F1265726Test 01_CE_Strenth of Material and Stryctural Analysis

7/31/2019 F1265726Test 01_CE_Strenth of Material and Stryctural Analysis

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CIVIL ENGINEERING STRENGTH OF MATERIALS & STRUCTURAL ANALYSIS [15]

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MADE

EASY1.1.1.1.1. (d)

2.2.2.2.2. (c)

3.3.3.3.3. (a)

4.4.4.4.4. (d)

5.5.5.5.5. (c)

6.6.6.6.6. (b)

7.7.7.7.7. (d)

15.15.15.15.15. (b)

16.16.16.16.16. (b)

17.17.17.17.17. (d)

18.18.18.18.18. (a)

19.19.19.19.19. (a)

20.20.20.20.20. (b)

21.21.21.21.21. (d)

22.22.22.22.22. (d)

23.23.23.23.23. (a)

24.24.24.24.24. (a)

25.25.25.25.25. (c)

26.26.26.26.26. (c)

27.27.27.27.27. (d)

28.28.28.28.28. (a)

29.29.29.29.29. (d)

30.30.30.30.30. (b)

31.31.31.31.31. (c)

32.32.32.32.32. (d)

33.33.33.33.33. (d)

34.34.34.34.34. (a)

35.35.35.35.35. (a)

8.8.8.8.8. (a)

9.9.9.9.9. (a)

10.10.10.10.10. (c)

11.11.11.11.11. (d)

12.12.12.12.12. (a)

13.13.13.13.13. (a)

14.14.14.14.14. (d)

GATE-2011

(TEST SERIES-1)

CE: Civil Engineering

(Strength of Materials & Structural Analysis)

SlSlSlSlSl.....NNNNNooooo .: 121210.: 121210.: 121210.: 121210.: 121210

A N S W E R SA N S W E R SA N S W E R SA N S W E R SA N S W E R S

Note: If you find any discrepancies in the Answer key/Solution, you

are requested to write us at [email protected]

within 24 hours, with the correct or suggested solutions from your

end. Please specify your Name, Branch of Engineering and phoneno. for future reference with the E mail. Also you are requested to

write us about the Feedback of the question paper at

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EASY

[16] GATE - 2011 (TEST SERIES - 1) CIVIL ENGINEERING

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EXPLANAEXPLANAEXPLANAEXPLANAEXPLANATIONSTIONSTIONSTIONSTIONS

Q.1Q.1Q.1Q.1Q.1 (d)(d)(d)(d)(d)

A

B

Shear strainC

Q.2Q.2Q.2Q.2Q.2 (c)(c)(c)(c)(c)

6EI

L2

6EI

L2

A

VA

B

VB

6EI

L2

6EI

L

2

Taking moments about B i.e. MB

= 0

2 2 2 2

6EI 6EI 6EI 6EI

L L L L

+ + + V

A L = 0

VA

= 324EI

L

As there is no external load, therefore both the reactions will be equal

and opposite which will give a constant shear force of magnitude

3

24EI

L

Q.3Q.3Q.3Q.3Q.3 (a)(a)(a)(a)(a)

I1

=3bd

12

I2

=

23bd dbd

12 2

+ =3 1 1bd

12 4

+ =3bd

3

1

2

I

I =1

4

Note: If you f ind any discrepancies in the Answer key/Solution, you are requested to write us at [email protected]

within 24 hours, with the correct or suggested solutions from your end. Please specify your Name, Branch of Engineering andphone no. for future reference with the E mail. Also you are requested to write us about the Feedback of the question paper at

[email protected].


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Q.4Q.4Q.4Q.4Q.4 (d)(d)(d)(d)(d)

For no change in volume, the condition is

x + y + z = 0 where all the stresses are tensile. 20 +

y+ 20 = 0

y

= 40 N/mm2 which means yis compressive.

Q.6Q.6Q.6Q.6Q.6 (b)(b)(b)(b)(b)

Total downward load at the base = 20 + 10 (20 20) 103 = 24 kN

Total area at the base = 40 40 = 1600 mm2

Bottom stress =324 10

1600

=

240

16= 15 N/mm2.

