BPSC Main Exam 2019 ASSISTANT ENGINEER. Test 11. CIVIL ENGINEERING Subjective Paper-I. Detailed Solutions. Detailed Solutions

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Bartholomew Lucas
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1 etailed Solutions SC Main Exam 19 SSISTNT ENGINEER CIVI ENGINEERING Subjective aper-i Test 11 Q.1 (a) Solution: (i) Calculation of maximum moment at 8 m. For maximum moment Case 1: Case : Case : Case 4: etailed Solutions > > > < So, maximum moment will be achieved by keeping kn load at x 8 m. Influence line diagram by Muller-reslau principle kn kn 8 kn 6 kn m m m m Maximum moment knm m

2 6 SC 19 : MINS TEST SERIES bsolute maximum bending moment 15 kn kn 8 kn 6 kn m m m istance of resultant from end, F R kn x F R x Q m. m 49 The absolute maximum bending moment will occur under the load kn. Hence, for condition of maximum bending moment, the load system should be so placed on the span, that resultant of all wheel loads kn are equidistant from middle point of the girder beam. istant between resultant load and kn load.. m Hence, for condition of absolute maximum bending moment, the kn should be placed.1 m on the left side of the centre of the beam. F R 49 kn 15 kn kn 8 kn 6 kn 7.9 m m m V 9.9 m C (Centre) V Taking moment about end, we have V V kn or V ( ) kn 4.45 kn bsolute maximum bending moment for girder M under kn load, knm Q.1 (b) Solution: rea of beam section mm Section modulus (z) rea of steel bars ( s ) mm mm 6 Eccentricity (e) mm restressing force () N kn

3 Test No : 11 CIVI ENGINEERING N/mm e N/mm z Total stress the beam soffit N/mm et M maximum M for zero stress at the beam soffit M 6.4 z M knm Q.1 (c) Solution: 1. Welded connection leads to saving in expenditure on material and labour.. Welding permits architects and structural engineer complete freedom of design.. Welded joints are better for fatigue, impact and vibration as compared to bolted joint. 4. Rigidity of joint also leads to reduce deflections. 5. The compactness and rigidity of welded joints permits design assumption to be realised more accurately. 6. It leads to rapid construction, thus saving in time of project. 7. It leads to rigid connection which in turn leads to reduced beam depth and weight. Section - Q. (a) Solution: y unit load method, R + R R 6 kn R kn Δ Mmdx EI (i) M x [ x 4 m] and M 4 x [4 x 8] Moment due to unit load (ii) m.5x [ x 4 m] and (.5x+ 4) [ 4 x 8] Δ ( ) ( + ) 4 8 ( x).5 x (4 x).5x 4 dx + dx EI EI EI EI EI

4 8 SC 19 : MINS TEST SERIES Q. (b) Solution: 1 m T T 5 C 75 5 m m et T and T be the Torsional Reactions at and respectively as shown in figure. Using equation of equilibrium, ΣT T T T + T 5 kn-m We know, angle of twist at and, i.e., θ and θ 5 75 T T T T + 5 T + 5 T T 5 θ θ θ θ C + θ C + θ T ( T + 5) ( T ) θ θ + + GI GI GI C C T 5 ( T + 5) 4 ( 5) + T + GI GI GI 5T + + 4T + T 75 T kn-m ( ) sign signifies T is anticlockwise when viewed from left. Now, T + T 5 T kn-m T is clockwise when viewed from left. T 5 C 75 T T 5 75 C T et at distance x from C be angle of twist θ x θ x θ θ C + θ Cx T T 5 T 5 T

5 Test No : 11 CIVI ENGINEERING 9 T C + (5 T ) x GI GI ( ) x x m from C. distance from left end x m Q. (a) Solution: Shear force sign convention ositive shear force etween and : et w kn/m be the udl. Shear force just left of S.F. S.F. w a w a et V be the upward force at, S.F. + + V / V 7 et 1 be the downward load at C, S.F. C et be the downward load at E, S.F. E et V be the upward force at, S.F. V V + 8 oading iagram: a 1 C E V 7 Checking for equilibrium condition, ΣF Y 7 8 a+ + a ΣM a 8 a a+ a 4a a V 8.a.a + 8a 8a

6 1 SC 19 : MINS TEST SERIES ending moment equation between and, x x. x a a M a a a. ending Moment between E and, x x M E M + a a ending Moment equation between and C, 8 x+ ( x a) 8 8 x + ( x a ) x a M a 8 M C + a a a ending Moment between and C, a 7 a + x + x a + 7 a x x x a M + a M C a a a ending Moment iagram: Sign Convention Sagging ending Moment ositive.m. + 7 / C 8 / E a / a / 8 oading iagram X X x X X 8 x a 7 x X X x a X X E E a.m. Curve a 1 Curve 1 Curve a 1 Curve

