# Strength of material part -1

Strength of material part -1
Question:-
Option (A)
25 T concentrated load at free end
Option (B)
20 T concentrated load at free end
Option (C)
5 T concentrated load at free end 2 T/m load over entire length
Option (D)
10 T/m udl over entire length
Correct Option:
D
Question Solution:

RA = 10 x 5 = 50T
M = 10 x 5 x (5/2) = 125 T.m
S.F diagram

For similar triangle
(50T/5)=(F/x) = (F/2)
\ F = 20T (at 2m from free end) and M ( at 2m from free end)
= (125 – 50 x 3 + 10 x 3 x(3/2)) T = 20 T-m
Question:-
Option (A)
M
Option (B)
ÖM
Option (C)
M2
Option (D)
1/M
Correct Option:
A
Question Solution:

(m/l)= (sb/ymax)
{M/(1/12)bh3} = {sb/(h/2)}
\ M = sxb(1/6)bh2
\ baM
Question:-
Option (A)
Option (B)
Option (C)
Option (D)
Correct Option:
D
Question Solution:
Question:-
Option (A)
zero
Option (B)
100 kg-m
Option (C)
150 kg-m
Option (D)
200 kg-m
Correct Option:
A
Question Solution:
Bending moment at internal hinged is always zero
Question:-
Option (A)
0.425 x 10-3
Option (B)
0.5 x 10-3
Option (C)
0.585 x 10-3
Option (D)
0.75 x 10-3
Correct Option:
A
Question Solution:
s1 = (pd/2t) = 100N/mm2
s2 = (pd/4t) = 50N/mm2
Hoop strain = (pd/4tE)(2-m)
= (50/(200x1000))x(2-0.3) = 0.425 x 10-3
Question:-
Option (A)
Three time its shear modulus
Option (B)
For times its shear modulus
Option (C)
equal to its shear modulus
Option (D)
indeterminate
Correct Option:
A
Question Solution:
E = 2G (1+m)
Put m = 0.5
\ E = 3G