steel structure - solutions - batch ab

1. (d)

2. (d)

3. (b)

4. (c)

5. (d)

6. (c)

7. (a)

8. (c)

9. (b)

10. (d)

11. (b)

12. (c)

13. (d)

14. (d)

15. (b)

16. (a)

17. (a)

18. (a)

19. (c)

20. (d)

21. (c)

22. (b)

23. (c)

24. (c)

25. (b)

26. (d)

27. (b)

28. (c)

29. (a)

30. (a)

Steel StructureDate : 14/09/2015

Civil Engineering

CE

CLASS TEST - 2015

Serial No. : JP_CE_AB_140915_Steel

ANSWERS

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CTCE15 Delhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata

7 Steel Structure www.madeeasy.in | Copyright :

EXPLANATIONS

1. (d)The riveted joint may fail in any of thefollowing manners.1. Tearing of plate between rivet holes2. Shearing of rivet3. Bearing of plate or rivet4. Edge cracking

2. (d)Minimum overlap = 4 thickness of

thinner plate= 4 5 = 20 mm < 40 mmSo, minimum overlap = 40 mm

4. (c)For a simply supported beam carrying auniformly distributed load, length of plastic

zone = 15 5 33 3

L m

5. (d)No. of plastic hinges for failure = (r + 1)Where, r = degree of indeterminacy

6. (c)Plug welds are not designed to carry loads.

7. (a)Slenderness ratio for web

= 1w

h 3 2903

t 8

= 63

8. (c)

fcd =0

0.66 250150MPa

1.1y

m

f

where, = stress reduction factor

9. (b)Lacing of compression member is designedto resist a transverse shear,V = 2.5% of axial force in the member

V =2.5

2000100

V = 50 kN

10. (d)Effective throat thickness

= 58

thickness of thinner part connected

strength of single-V butt joint=y

wmw

fA

= 250 5

150 12 225 kN1.25 1000 8

11. (b)

6 m

Beam mechanism,6 20 load factor

32

=100( + 2 + ) load factor = 2.22

12. (c)Gross diameter of bolts = 22+2 = 24 mmFor most critical area, consider differentsections as shown

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37.5

37.5

4

5

6

7

2

3

1

50 70 50 50

(All dimensions are in mm)

Section 1-2-3

Anet =2s

t b n d4g

= 1.2 (15 1 2.4 + 0)= 15.12 cm2

Section 4-5-2-6-7

Anet =25

1.2 15 3 2.4 24 3.75

= 13.36 cm2

Most critical area = 13.36 cm2

Maximum tension in the flat=1

0.9 um

fA

=

20.9 410 13.36 101.25 1000

= 394.39 kN

13.(d)

1 m

1 m

0.3 m

0.3 m

Thickness of slab base,

t = 2 2 m0y

2.5wa 0.3b

f

a = max(, ) and b = min (, )

=1 0.3

0.352

= 0.35 a = b = 0.35 = 350 mmw = intensity of pressure

=2000

1 = 2000 kN/m2

= 2 N/mm2

t = 2 22.5 2 350 0.3 350 1.1250

= 43.43 44 mm

14. (d)Shear strength of one bolt,

Vsb = (An nn + As ns)ub

mb

f

3

Vsb =20.78 20 2 0

4

3400 103 1.25

Vsb = 90.55 kN

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Bearing strength of one bolt

Vpb =u

bmb

f2.5k dt

t = minimum of combined thickness of coverplates or thickness of main plate= 14 mm

kb = minimum of 0 0

e p, 0.25

3d 3d ,

ub

u

f,1.0

f

As pitch (p) and edge distance (e) are notgivenwe take kb = 0.5

Vpb = 2.50.52014 3410 10

1.25

= 114.8 kN

15. (b)Design shear capacity,

Vd = 03

yw

m

fh t

= 3250 75 4.4 10

3 1.1

= 43.3 kN

17. (a)IXX of ISHB 250 = 7983.9 10

4 mm4

300 mm

X X250 mm

aa

20 mm

NA

IXX for plates = 2 [Iaa + Ay21]

