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Mila [183]
3 years ago
11

Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin

ed plane (b) principal and maximum shear stresses and identify all the terms in the expression including units.
Engineering
1 answer:
Alborosie3 years ago
6 0

Answer:

a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

b) principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

Explanation:

Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

бy =  tensile stress in y direction

Txy = y component of shear stress acting on the perpendicular plane to x axis

бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

A) Normal and shear stresses on inclined plane

Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

B) principal and maximum shear stresses

principal stress :

б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

maximum shear stress:

Tmax  = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy

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Write the definitions for engineering stress, true stress, engineering strain, and true strain for loading along a single axis.
shusha [124]

Answer and Explanation:

Engineering Stress of a material is defined as the applied load or force divided by the original cross sectional Area of the material.

σ(engineering) = F/(Ao)

True Stress is defined as the applied load or force divided by the actual cross-sectional area (the changing area with respect to time) of the material at that point in time. It's an instantaneous stress.

σ(true) = F/A

Engineering strain is a measure of how much a material deforms under a particular load. It is the amount of deformation in the direction of the applied force divided by the initial length of the material.

ε(engineering) = Δl/lo

True Strain measures instantaneous deformation. It is obtained mathematically by integrating strain over small time periods and summing them up. Hence,

ε(true) = In (lf/lo)

The calculations,

First step, 10m to 10.1m, Δl = 0.1m, lf = 10.1m, lo = 10m

ε(engineering) = 0.1/10 = 0.01

ε(true) = In (10.1/10) = 0.00995

Second step, 10.1m to 10.2m, Δl = 0.1m, lf = 10.2m, lo = 10.1m

ε(engineering) = 0.1/10.1 = 0.0099

ε(true) = In (10.2/10.1) = 0.00985

Overall, 10m to 10.2m, Δl = 0.2m, lf = 10.2m, lo = 10m

ε(engineering) = 0.2/10 = 0.02

ε(true) = In (10.2/10) = 0.0198

QED!

5 0
3 years ago
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