A mild steel specimen is under uniaxial tensile stress - Young-s modulus and yield stress for mild steel are 2 - 105 MPa and 250 MPa respectively- The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is

Young's modulus, E= 2 × 10^5 MPa Yield stress, f_y = 250 MPa Strain energy per unit volume = 12 stress×strain nbsp; = 12f_y ×f_yE = 12×(250)^22 × 10^5 =0.156 ...

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

Byju's Answer

Standard XII

Physics

Distance

A mild steel ...

Question

A mild steel specimen is under uniaxial tensile stress . Young's modulus and yield stress for mild steel are $2 \times 10^{5} M P a$ and $250 M P a$ respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is

156 N m m / m m^{3}

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

156 N m m / m m^{3}

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

1.56 N m m / m m^{3}

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

0.156 N m m / m m^{3}

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

Open in App

Solution

The correct option is D $0.156 N m m / m m^{3}$
Young's modulus, $E = 2 \times 10^{5} M P a$
Yield stress, $f_{y} = 250 M P a$
Strain energy per unit volume
$= \frac{1}{2} stress \times strain$
$= \frac{1}{2} f_{y} \times \frac{f_{y}}{E}$
$= \frac{1}{2} \times \frac{(250)^{2}}{2 \times 10^{5}}$
$= 0.156 N / m m^{2}$
$= 0.156 N m m / m m^{3}$

Suggest Corrections

Similar questions

Q. An ordinary mild steel bar has been prestressed to a working stress of 200 MPa. Young's modulus of steel is 200 GPa. Permanent negative strain due to shrinkage and creep is 0.0008. How much is the effective stress left in steel?

Q. At a point in a steel member, the major principal stress is 200 MPa (tensile) and the minor principal stress is compressive. If the uniaxial tensile yield stress is 250 MPa, then according to the maximum shear stress theory, the magnitude of the minor principal stress (compressive) at which yielding will commence is

Q. The minimum strain at failure in tension steel having yield stress

f_{y} = 415

MPa and Young's modulus

E_{s} = 200

GPa, as per Limit State Method of Design, is

Q. In the engineering stress-strain curve for mild steel, the Ultimate Tensile Strength (UTS) refers to

Q. A

2 m

long axially loaded mild steel rod of

8 m m

diameter exhibits the load-displacement

(P - δ)

behaviour as shown in the figure.

Assume the yield stress of steel as

250 M P a .

The complementary strain energy (in N-mm) stored in the bar up to its linear elastic behaviour will be

Join BYJU'S Learning Program

A mild steel specimen is under uniaxial tensile stress . Young's modulus and yield stress for mild steel are 2×105 MPa and 250 MPa respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is

The correct option is D 0.156 Nmm/mm3Young's modulus, E=2×105 MPa Yield stress, fy=250 MPa Strain energy per unit volume =12 stress×strain =12fy×fyE =12×(250)22×105 =0.156 N/mm2 =0.156 Nmm/mm3

A mild steel specimen is under uniaxial tensile stress . Young's modulus and yield stress for mild steel are $2 \times 10^{5} M P a$ and $250 M P a$ respectively. The maximum amount of strain energy per unit volume that can be stored in this specimen without permanent set is