=3

2300 202 300 20 13512

= 21910 104 mm4

IXX = 7983.9 104 + 21910 104

= 29893.9 104 mm4

Area of built up section= 6971 + 2 300 20

= 18971 mm2

rxx =4

XXI 29,893.9 10A 18971

= 125.52 mmNow, IYY of ISHB = 2011.7 10

4 mm4

IYY of plates =3300

2 2012

= 9000 104 mm4

IYY = 2011.7 104 + 9000 104

= 11,011.7 104 mm4

ryy =4

YYI 11,011.7 10A 18,971

= 76.187 mm

18. (a)Nominal diameter of the rivet,

= 20 mmGross dia, d = 20 + 1.5 = 21.5 mmArea of connected leg,

A1 =10

100 21.5 102

= 735 mm2

Area of outstanding leg,

A2 =10

75 102

= 700 mm2

Now, K = 11 2

3 A 3 7353A A 3 735 700

= 0.759 Anet = A1 + KA2

= 735 + 0.759 700

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= 1266.32 mmStrength= Anet at

= 150 1266.32= 189.95 kN

19. (c)Force due to direct load,

Fa =W 25000n 10

= 2500 NTorsional shear in any rivet distant r fromthe centroid G of rivet group,

Fm = 2M r

r

where, r2 = x2 + y2

= 10 502 + 4 1202 + 4 602

= 97000 mm2

Since,the stress will be maximum in top andrightmost rivert.

Fm = 2Per

r

=2 225000 500 120 50

97000

16752.58 Nwhere ra is distance of rivet A from G

Fm

Fd

A

G

r

50 5cos

130 13

R = 2 25

2500 16752.58 2 2500 16752.5813

= 17863.8 N

20. (d)For Fe 410,

fu =410 N/mm2

fy =250 N/mm2

Partial safety factor for material=m1 = 1.25

Partial safety factor for dead & live load=f = 1.5

Factored tension(T ) =1.5 (22 + 55) = 115.5 kN

Net area required on the basis of netsection fracture

An =1

0.9m

u

Tf

=115.5 1000 1.25

0.9 410

=391.26 mm2

=392 mm2 (say)

21. (c)To convert the beam into a mechanism,three plastic hinges are required.Case-ITwo hinges at supports, and one hinge atthe point where cross-section changes.

2

A CB

The plastic hinge will form at B in limb BCand its value will be Mp.External work done = Load Deflection

= uL

W3

Internal work done= 2 Mp + Mp( + ) + Mp= 5 Mp

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uL

W3 = 5 Mp

Wu =p15 M

L

Case-II:Two hinges at supports and one below theconcentrated load.

1

1 = L / 3

= 1L3

=2L / 3

= 2L3

1L3 =

2L3

1 = 2 External work done

= u 1 u2 2

W W 23 3

= u2

W L3

Internal work done= 2 Mp1 + 2 Mp ( + 1) + Mp= 2 Mp2 + 2 Mp( + 2) + Mp= 11 Mp

u2

W L3

= 11 Mp

Wu =pM16.5

L

Collapse load = Min of 2 = p15 M

L

22. (b)Let thickness of weld throat = t mmVertical shear stress at weld,

fs =3W 60 10

2 d t 500 t

= 120 MPat

Horizontal shear stress due to bending atextreme fiber,

fb = 26 W e

2 t d

=3

2

6 60 10 150

2 t 250

=432

MPat

Resultant stress at extreme fibre,

fr =2 2s bf 3 f

=2 2120 432

3t t

=757.81

MPat

757.81

t=

3u

mw

f

757.81t

=410

3 1.25

or t = 4 mm

size of weld =4

0.7 = 5.72 mm

6 mm

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23. (c)Bolt grade 5.4Here,

5 = 1

100 ultimate tensile strength (UTS)

UTS = 5 100 = 500 MPa

and 0.4 = Yield strengthUTS

Yield strength = 0.4 500 = 200 MPa

24. (c)Moment, M = 100 103 15 10

= 15 106 N-mm

Approximate length of weld = b

6Mt f

where, fb = y

m0

f 2501.1

= 227.27 MPa

Approximate length

=66 15 10

12 227.27

= 182 mm

25. (b)For collapse in BC

LP

2

= MP( + 2 + )

P = P8M

LFor collapse in CD

LP

2

= MP ( + 2)

P = P6M

L

Collapse load = P6M

L

28. (c)Maximum bending moment,

M = 2 240 5

8 8w

l

= 125 kN-m

Required section modulous,

Zreq = 6125 10

0.66 0.66 250y

Mf

= 757575.75 mm3 = 757.576 cm3

29. (a)

Wide flange beambending

about weak axis

Shape factor1.5

b

d

Rectangular section

1.5 Ratio = 1.0

steel structure - solutions - batch ab

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