Question

1. A circular rod of diameter 20 mm and 500 mm long is subjected to a tensile force 45
KN. The modulus of elasticity for steel may be taken as 200 kN/mm'. Find stress, strain and elongation
of the bar due to applied load.
Solution
Load A = 45 KN = 45 x 1000 N
E = 200 kN/mm² = 200 x 10' N/mm²
L = 500 mm
Diameter of the rod d = 20 mm
Therefore, Sectional area
A =-
1 - red - 7 x 20²
4
4 207
= 314.159 mm
P 45 x 1000
Stress p =
= 143.24 N/mm² (Ans)
A 314.159
P 143.24
Strain e =
=0.0007162 (Ans)
E 200 x103
PL 45 x 1000 500
Elongation A=
3=0.358 mm (Ans)
AE 314.159 x 200 x 10​

Answer 1

Answer:

628312 is the answer it is the answer

it is the answer

Answer 2

Answer:

143.3 N/mm²

7.155 * 10⁻⁴

0.3578 mm

Step-by-step explanation:

Load = tensile force = F =   45  kN = 45 * 10³ N

modulus of elasticity = E = 200 kN/mm² = 200 * 10³ N/mm²

Rod Length = 500 mm

Rod Radius = Diameter/2 = 20/2 = 10 mm

Cross sectional area = πR²  = π *10² = 314  mm²

Stress = F/A  = 45 * 10³/ 314  =  143.3 N/mm²

E = Stress / Strain

=> Strain = Stress / E  =   143.3 / 200 * 10³  = 0.7155 * 10⁻³

= 7.155 * 10⁻⁴

Strain = Elongation / Length

=> 7.155 * 10⁻⁴ = Elongation / 500

=> Elongation = 0.3578 mm