Q Bank on DESIGN for Civil Engineer LDC Exam MCQ New Pattern

Q1. Determine the forces in all members of a truss with the loading and support system as given below 

Answer  : Roller Support at end “B” is frictionless and provides a reaction Rb at ,right angles to the roller base. The Reaction at the hinged support at end “A” can have two components acting in the horizontal and vertical directions. Since the load of 30KN acts vertically downward, the horizontal components of reaction at “A” is zero and these will be only vertical reaction Ra. 

Considering the free-body diagram of the whole truss is as given below :

The triangle ABC is a right angled triangle with angle ACB = 90 

AC = AB cos60 = 5 x 0.50 = 2.5 m

Distance of line of action of 30 KN force from A,

AD=AC cos60 = 2.5 x 0.5 = 1.25 m 

Taking moment about A, we obtain 

Rb x 5 = 30 x 1.25 Rb = 30 x 1.25 = 7.5 KN 5 Ra = 30 - 7.50 = 22.50 KNJoint “A” - 

Consider the free body diagram of joint A with the direction of force assumed as shown in the figure given below :

Equation of equilibrium can be written as 

E Fx = 0 

F2-F1 cos 60 = 0 

Efy = 0 F1 sin 60 - Ra = 0 

Hence F1 = Ra = 22.50 = 25.97 KN (Compressive) 

Sin 60 0.866 F2 = F1 cos60 = 25.97 x 0.50 = 12.99 KN (Tensile)

The Force F1 is acting towards the pin which means that the member AC is in compression. 

The Force F2 is pulling the joint A which means that the member AB is in tension. 

Joint “B” - Consider the free body diagram of joint B. The force F2 in Member AB has already been calculated above and formed to be tensile. Hence Force F2 will pull the joint B and will be directed away from its as given in the figure : -

From the equation of equilibrium, 

Efx = 0 

F3 cos 30 -  F2 = 0 

or F3 = F2 = 12.99 = 15 KN (compressive)

cos 30 0.866 The force F3 is acting towards the joint B which means that the member BC is compression Conclusions :- 

Force in the members :- 

AB = F2 = 12.99 KN (Tensile)

AC = 25.97 KN (Compressive)

BC = F3 = 15.00 (Compressive)

Q.2. Determine the reactions and construct the shear force and bending moment diagrams for the beam loaded as shown in figure given below. Also find the point of contraflexture if any.

Answer : A point of contraflexture is a point where bending moment is zero from conditions of static equilibrium.

EV = 0 and EM = 0, we have 

Equation 1 = Ra + Rb = 2x 2 + 10 + 2 = 16 

 Equation 2 = -2 x 2 x 10 + Ra x 9 - 10 x 5 + Rb x 1 = 0

Equation 3 = 9Ra + Rb = 90 

The udl is considered to be concentrated at its CG. 

From equation 1 and 3 

Ra = 9.25 KN and Rb = 6.75 KN 

Calculation of shear force :

At D = 0 

Just Left of A = -2 x 2 = 4 KN 

Just right of A = -4 + 9.25 = 5.25 KN 

Just left of C = 5.25 KN 

Just right of C = 5.25 - 10 = -4.75 KN

Just left of B = -4.75 KN 

Just right of B = -4.75 + 6.75 = 2 KN 

Just left of E = 2 KN 

Just right of E = 2 - 2 = 0 

Calculation of bending moment : - 

Md = 0 

At distance x from D (within portion DA) 

Mx = -2 x X x/2 = -x 

hence M (at x = 1m) = 1 and

M (at x=2m) = -4 

Ma = -4  KNm

Mc = -2 x 2 x 5 + 9.25 x 4 = -20 +37 = 17 KNm

Apparently there is a point of contraflexture between A & C as bending moment charges sign between A and C. 

Bending moment at x between A & C with x measured from D 

Mx = -4(x-1) + 9.25 (x-2) = 5.25x - 14.50 

5.25x - 14.50 = 0 for point of contraflexture

That gives x = 14.50 / 5.25 = 2.75 m

Mb = -2 x 1 = -2 KN m (considering the segment EB from right hand side). 

Since bending moment at C is +ve and at B is -ve, there is also a point of contraflexture between C and B 

Bending moment at distance x measured from end E towards left

Mx = -2x + 6.75(x-1) = 4.75x - 6.75 

or 4.75x - 6.75 = 0 for the point of contraflexture

That gives x = 6.75 / 4.75 = 1.42 m 

The shear force and bending moment diagram for the entire beam is as under 

Q.3. Determine the product of inertia of the channel section as given below with respect to its centroidial axis. All dimensions are in mm. Comment on the result.

Answer  : The section consists of three rectangular segments which have been numbered as 1, 2 and 3. 