Q.7Q.7Q.7Q.7Q.7 (d)(d)(d)(d)(d)

DK = [3(j + j ) re] + rr mwhere j = total number of rigid joints = 12

j = total number of hybrid joints = 2

re

= total number of external reactions = 3 + 2 + 2 = 7

rr

= mj 1 = (2 1) + (2 1) = 2

m = total number of axially rigid members = 9

DK = [3(12 + 2) 7] + 2 9 = 35 7 = 28

Q.10Q.10Q.10Q.10Q.10 (c)(c)(c)(c)(c)

As we know,

M

I=

E

R

MEI

= 1R

= Curvature

Q.11Q.11Q.11Q.11Q.11 (d)(d)(d)(d)(d)

K11

= Force required along degree of freedom 1 to produce unit

displacement in the direction of degree of freedom 1. And degree of

freedom 2 must be locked.

K11

D=1

11 3

12EIK (standard result)

L=

Q.Q.Q.Q.Q.1212121212 (a)(a)(a)(a)(a)

Vertical reaction at both support = P/2 (by symmetry)

Now, taking moment about hinge at crown from LHS = 0



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P R R

H R2 2 2

= 0

H =( )

P

2 2 1

Q.13Q.13Q.13Q.13Q.13 (a)(a)(a)(a)(a)

The elements of flexibility matrix are not necessarily dimensionally

homogenous as they represent either translation or rotation. For different

released structures, different flexibility matrices are obtained.

Q.16Q.16Q.16Q.16Q.16 (b)(b)(b)(b)(b)

Pure bending is said to be at the point on the beam/structure, where

there is no shear force.

A DB C

40 kN-m

BMD

A

D

B

C

SFD

10

AB C

D

10 kN 10 kN

10 m

4 m 1 m 1 m 4 m

There is only bending in BC and no shear.

Q.17Q.17Q.17Q.17Q.17 (d)(d)(d)(d)(d)

Let = y =KLS

AE

dy

ds=

K 1KLS

AE

total strain energy = S dy =K 1KLS

S dsAE

= KKL

S dsAE

=

K 1KL S

AE K 1

+

+



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Q.18Q.18Q.18Q.18Q.18 (a)(a)(a)(a)(a)

CA

5 kN

B D

10 kN

VAVB

3 m 3 m 3 m

Taking MA

= 0

10 3 + 5 9 = VB

6

VB

= 12.5 kN

VA = 10 + 5 12.5 = 2.5 kNBy moment area mehod, we have

A CD

B+

7.5

EI

15

EI

C

= A

LAC

C/A

=

7.5 1 7.5 13 3 3

EI 2 EI 3

=

11.25

EI

Q.19Q.19Q.19Q.19Q.19 (a)(a)(a)(a)(a)

dx

x

dl = x0T dx

l

Total elongation prevented =2

20

Tx. dx

l

l

=

3

2

T T

3 3

=

l l

l

Strain prevented = T3

ll

= T3

Compressive stress induced =T E T

E3 3

=

Q.20Q.20Q.20Q.20Q.20 (b)(b)(b)(b)(b)

The reaction at left support = 25 kN.



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Cutting a section through member AB and other horizontal members

of the same panel and considering left part of the cut section. The

force in the member AB will be tensile.

45

B

A

25 KN

ABF .cos 45 = 25 kN

FAB

= 25 2 kN tension

Q.21Q.21Q.21Q.21Q.21 (d)(d)(d)(d)(d)Stiffness matrix can be obtained by making second coordinate i.e.

rotation in the direction of 2 as zero and considering unit rotation in

the direction of coordinate 1.

D = 02D = 1.0

1

B

A

CE

4 m 4 m 8 m

The beam AB has rotational stiffness,( )4E 2I

2EI

4

=

and beam BC has rotational stiffness,( )4E I

EI4

= .

So moment required in the direction of 1 to produce unit rotation will

be 2EI + EI = 3EI. Thus K11

= 3EI.

