8/17/2019 Answers to Assgn 2.pdf

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Solution to Assignment No. 2

(b) Assume good bond, f  bd  = 0.315 f ck 2/3

 = 0.315x352/3

 = 3.37 N/mm2 

cd  = min (a/2, c1, c) = min (105/2, 47.5, 97.5) = 47.5 mm < 3  = 75 mm

  1 = 1.0 ;  2 = 1 – 0.15(cd  -  )/  = 1 – 0.15(47.5 -25)/25 = 0.865 < 1

 b,rqd  = [500/(4.6x3.37)]x25 =806 mm A

s,req  = 952x10

3/(500/1.15) = 2190 mm

2; A

s,prov = 2450 mm

2 

  bd  = 1.0 x 0.865 x 806x(2190/2450) = 623 mm > 0.3 b,rqd  = 242 mm

> 10  = 250 mm

Force provided by 3H25, T  = 3x490x(500/1.15)x10-3

 = 640 kN

Equating T  = (180 x –22.5 x2)/(0.9x0.42)+ 0.5(180 – 45 x) where x < 4 m gives

59.5 x2 – 453.7 x +550 = 0 from which  x = 1.512 m.

 the upper 2H25 bars can be curtailed at 1.512 – 0.623  0.84 m from thecentre of support.

(c)    = 2450/(575x420) = 0.010 >   0 = 10-335 = 0.00592 ;   ’ = 0 ; K = 1

/d = 1x[11 + 1.535x0.00592/0.010] = 16.2To be modified by:

310/s = As,prov / As,req  = 2450/2190 = 1.12Flange breadth/web breadth = 575/250 = 2.3

 allowable /d = 16.2 x 1.12 x [1 – 0.2x(2.3 – 1)/(3 – 1)] = 15.8

< actual /d = 8000/420 = 19.0 ; hence not satisfactory

Increase depth of beam.

500 mm

For the upper 2H25 bars:c1 =60 – 12.5 = 47.5 mm

c = 50 + 47.5 = 97.5

a = 250 – 2x47.5 – 2x25 = 105 mm d  for 5H25 = 500 – 80 = 420 mm

(a) At mid-span:

 M  = 45x82/8 = 360 kNm; V =0 kN

T =360x103

/(0.9x420)=952 kNAt ¼ - span:

 M  =180x2-45x22/2=270 kNm; V =90kN

T =270x103/(0.9x420)+0.5x90=759 kN

At support:

 M  = 0 kNm; V =180kN ; T =0.5x180=90 kN

250 mm

120 mm

575 mm

50

60 60

2H25 +

3H25

180 kN

90 kN

1.35Gk + 1.5Qk = 45 kN/m

180 kN8 m

952 kN759 kN