A body is subjected to two normal stresses 20 kn/m2 (tensile) and 10 kn/m2 (compressive) acting perpendicular to each other. The maximum shear stress is
Answers
Answer: 15 kN/m²
Explanation:
Given data:
Tensile stress acting on the body, σ1 = 20 kN/m²
Compressive stress acting on the body, σ2 = 10 kN/m²
To find: the maximum shear stress
Solution:
Since the tensile and compressive stresses are in opposite direction, therefore the sign convention of these stresses should also be in opposite direction.
The formula for the maximum shear stress,
Maximum shear stress(τ) = [Tensile stress(σ1) – Compressive stress(σ2)] / 2
Thus, based on the above formula, and substituting the values
The maximum shear stress (τ) ,
= [20 – (- 10)] / 2
= 30/2
= 15 kN/m²
Answer:
The maximum shear stress is 15 kN/m²
Explanation:
The most shear pressure may be placed on the beam's impartial axis, in order to be the midpoint of the length (h). At the impartial axis of a cross-section, the regular pressure and stress price are same to zero.
This concept additionally applies to triaxial states of pressure which predicts that yielding will arise each time one-1/2 of the algebraic distinction among the most and minimal pressure is same to one-1/2 of the yield pressure. Thus, for a triaxial nation of pressure where. σ1 > σ2 > σ3, the most shear pressure is (σ1 > σ3)/2.
Tensile pressure performing at the body, σ1 = 20 kN/m²
Compressive pressure performing at the body, σ2 = 10 kN/m²
To find: the most shear pressure
Since the tensile and compressive stresses are in contrary direction, consequently the signal conference of those stresses ought to additionally be in contrary direction.
The components for the most shear pressure,
Maximum shear pressure(τ) = [Tensile stress(σ1) – Compressive stress(σ2)] / 2
Thus, primarily based totally at the above components, and substituting the values
The most shear pressure (τ) ,
= [20 – (- 10)] / 2
= 30/2
= 15 kN/m²
#SPJ2