 Rectangular 1 -a1 = 40 x 10 = 400 mm

x1 = 40/2 = 20 mm (From Y.Y)

Rectangular 2 -

a2 = 100 x 10 = 1000 mm

x2 = 10 / 2 = 5mm (From Y-Y)

Rectangular 3 :- 

 a3 = 40 x 10 = 400 mm 

 x3 = 40 / 2 = 20 mm (From Y-Y)

x = a1x1 + a2x2 + a3x3 

a1 + a2 + a3= 400 x 20 + 1000 x 5 + 400 x 20 = 21000 = 11.67 mm (from y-y) 400 + 1000 + 400 1800 

 Further the given channel section is symmetrical about x axis and therefore Y = 100 + 10 + 10 = 60mm (from x-x) 2 Product of inertia about centroidal axis = a1 (x1 - x)(y1-y) + a2 (x2-x)(y2-y) + a3 (x3-x) (y3-y) = 400 (20 - 11.67) (115-60) + 1000 (5-11.67) (60-60) + 400 (20-11.67) (5-60)= 183260 + 0 - 183260 = 0 Comment The section is symmetrical about x axis. Hence the product of inertia has to be zero.

Q.4. Design an singly reinforced simply supported beam having spam of 5 mm it is carrying a load of 10000 N/m including itself wt. 

Assumed 

Concrete = M20 grade steel = Tor Allowable bending stress in steel 210 N / mm 

Answer - 

Let us design the beam using M20 grade of concrete and with for steel having allowable stress in bending as 210N/mm

From IS456 : 1978 allowable compressive stress in concrete for M20 concrete, 7.0 N/mm 

Modular ration, m = 280 / 3 6bc

= 280 / 3 x 7 = 13.33 = 13.00 

Bending moment = wl /8 = 10000 x 5 x 5 = 31250 N - m 

= 31250000 N-mm

Neutral axis cofficient = N = 1 

1 + 6st / m 6bc

N = 1 = 0.302 

1 + 210 / 13 x 7 

Now let d/b = 2.5 or d = 2.5b 

M = kbd 

Where K = 1/2 6bc NJ = 1/2 x 7.0 x 0.302 x 0.899 = 0.95 

J= 1 - N/3 = 1-0.302 / 3 = 0.899 

M = kbd = kb x (2.5b) = 6.25 kb = 6.25 x 0.95 b = 5.9375 b 

We have already calculated that the due to load, beam is subjected to bending moment of 31250 000N - mm. Evaluating external moment with the moment of resistance 

We get, 

5.9375 b = 31250000 

or b = 5263157.90 

or b = 173.95 mm say 180 mm 

d = 2.5b = 2.5 x 180 = 450 mm 

Area of Steel = M/6stJd = 31250000 210 x 0.899 x 450= 367.84 mm 

Let us adopt 12 0 bars Area of each bar = A / 4 x 12 = 113.10 mm 

No. of Bars required = 367.84 = 3.25

113.10 

Provide 4 numbers of 12 0 bars. 

Provide 30 mm of effective cover design section of the beam will be as given :- 

Conclusions : 

 Width of beam = b = 180 mm

Depth of beam = d = 480 mm 

Steel Reinforcement = Ast = 4-12 0 

Q. No.4 Draw the Shear Force and Bending Moment Diagrams for the beam loaded as shown in figure below :- 

ANS.   1 SIGN CONVENTION USED AS 

1 Anticlockwise moments on LH side & clockwise moments on RH side are taken as ( -ve).

2 Anticlockwise moments on RH side and clockwise moments on LH side taken as 

(+ve). 

CALCULATION OF REACTIONS

Taking moment about C & putting ƩMc = 0

-10x7+RB x5-10x4-2x3x1.50 =0 

OR  5RB =70+40+95

RB = 119

RB = 119/5 = 23.8 t 

Considering ƩV = 0

-10+ RB-10-2x3+Rc = 0

OR  RB+Rc = 10+10+6

RB+Rc = 26  tonne 

Putting value of RB as calculated above.

OR23.8 + Rc = 26ORRc = 26-23.8

OR  Rc = 2.2 tonne 

CALCULATION OF SHEAR FORCES:- 

S.F at A = -10 tonne (-ve) 

S.F at B on LH Side = -10 t (-ve)

S.F at B on RH side = -10+23.8 = 13.8t (+ve)

S.F at D on LH side = -10+23.8 = 13.8t (+ve)

S.F at D on RH side = -10+23.8-10 = 3.8t (+ve)

S.F at E = -10+23.8-10 = 3.8t (+ve)

S.F at C = -10+23.8-10-2x3

 =  -10+23.8-10 -6 

 = 23.8  -26 

 = -2.2 t (-ve)

As the  S.F at E is 3.8 t (+ve) and the S.F at C is -2.2 t (-ve).

Let the point of zero Shear Force be at a distance of x meters from the end C.

Considering the forces on the RH side of the point of zero S.F

2.2 -2x = 0

OR2x = 2.2

ORx = 2.2/2= 1.10m from the end C

CALCULATIONS FOR BENDING MOMENTS:- 

B.M at A = MA = 0B.

M at B = MB = -10x2 = -20 tm (-ve )

B.M at D = MD = -10x3+RBx1

 = -30+23.8x1 

 = -30+23.8MD 

 = -6.2 tm (-ve)

B.M at E = ME = -10x4+RBx2-10x1 

 = -40+23.8x2-10 

 = -50+47.6ME 

 = -2.4tm (-ve)

B.M at 1.10m from the end C, (Considering forces on the RH side)

= 2.2x1.1-2x1.1x(0.55)= 2.42-1.21= 1.21 tm (+ve)