7 Test No : 11 CIVI ENGINEERING 11 Q. (b) Solution: Components of a Steel late Girder: (a) Web plate (b) Flange angles with or without cover plates for riveted/bolted plate girders and only flange plates for welded plate girders. (c) Stiffeners-earing, transverse and longitudinal (d) Splices for web and flange Flange cover plate Flange cover plate Horizontal stiffener Flange angle Horizontal stiffener Flange angle Vertical stiffener Web plate acking Vertical stiffener Web plate acking Web splice Web splice earing stiffener earing stiffener Typical elements of a riveted/bolted plate girder Typical elements of welded plate girder Functions of Various Components of a Steel late Girder: Flanges: Flanges of required width and thickness are required to resist the flexural moment due to dead and superimposed loads coming on the plate girder by developing compressive force in one flange and tensile force in the other flange i.e. by developing an opposite resisting couple. Web: Web of required depth and thickness is required to keep the flange plates at the requisite distance and to resist shear in the beam. Stiffeners: Stiffeners are provided to safeguard the web against local buckling failure. These stiffeners are provided either in the longitudinal direction called as horizontal stiffeners or in the transverse direction called as vertical stiffeners or in both the directions. Section - Q.4 Solution: Fixed end moments Moment equations FEM FEM FEM C FEM C FEM C FEM C M M EI δ + +θ.5ei θ.75ei δ (i) EI δ + θ + EI θ.75ei δ (ii)

8 1 SC 19 : MINS TEST SERIES EI + θ +θc 1.EI θ +.67 EI θc...(iii) M C ( ) EI + θ C +θ 1.EI θ C +.67 EI θ...(iv) M C ( ) M C EI δ + θ C + EI θc.75ei δ...(v) 4 4 EI δ M C + +θc.5ei θc.75ei δ...(vi) 4 4 We know that, joint equilibrium at joint, M + M C.EIθ +.67EIθC.75EI δ...(vii) M C θ θ C...(viii) M C EIθ C.75 EIδ θ C.75δ...(ix) Shear equation, M + M M C EIθ.75EIδ.5EI θc.75eiδ EIθ +.5EI θc 1.15EIδ+ 4 y equation (ix) and (x), we get 1.5EIθ.975EIδ+ 4 y equation (vii) and (viii), we get EIθ.75EIδ y equation (xi) and (xii), we get EI θ EI δ 6.95 M knm M 11.4 knm M C 11.4 knm M C 11.4 knm...(x)...(xi)...(xii)

9 Test No : 11 CIVI ENGINEERING knm 11.4 knm 1 kn kn m kn.81 kn.81 kn 4 m 4 m knm 7.14 kn 11.4 knm.85 kn.14 kn kn.85 kn Shear Force iagram 11.4 knm 11.4 knm knm 11.4 knm Moment iagram

10 14 SC 19 : MINS TEST SERIES Q.5 (a) Solution: et R and R be the support reactions and M moment at fixed support of the beam. M C R R C We know, R + R C Taking moment about from RHS, R C R C ( ) upward R ( ) downward Taking moment about hinge from HS; R M M ( nticlockwise) eflection under the load is given by δ U U C U + For the span C (x measured from ) R x M C 1 x x x U C M 1 x Mx EI dx 1 EI x x dx U C M 1 x Mx EI

11 Test No : 11 CIVI ENGINEERING 15 U C 1 1 x x x x + x dx + EI 4 EI 4 U C 1EI For the span C (x-measured from ) x x U C / 1 Mx Mx EI dx U C Thus, deflection under the load 1 EI / ( ) 4EI x x x dx EI δ + 1EI 4EI 8EI Rotation at internal hinge pply a fictitious couple M at. The rotation of the internal hinge is given by, U θ M M Reaction at support C by taking the moment about R C + M For the span C (x measure from ) and M x x M U C M For the span C (x-measured from ) and M x x M U co M /

12 16 SC 19 : MINS TEST SERIES For the span C (x from C) M x θ + x Rc x x U C M 1 x x dx EI M + 8EI 6EI U M EI M 8 Q.5 (b) Solution: 1.5 W W W/ W C M 1.5M M Using kinematic theorem, the collapse load for the beam will be the minimum of the collapse loads for the various spans. Span The end is simple support and end is intermediate support. Therefore, the beam acts as propped cantilever with as fixed end. lastic hinges will form at centre of and at. y principal of virtual work, External work done Internal work done oad deflection Moment rotation W 1 θ M (θ + θ) + M θ W θ 4 M θ θ E W / Δ θ θ M

