Question
Download Solution PDFFor an RCC beam of width 230 mm and effective depth 300 mm subjected to a shear force of 69 kN due to design loads, what will be the nominal shear stress in the beam?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept
As per the IS Code 456: 2000, clause 40.1
Nominal Shear Stress
The nominal shear stress in beams of uniform depth shall be obtained by the following equation:
\(\tau_v=\frac{V_u}{bd}\)
where
Vu = shear force due to design loads;
b = breadth of the member, which for flanled section shall be taken as the breadth of the web bw and
d =effective depth.
Given:
b = 230 mm
d = 300 mm
Design shear force, Vu = 69 KN = 69,000 N
\(\tau_v=\frac{V_u}{bd}\)
Nominal shear stress = 69,000 / (230 x 300)
= 1 N/mm2
Nominal shear stress = 1 N/mm2
Last updated on May 28, 2025
-> SSC JE notification 2025 for Civil Engineering will be released on June 30.
-> Candidates can fill the SSC JE CE application from June 30 to July 21.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.
-> The Staff Selection Commission conducts the SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.