Question
Download Solution PDFThe average shear stress in a rectangular wooden beam 100 mm wide, 250 mm deep and 3 m long carrying a uniformly distributed load with an intensity of 40 kN/m is found to be 2.4 Mpa, calculate the maximum shear stress.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Nominal or average shear stress (τavg):
\({{\rm{τ }}_{{\rm{avg}}}} = \frac{{{\rm{Shear\;force}}}}{{{\rm{Cross}} - {\rm{section\;area}}}}\)
The relation between maximum shear stress (τmax) and average shear stress (τavg) for different shapes is given in the below table
|
Section |
τmax/τavg |
ΤN.A/τavg |
|
Rectangular or square |
3/2 |
3/2 |
|
Circular |
4/3 |
4/3 |
|
Triangle |
3/2 |
4/3 |
|
Diamond |
9/8 |
1 |
Calculation:
Last updated on May 28, 2025
-> The DDA JE Recruitment 2025 Notification will be released soon.
-> A total of 1383 vacancies are expected to be announced through DDA recruitment.
-> Candidates who want a final selection should refer to the DDA JE Previous Year Papers to analyze the pattern of the exam and improve their preparation.
-> The candidates must take the DDA JE Electrical/Mechanical mock tests or DDA JE Civil Mock tests as per their subject.