13 W 4M 1M Span C lastic hinge will form at, C and under the load 1.5 W External work done Internal work done. 1.5 W Δ M θ M (θ) + M θ 1.5W θ 5 M θ W u 6.67 M Test No : 11 CIVI ENGINEERING W / / M θ 1.5 M θ C M Δ F θ Span C Δ θ θ 1 θ θ 1 lastic hinge will form at C and under the load W. External work done Internal work done W Δ M θ +M (θ + θ 1 ) W θ M θ + M θ 1 W θ 1 M θ 1 + M θ 1 14M 4.67M W (6 M + M) The collapse load is least of three ultimate loads. Hence, collapse load, W 4.67 M C W / / M θ θ 1 Δ θ + θ1 M Section - C Q.6 (a) Solution: Given: S 1; w 4%; G.7 ssuming γ w 1 kn/m S.e. w G.4.7 e 1 Void ratio, e e 1.8 orosity, n 1+ e.8 n 51.9%.519

14 18 SC 19 : MINS TEST SERIES Saturated unit weight, γ sat γ sat ry unit weight, γ d G+ e γ 1 + e w kn/m Gγ w e kn/m Q.6 (b) Solution: Given: H 1 m; ΔH mm; at, t year, Δh 15 mm U Δ h 1 ΔH % < 6% T v π 4 U C v d t π (.4578) ssuming two-way drainage, d H 6 m If Δh 7 mm, It will take another, C v C v m /year Cv t d U Δ h % > 6% ΔH T v log 1 (1% U) t (7.69 ) 4.69 years 7.69 years Q.7 (a) Solution: The knowledge of subsoil conditions at a site is a prerequisite for safe and economical design of substructure elements. The field and laboratory studies carried out for obtaining the necessary information about the subsoil characteristics including the position of ground water table, are termed as soil investigation or exploration. It is generally carried out in two stages, namely, preliminary and detailed.

15 Test No : 11 CIVI ENGINEERING 19 reliminary exploration consists of the geological study of the site and the site reconnaissance. uring preliminary investigations, geophysical methods and tests with cone penetrometers and sounding rods can be very useful. etailed investigation follows the preliminary investigation and is normally carried out to determine the nature, sequence and thickness of various subsoil layers, their lateral variation, physical properties and the position of ground water table. oring and detailed sampling are usually undertaken to obtain this information. Various in situ tests also form a part of the detailed investigation programme. ST: The test employs a split spoon sampler which consists of a driving shoe, a split-barrel of circular cross-section which is longitudinally split into two parts and coupling. IS 11 :1981 gives the standard procedure for carrying out test. It is briefly stated below: (a) The borehole is advanced to the required depth and the bottom cleaned. (b) The split-spoon sampler, attached to standard drill rods of required length is lowered into the borehole and rested at the bottom. (c) The split-spoon sampler is driven into the soil for a distance of 45 mm by blows of a drop hammer of 65 kg falling vertically and freely from a height of 75 mm. The number of blows required to penetrate every 15 mm is recorded while driving the sampler. The number of blows required for the last mm of penetration is added together and recorded as the N value at that particular depth of borehole. The number of blows required to effect the first 15 mm of penetration, called the seating drive, is disregarded. (d) The split-spoon sampler is then withdrawn. The soil sample collected inside the split barrel is carefully collected so as to preserve the natural moisture content and transported to the laboratory for tests. Q.7 (b) Solution: Given: h 1 45 cm at t ; h 44 cm at t 5 min; h 4 cm we know, K a 1 ln h t h 1 45 ln 5 44 a 1 ln h t h a 1 ln h t h 1 45 ln t 4 t min Time taken for head to drop from 44 cm to 4 cm, t t t min Q.7 (c) Solution: (i) 1 m; f 1 m; γ 18 kn/m q n CN c + q N q +.5 γ N γ + ( ) + ( ) 1 kn/m

16 SC 19 : MINS TEST SERIES q nu q n γ f 1 (18 1) 115 kn/m q nu q ns kn/m F Gross bearing pressure 1 kn/m Net safe bearing pressure 45 kn/m (ii) 1 m; f 1 m; γ 18 kn/m q n 1. CN c + q N q +.4 γ N γ + ( ) + ( ) kn/m q nu q u γ f (18 1) kn/m q nu q ns kn/m F Gross bearing pressure kn/m Net safe bearing pressure 79. kn/m

Design of Beams (Unit - 8)

Design of Beams (Unit - 8) Contents Introduction Beam types Lateral stability of beams Factors affecting lateral stability Behaviour of simple and built - up beams in bending (Without vertical stiffeners)