Civil Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 1 (Q.No. 25)
25.
A rectangular beam 20 cm wide is subjected to a maximum shearing force of 10, 000 kg, the corresponding maximum shearing stress being 30 kg/cm2. The depth of the beam is
Discussion:
26 comments Page 3 of 3.
Chintu said:
8 years ago
Excellent explanation @Arpita.
Bibek dey said:
9 years ago
Since, maximum shearing stress = 1.5 * average shearing stress.
Average shearing stress = V/bd (V=shear force). 30 = 1.5 * 10000(20/d).
So, d = 25 cm.
Average shearing stress = V/bd (V=shear force). 30 = 1.5 * 10000(20/d).
So, d = 25 cm.
(5)
Krishna said:
9 years ago
Convert the given max to average shear stress and then proceed using,
T(max) = 2/3 ( T(avg)).
T(max) = 2/3 ( T(avg)).
Abhay said:
10 years ago
Maximum shearing stress σ max = 3P/2(bd).
where P = Shearing force.
30 = 3*10000/(2*20*d) => d = 25 cm.
where P = Shearing force.
30 = 3*10000/(2*20*d) => d = 25 cm.
Arpita saxena said:
1 decade ago
Max. shear stress = 1.5*avg.shear stress.
Avg.shear stress = S.F/(B*d).
Avg.shear stress = 10000/(20*d).
30 = 1.5*10000/(20*d).
d = 25cm.
Avg.shear stress = S.F/(B*d).
Avg.shear stress = 10000/(20*d).
30 = 1.5*10000/(20*d).
d = 25cm.
(3)
Mounika Reddy said:
1 decade ago
T = SQ/Ib.
S=10000, T=30, Q=bh^2/8, I=bh^3/12.
Then h = 25cm,
or
Tmax = (3*Vmax)/(2*A).
h = 25cm, how will be 15cm.
S=10000, T=30, Q=bh^2/8, I=bh^3/12.
Then h = 25cm,
or
Tmax = (3*Vmax)/(2*A).
h = 25cm, how will be 15cm.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers