Strength of materials Paper 1 Solution

STRENGTH OF MATERIALS
(STM - I : SOLUTIONS)

1) Maximum bending moment depends upon to the magnitude & position of UDL &
concentrated point load. So ans (d)

2) Modulus of resilience  Area under  & upto elastic limit

= 1 x 250 x 0.002
2
= 0.25 MPa

3)

1  PL3
3EI

1  12PL3
3Ebd 3

I  bd 3
12

2  3E 12PL3 12PL3
x 2b x  d3  163Ebd 3

2  1  1
16

2  6.25%  2

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4)

1  (D 4P  a)  4P a2 )
 a)(D (D2 

Using Mean dia. D

2  4P
D2

 1   2 x 100%
1

So Error

4P 4P
  x (D2 -a2 )  D2

           x 100%
4P

(D2  a2)

 D2  D2  a2 x D2  a2 x 100%
(D2  a2) x D2 1

Error  a2 x 100% =  10a 2
D2  D 

5) t1 = 2t2 , d1 = 3d2
P1 = P2 = P

1  Pd1
4t1

2  Pd2
4t2

1  d1 x t2  3d2 x t2  3
2 t1 d2 2t2 d2 2

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6)

B  BA  AO

 0   AO

 No Torque in member BA so No twist.

 AO  T.L / 2  TL
GJ 2GJ

7)

1  WL3
48EI

2  5 WL4 5 WL3
384 EI  384 EI

1  24
2 15

8) (c)

9) (c)

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10) Euler’s theory is useful for long column only.

11) P1 = P2
N1T1 = N2T2

N.T  N .T2
2

T2 = 2T

12) Solid shaft  Hollow shaft

s = H

13) 1   2  3  0

14)

Z circle  D3 , Z Rect = bd 2
32 6

D2 = d2 + b2 , d2 = D2 - b2

(Z rec)N.A. = b(D2  b2 )
6

dzrec  0, for efficient section
db

D2  3b2 0
6

D2 = 3b2

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D  3b , b  1
D 3

d2 = 2b2

d  2
b

15)   16T , b = 32M
d3 d3

 16T  T
b  32M 2M

16) (c)

At top Fibre

ONLY tensile stress outing will be zero

( )shear stress At atove the N.A. both  max &  acts so

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17)

at point B 1 total deflection is zero

WL3 (wa)L2
3EI  2EI

a  2
3L

18) So, according to Maxwe-ll Reciprocal thermal both statements are correct but R is not
correct explanation of A.

19)

 AB 1
 BC 2

20)

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1  (P  R)(3a)3
3EI

2  Ra3
3EI

1  2

(P  R) x 27a3  R x a3
3EI 3EI

R  27P
28

So  = 27P x a3 = 9Pa3
28 x 3EI 28EI

21)

b  M  E
Y I R

Y  0.5  0.25
2

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b  200 x 103 x 0.25  250MPa
200.25

22) Tc = 0.662 W,

Ts = 0.338 W

23)

t  y   x
t E E

= 40  0.3 x 60
200 x 103 200 x 103

 t  0.022  0.03 mm

24)

MB  P , MC  P , MD  P
4 8 4

Looking the given ans & given diagram it is clear that at a there is no reaction & there is a
support so at A, a force of Maq. Is acting so that

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MB  P  P
4 4

Mc   P x 2.5 + P x 1.5 = P
4 2 8

Because Reaction at RA < RB

RA  RB  Total Vertical load
2

RA  RB  P
2

25)  spring  2
0

zfy = 0 ,
RA + RC = 10 KN
RA = RC = 5 KN

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Simple supported beam with central load
K. = RC
500. = 5000
 = 10 mm

26)
M = 5x2 + 20x – 7

dM  10x  20
dx

 dM  S.F.
dx

S.F  at x  2

= 40 N

27)

RA + RB = 2 KN

 MA  0 ,

R x 4 – 1x2 – 1 x 6 = 0
Rx4=8
R=2
R=0

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B  A  1 1 x 2m x2 x kN.m
EI  2

2
EI

28) Cantilever beam subjected to moment as free end

Deflection  ML2
2EI

29)

width = bx = x.b
L

I  bx t 3  xbt 3
12 12L

30)

U  t M 2 dx
0 x

2 EI x

  1 Mx 2M x dx
EI 2P

Use Mx = Px

 2M x  x and use I = xbt 3
2P 12l

  12 l Px.x dx
E 0 bxt 3

12P l xdx
EBt 3 0


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 12Pl  x2 l  12 Pl 3  6 Pl 3
EBt 3   2 EBt 3 Ebt 3
 2 0

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