Derive an expression of buoyant unit weight of soil as a function of void ratio,

specific gravity and unit weight of soil. (10)

OR5. (d) A saturated clay has a water content of 39.3% and its mass density of 1.84 gm/cm3

Determine the porosity of soil and specific gravity of soil skeleton. (15)

(e) Describe the effects compacting clay on dry or wet of optimum moisture content.

"There are two ways of specifying compaction of earthworks", Explain. Also give your

opinion with justifications, with one you will adopt? (10)

(f) How do you estimate the void ratio of soil in the field? Describe the importance of the

term Density Index and how it can be used to describe the state of compaction of soil.

Also state for what type of soil this term is usually used. (10)

6. (a) Determine the total stress (vertical) 8.0 m below the point "A" due to line loads sho",,"

in Fig. 2. ~~

(b) How do you construct field e-Iog(p) curve using Schmertmann's procedure for Over

Consolidated (OC) clay. (10)

OR

6. (c) What is time factor? How it is related to the average degree of consolidation? Also,

state how it can be used in calculating settlement rate of a foundation? (10)

(d) A "T" shaped foundation is loaded with a uniform load of 100 kN/m2 as shown in Fig.

3. Estimate the vertical pressure at a point which is 6.0 m below the point "0". (25)

7. (a) A footing is placed on silty clay stratum, with properties shown in Fig. 4. Then a 4

feet fill layer is added as shown. Calculate: (30)

(i) Settlement of The clay layer, and

(ii) Time require to attain 8.5 inch of settlement.

(b) How do you mitigate piping in hydraulic structures? (5)

8. (a) Draw the uplift pressure diagram on the bottom face of the concrete spillway, Fig. 5

(at Points: A, B, C, E, I and K). Also calculate the safety factor with respect to piping,

assuming saturated unit weight of the base material is 20 kNm3 (20)

(b) Calculate the coefficient of permeability (k) of a soil sample 6 cm in height and 50

cm2 in cross sectional area, if a quantity of water equal to 430 cm3 percolates in 10

minutes under a constant head of 40 cm. Comment on the value of k. Also, calculate

seepage velocity if the dry weight of the sample is 500 gm and Os of the sample is 2.65. (15)

J

5-:::: -

3m-

Ej' f

---

'1,;

I' 10m ',,'

r 50 kN/mI EI M

I~,--~

',\ .3 m-jA ,,E

,CXJ

,

1 60 kN/m

F!.5 :Jv

Silty clay

New filly =135 pef

,SIAVl5Yy= 102 pef (ip.situ)

2°'. . G=2.6(l) = 2 /01 - .

C = 0.3 - 2c' = 5 X 104 em isec.

No;mally consolidated

.'.: '.. :' "'::'::'.::.~:'~':-10' .. :::sr: : :>:.:::.:: sa~.d&GraVel _'.' '.: ..•..... ',:.. :.'..: :':.;.' .'~:..', " .. ' .' .. '~ '. --------

--'-............_------./...._.~

.' ... ,.,.

. .

. ..

. " ...

I100

eJ. 193.3 ft

':':0': ........•:

,. ':,Scale (ft) .

,o

" .._ .. -..

fif.! g-& oj.£. '.Concrete spillway.Reservoir

~'7 el. 260.3 ft .'

., ....

.41.lJliil;11.mIn'-,

Ciayay SQil~.:

~ .

,Silty sot1&

. SJ:f{Clay~af"r.Jal&(MOl'''thanSs% passing ~o.'200 .

. Sloye) ..' .A4 .A-5 ! A-6. A.7.

.A;NiA,;7"6 ,

:...:-", .....

.A-IlA.U' . A.2;o: '. A.2.7

.~.

:,' •. j

.-5ttntn '~..'~. ~. .'.~.. ._._1Qmax::.35max 3Smax. .3Sm:IX 35tnl1X36mTn. :S6:.niin .36min '. 36niln'.. ;

-:-.~---

1\4 . . A-3 ". j .

Moll '.A.1"b k~.',. .

5O-max- .:-.3Gft\aX. iSO'1l\;l)t

15max -:25'max

Grc.up Claulflciltlon!

SlOV9AniltYo\llil;i .Pereant:. Pau'll'ig.~.10lob. 40.lob: 200Olal'lietGilsllcii'Offioai:i.ion j>i>S••llfli No, AD

Uq\lJd.ll~~' .. '. ••• .;.\ "~mu4lmln '~O!'MX41mia. 40max. '41.nln. 40maj(-Pfaslli:lt;y llidciX. ~'max.' . :N.P.'1Gl\,~-iii ~ 11 II'IIn ~1 fui~,ill.ba1l10,ma\c 11 mIn.Ui~a:liyil9&c{;~£n~~. ..' ";"li. ' .'~~~~:u cC~StitllOf1t~~btid $2nd, Sllty.Cl'f c1aY-&r~&l,alldWid

GGnar<iJ.,.RalIng';ji; . -e.>~j,,, . t"':' ..;_.. .b . . .' . .••••••••••ll,n "'II""".su gradll' . .." .,' "./ .. '

• P!a~11Clt)'lndtiXOfA.7-S~Ubgroup•.mll~"'a1to.~r'mll~n LL minUs 30. ": l

• pt"StlCJfYlnd""pfA,;-6.~llb9~isgf&atllr~u.rlJi'llnfl&Sl);. . '.' I .c.~y-&1. ,I\~5HTO soil clasSitic(l'liionsystem~atter Atk~S, ~997)

I ,-' I_

. Ganllral Classl~l:atlon

.::".." .

"

l-3rr-i/ce date: 0211112019 bangladesh university

Documents