You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Strength of Materials

Maximum Principal Strain Theory

The limit of ...

Question

The limit of proportionality in the material of a structural steel member, when subjected to simple tension, is 280 $N / m m^{2}$ . The principal stresses in the member are $σ_{1} = 122 N / m m^{2}$ (Tensile) and $σ_{2} = 60 N / m m^{2}$ (Compressive). $μ = 0.3$ . According to maximum strain theory, the factor of safety is

2.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 2
According to maximum strain theory,
$σ_{1} - μ σ_{2} \leq \frac{σ_{y}}{F}$

$\Rightarrow 122 - 0.3 \times (- 60) \leq \frac{280}{F}$

$\Rightarrow F \leq \frac{280}{140}$

$\Rightarrow F \leq 2$

Suggest Corrections

Similar questions

Q. A member is made of structural steel;. When it is subjected to simple tension, the limit of proportionality is 280

N / m m^{2}

. If the principal stresses

p_{1} a n d p_{2}

developed in the member are 100

N / m m^{2}

(tensile) and 40

N / m m^{2}

(compressive) respectively and Poisson's ratio is 0.30, then the factor of safety according to maximum shear stress theory would be

Q. An element of a structure is subjected to two principal stresses

σ_{1} a n d σ_{2}

σ_{1} = 200 N / m m^{2}

(Tensile)

σ_{2}

is Compressive
The yield stress both in simple tension and compression for the material is 240

N / m m^{2}

. Poisson's ratio

μ = 0.25

; what is the value of

σ_{2}

N / m m^{2}

as per maximum normal strain theory at which the yield of material will commence?

Q. A structural element is subjected to a two-dimensional stress system, wherein

σ_{1} = 225 N / m m^{2}

(tensile) with

σ_{2}

being compressive. The yield stress in both simple tension

(σ_{y})_{t}

and simple compression

(σ_{y})_{c}

is 250

N / m m^{2}

and

μ

= 0.25. What is the value of

σ_{2}

, according to Maximum Strain Theory?

Q. In a two-dimensional stress system, the principal stresses are

σ_{1} = 200 N / m m^{2}

(tensile) and

σ_{2}

(compressive). The yield stress in both simple tension and compression is 250 N/mm^{2}, with

μ = 0.25

. What will be the value of

σ_{2}

according to the maximum normal strain theory?

Q. Principal stresses at a point are 80

N / m m^{2}

and 40

N / m m^{2}

, both tensile. The yield stess in simple tension for this material is 200

N / m m^{2}

. The value of factors of safety according to maximum principal stress theory and maximum shear stress theory, respectively, are

Join BYJU'S Learning Program

The limit of proportionality in the material of a structural steel member, when subjected to simple tension, is 280 N/mm2. The principal stresses in the member are σ1=122 N/mm2 (Tensile) and σ2=60 N/mm2 (Compressive). μ=0.3. According to maximum strain theory, the factor of safety is

The correct option is C 2According to maximum strain theory, σ1−μσ2≤σyF ⇒122−0.3×(−60)≤280F ⇒F≤280140 ⇒F≤2

The correct option is C 2
According to maximum strain theory,
$σ_{1} - μ σ_{2} \leq \frac{σ_{y}}{F}$

$\Rightarrow 122 - 0.3 \times (- 60) \leq \frac{280}{F}$

$\Rightarrow F \leq \frac{280}{140}$

$\Rightarrow F \leq 2$