# Mechanical Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 7)
7.
A body is subjected to a tensile stress of 1200 MPa on one plane and another tensile stress of 600 MPa on a plane at right angles to the former. It is also subjected to a shear stress of 400 MPa on the same planes. The maximum normal stress will be
400 MPa
500 MPa
900 MPa
1400 MPa
Explanation:
No answer description is available. Let's discuss.
Discussion:
56 comments Page 1 of 6.

Rashid Khan Kakar said:   2 years ago
It's subject to tow perpendicular normal stresses on oblique section.
(1)

Yugal Chaudhari said:   5 years ago
Use @Kamal talukdar formula. It is correct.

Kamal Talukdar. said:   5 years ago
1200+600/2+1/2 * (1200-600)^2+4 * (400)^2.
= 1800/2+1/2 * (600)^2+4 * (400)^2,
= 900+1/2 * 360000+640000,
= 900+1/2*1000,
= 900+500.
= 1400MPa.
(17)

Bangaram said:   5 years ago
1200+6000/2+1/2 ( (1200-600) + 4 (400) ) =1400mpa.

So it is by using formula of maximum normal stress.
(4)

Mustaf HD said:   6 years ago
Normal stress= (x+y)/2-[{(x-y)/2}cos2@]-(z*sin2θ).

Z-shear stress.
θ perpendicular means 90.
900-(-300)-0=1200.
The answer should be 1200.
(3)

Kaushal roy said:   6 years ago
Two different stress act on a member design should be based on max principal stress so here we find max principal stress.

Anwarul haque said:   6 years ago
Fastest Method is Mohrs circle Method.
(3)

Sagar bankar said:   6 years ago
Maximum normal stress : σ 1=1200 Mpa , σ2=600 Mpa, τ= 400 Mpa.

Maximum normal stress =(σ 1+σ2/2) +(1/2) √[(σ 1-σ2)^2 +4(T)^2].
=(1200+600/2) +(1/2) √ [(1200-600)^2 +4(400)^2],
=(1800/2)+(1000/2),
=1400 MPa.
(19)

Shiva said:   6 years ago
Maximum normal stress : σ 1=1200 Mpa , σ2=600 Mpa, τ= 400 Mpa.

Maximum normal stress formula= (σ1 + σ2)/2 + √(σ1 + σ2)/2 + τ 1500 Mp.
(2)

Afrooj said:   6 years ago
Thanks for all for the explanation.