Figure Q6 (b) Sketch a typical stress-strain curve for mild steel, aluminium alloy and polyethylene, and discuss the most likely failure mechanism for each material when it is under tensile load.

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Chapter1: Introduction
Section: Chapter Questions
Problem 1.5.5P: The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen...
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Please do not rely too much on chatgpt, because its answer may be wrong. Please consider it carefully and give your own answer. You can borrow ideas from gpt, but please do not believe its answer.Very very grateful!Please do not rely too much on chatgpt, because its answer may be wrong. Please consider it carefully and give your own answer. You can borrow ideas from gpt, but please do not believe its answer.Very very grateful! Just solve for question (b)!!!Thanks!
(a) A cylindrical sample as shown in Figure Q6 is under tensile force F. The tensile stress
produces a shear stress T on a plane inclined at an angle to the horizontal. Prove that
the shear stress on the inclined plane is at a maximum when is 45°.
AF
Circular cross-section
area A, radius r
Inclined plane with
shear stress T
Figure Q6
(b) Sketch a typical stress-strain curve for mild steel, aluminium alloy and polyethylene, and
discuss the most likely failure mechanism for each material when it is under tensile load.
Transcribed Image Text:(a) A cylindrical sample as shown in Figure Q6 is under tensile force F. The tensile stress produces a shear stress T on a plane inclined at an angle to the horizontal. Prove that the shear stress on the inclined plane is at a maximum when is 45°. AF Circular cross-section area A, radius r Inclined plane with shear stress T Figure Q6 (b) Sketch a typical stress-strain curve for mild steel, aluminium alloy and polyethylene, and discuss the most likely failure mechanism for each material when it is under tensile load.
(a)
shear force:
inclined plane Area:
shear stress:
T=
介
F = F sin 0.
dT
de
-
As
Fs F
As A
For maximum value of T,
F
0,⇒ x 2 x cos 20 = 0
2A
=
A
Cos
-sin cos 0:
cos 2000 = 45°
F
2.4 sin 28.
2A
FCOSO
AF
4 FSing
F
VA
Cose
Transcribed Image Text:(a) shear force: inclined plane Area: shear stress: T= 介 F = F sin 0. dT de - As Fs F As A For maximum value of T, F 0,⇒ x 2 x cos 20 = 0 2A = A Cos -sin cos 0: cos 2000 = 45° F 2.4 sin 28. 2A FCOSO AF 4 FSing F VA Cose
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