Numerical - Shear stress, given load, diameter

Q.  What is the value of shear stress acting on a plane of circular bar which is subjected to axial tensile load of 100 kN? (Diameter of bar = 40 mm , θ = 42.3o)
- Published on 23 Sep 15

a. 58.73 Mpa
b. 40.23 Mpa
c. 39.60 Mpa
d. Insufficient data

ANSWER: 39.60 Mpa
 

    Discussion

  • Sravanthi   -Posted on 25 Nov 15
    Given: Axial tensile load = 100 kN, Diameter of bar = 40 mm , θ = 42.3°

    Formula: Shear stress = (σx /2) sin 2θ

    Solution:

    - Stress acting in the direction of force = P /A

    = 100 x 103 / [(π / 4) 402]

    = 79.577 Mpa

    - Shear stress = ( 79.577/2) sin 2 (42.3)

    = 39.60 Mpa

Post your comment / Share knowledge


Enter the code shown above:
 
(Note: If you cannot read the numbers in the above image, reload the page to generate a new one.)