The moment generated at point C in the direction of coordinate 2 is

1/2 EI = 0.5 EI as the carry over factor is half. So K21

= 0.5 EI.

Similarly make D1

= 0 and D2

= 1.0.

D = 1.02D = 01

B

AC

E

K22

=( )4E 2I4EI

2EI4 8

+ =

K12

= 0.5 EI

Stiffness matrix =11 12

21 22

K K 3EI 0.5EI

K K 0.5EI 2EI

=



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Q.23Q.23Q.23Q.23Q.23 (a)(a)(a)(a)(a)

R

h

Angle =Arc

Radius

2 =1 2L L T L L T

R R h

+ + =

+

( )1L 1 T

R

+ =

( )2L 1 T

R h

+ +

R h

4

+=

2

1

1 T

1 T

+ +

h

1R

+ =2

1

1 T

1 T

+ +

h

R=

( )2 1

1

T T

1 T

+

1

R=

( )( )

2 1

1

T T

h 1 T

+

=( ) ( )

( )1 2 1

1

L 1 T T T

2 h 1 T

+

+

=( )2 1L T T

2h

Q.25Q.25Q.25Q.25Q.25 (c)(c)(c)(c)(c)

E = 2G(1 + ) = 3K (1 2)

2 2

1 2

+ =

3K

G

3

1 2 =3K G

G

+[Adding 1 on both sides]

3G

3K G+= 1 2



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2 =3G

13K G

+

=(3K 2G)

2(3K G)

+

Q.26Q.26Q.26Q.26Q.26 (c)(c)(c)(c)(c)

Wx

dx

dFx

= dMx2

and dM =Adx

g

dFx

=

2Axdx

g

xd =2x

dxg

x

=2 2x

C2g

+

Now at x = L, x

= 0

C = 2 2L

2g

x = ( )

2 2 2x L2g

Q.27Q.27Q.27Q.27Q.27 (d)(d)(d)(d)(d)

Tensile force = W + wL = 100 + 5(10 5) = 125 N

5 m

10 m

W = 100 N

Q.30Q.30Q.30Q.30Q.30 (b)(b)(b)(b)(b)

From the formula of strain energy

U =21

2 E



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dU

d=

E

2

2

d U

d=

1

E=

2

5

1N/mm

2 10[from 2nd curve]

dU

d= 5

1C

2 10

+ [but, C = 0 from 1st curve]

U =2

5

1

2 (2 10 )

for = 10 kN/mm2

U = 250 N/mm2

Q.31Q.31Q.31Q.31Q.31 (c)(c)(c)(c)(c)

E =

=E

=

3

5

10 10

2 10

= 0.05

Q.32Q.32Q.32Q.32Q.32 (d)(d)(d)(d)(d)

Absolute maximum shear stress is obtained when stresses are

principal stresses i.e. except the diagonal elements all other values

are zero. Therefore the value of n = 6

Q.33Q.33Q.33Q.33Q.33 (d)(d)(d)(d)(d)

If roots of the given equation are 1, 2 or 3

then 1

+ 2

+ 3

=( A)

1

= A

Now keeping the n elements as zero, the stress matrix will become

[] =

10 0 0

0 20 0

0 0 30

i.e. A = 1

+ 2

+ 3

A = 10 + 20 + 30 = 60

Q.34Q.34Q.34Q.34Q.34 (a)(a)(a)(a)(a)

Transverse strain at a-a =316 10

180

= 8.88 105 (compression)

Longitudinal strain at a-a, =58.88 10

0.25

= 35.52 105 (extension)



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Total force at a-a = 2px 400

Longitudinal stress at a-a = x2p 400

10 180

Longitudinal strain at a-a =x

5

2p 400

10 180 2 10

35.52 105 = 5 x2

10 p9

px

= 35.52 4.5

px

= 159.84 N/mm

Q.35Q.35Q.35Q.35Q.35 (a)(a)(a)(a)(a)

Total reaction at A = 2px L = 2 159.84 2000 = 639.36 kN


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F1265726Test 01_CE_Strenth of Material and Stryctural Analysis