Question

A solid round bar 3m long and 5cm in diameter is used as strut. Both ends of the struts are hinged. Determine the Crippling Load if the value of E =2x10^5 N/mm^2

Answer 1

Answer:

Step 1: Data

Length of the column = 3000mm

Diameter of the column = 50mm

condition = Both ends hinged

Crippling load = ??

E=2x105N/mm2

Step 2: Calculation of moment of inertia

I = π d 4 / 64

I= π (50)4 / 64

I = 0.306X106mm4

Step 3: Calculation of crippling load

Condition = Both ends hinged

P = Π 2E I/ L2

P = Π 2(2x105) (0.306X106 ) / (3000)2

P = 67.11 KN

...................

if helpful so mark it as a brainaliest

Answer 2

Answer:

Explanation:

Tip

• The buckling load of long columns is calculated using Euler's formula for the crippling load. The load that may be calculated using this technique is the maximum load that a column can support. Divide the ultimate load by the safety factor to determine the safe load (F.O.S)
• EULER’S FORMULA

$P=\frac{\Pi^{2} E I}{l^{2}}$

Where,

P= Buckling Load

E= Modulus of Elasticity of material,

I= Moment of Interia of column section

$\mathrm{l}$= Equivalent/ Effective length of the column

Given

Length of solid round bar=3000mm

Diameter of strut=50 mm

E=2*$10^{5}$N/$mm^{2}$

Find

Crippling load

Solution

Calculation of moment of inertia

$I=\pi d^{4} / 64$

$I=\pi(50)^{4} / 64$

$I=0.306 \times 10^{6} \mathrm{~mm}^{4}$

Calculation of crippling load

Condition = Both ends hinged

$P=\Pi^{2} E I / L^{2}$

$P=\Pi^{2}\left(2 \times 10^{5}\right)\left(0.306 \times 10^{6}\right) /(3000)^{2}$

$P=67.11 \mathbf{K N}$

#SPJ2