Rankine Formula MCQ Quiz - Objective Question with Answer for Rankine Formula - Download Free PDF

Concept:

Rankine's Gorden Theory / Rankine's theory

• Short column fails in crushing and long columns fail in buckling but in practice, the columns fail due to combined effect of crushing and buckling.
• This theory assumed a combined mode of failure.
• The Rankine's critical load is given by, $\frac{1}{P}=\frac{1}{P_c}+\frac{1}{P_e}$ where,
• P_c = crushing load, P_e = Euler's buckling load

Calculation:

Given,

P_c = 2000 kN and P_e = 2000 kN

$\begin{array}{l}\frac{1}{P}=\frac{1}{2000}+\frac{1}{2000}=\frac{1}{1000}\end{array}$

P = 1000 kN

Explanation:

Rankine formula is valid for both short and long column. Hence failure by, both crushing and buckling should be considered.

If P = Failure load

1. P_C = Crushing load = σ_C A
2. P_E = Buckling load = $=\frac{{\pi }^{2}E}{l^2}$

$P=\frac{{\sigma }_{C}A}{1+\propto {\lambda }^2}$

where ∝ = Rankine constant = $=\frac{{\sigma }_c}{{\pi }^{2}E}$, λ = Slenderness ratio, σC = Crushing stress, A = cross-sectional area

Rankine’s constant for the type of materials are:

Explanation:

For short or long columns Rankine’s Formula is used.

$\frac{1}{P}=\frac{1}{P_C}+\frac{1}{P_E}$

Where P is crippling load by Rankine formula; $P_C$ is crushing load; $P_E$ is crippling load by Euler’s formula.

For Short column: $P_E$ is very large so $\frac{1}{P_E}$  will be very small and negligible as compared to $\frac{1}{P_C}.$ so

$\frac{1}{P}=\frac{1}{P_C}\to P=P_C$

For Long column: $P_E$ is small so $\frac{1}{P_E}$ will be large as compared to  $\frac{1}{P_C}$ . Hence the value of  $\frac{1}{P_C}$ can be neglected. So

$\frac{1}{P}=\frac{1}{P_E}\to P=P_E$

Explanation:

Rankine formula is valid for both short and long column. Hence failure by, both crushing and buckling should be considered.

If P = Failure load

1. P_C = Crushing load = σ_C A
2. P_E = Buckling load = $=\frac{{\pi }^{2}E}{l^2}$

$P=\frac{{\sigma }_{C}A}{1+\propto {\lambda }^2}$

where ∝ = Rankine constant = $=\frac{{\sigma }_c}{{\pi }^{2}E}$, λ = Slenderness ratio, σC = Crushing stress, A = cross-sectional area

Rankine’s constant for the type of materials are:

Explanation:

For short or long columns Rankine’s Formula is used.

$\frac{1}{P}=\frac{1}{P_C}+\frac{1}{P_E}$

Where P is crippling load by Rankine formula; $P_C$ is crushing load; $P_E$ is crippling load by Euler’s formula.

For Short column: $P_E$ is very large so $\frac{1}{P_E}$  will be very small and negligible as compared to $\frac{1}{P_C}.$ so

$\frac{1}{P}=\frac{1}{P_C}\to P=P_C$

For Long column: $P_E$ is small so $\frac{1}{P_E}$ will be large as compared to  $\frac{1}{P_C}$ . Hence the value of  $\frac{1}{P_C}$ can be neglected. So

$\frac{1}{P}=\frac{1}{P_E}\to P=P_E$

Concept:

Rankine's Gorden Theory / Rankine's theory

• Short column fails in crushing and long columns fail in buckling but in practice, the columns fail due to combined effect of crushing and buckling.
• This theory assumed a combined mode of failure.
• The Rankine's critical load is given by, $\frac{1}{P}=\frac{1}{P_c}+\frac{1}{P_e}$ where,
• P_c = crushing load, P_e = Euler's buckling load

Calculation:

Given,

P_c = 2000 kN and P_e = 2000 kN

$\begin{array}{l}\frac{1}{P}=\frac{1}{2000}+\frac{1}{2000}=\frac{1}{1000}\end{array}$

P = 1000 kN

Explanation:

Rankine formula is valid for both short and long column. Hence failure by, both crushing and buckling should be considered.

If P = Failure load

1. P_C = Crushing load = σ_C A
2. P_E = Buckling load = $=\frac{{\pi }^{2}E}{l^2}$

$P=\frac{{\sigma }_{C}A}{1+\propto {\lambda }^2}$

where ∝ = Rankine constant = $=\frac{{\sigma }_c}{{\pi }^{2}E}$, λ = Slenderness ratio, σC = Crushing stress, A = cross-sectional area

Rankine’s constant for the type of materials are:

Explanation:

For short or long columns Rankine’s Formula is used.

$\frac{1}{P}=\frac{1}{P_C}+\frac{1}{P_E}$

Where P is crippling load by Rankine formula; $P_C$ is crushing load; $P_E$ is crippling load by Euler’s formula.

For Short column: $P_E$ is very large so $\frac{1}{P_E}$  will be very small and negligible as compared to $\frac{1}{P_C}.$ so

$\frac{1}{P}=\frac{1}{P_C}\to P=P_C$

For Long column: $P_E$ is small so $\frac{1}{P_E}$ will be large as compared to  $\frac{1}{P_C}$ . Hence the value of  $\frac{1}{P_C}$ can be neglected. So

$\frac{1}{P}=\frac{1}{P_E}\to P=P_E$

Concept:

Rankine's Gorden Theory / Rankine's theory

• Short column fails in crushing and long columns fail in buckling but in practice, the columns fail due to combined effect of crushing and buckling.
• This theory assumed a combined mode of failure.
• The Rankine's critical load is given by, $\frac{1}{P}=\frac{1}{P_c}+\frac{1}{P_e}$ where,
• P_c = crushing load, P_e = Euler's buckling load

Calculation:

Given,

P_c = 2000 kN and P_e = 2000 kN

$\begin{array}{l}\frac{1}{P}=\frac{1}{2000}+\frac{1}{2000}=\frac{1}{1000}\end{array}$

P = 1000 kN

Given,

$L_e=\frac{l}{2}=2000mm$

P_safe= 250 kN

Let External diameter = D

Internal dia. = 0.8 × D

Crushing stress, σ_c = 550 N/mm²

Now factor of safety $=\frac{Ciripplingload}{Safeload}or5=\frac{Ciripplingload}{250}$

∴ Crippling load, P = 5 × 250 = 1250 kN = 1250000 N

Area of column, $A=\frac{\pi }{4}\left[D^2-{\left(0.8D\right)}^2\right]$

$=\frac{\pi }{4}\left[D^2-0.64D^2\right]=\frac{\pi }{4}\times 0.36D^2=\pi \times 0.09D^2$

Moment of Inertia, $I=\frac{\pi }{64}\left[D^4-\left(0.8D^4\right)\right]=\frac{\pi }{64}\left[D^4-0.4096D^4\right]$

$=\frac{\pi }{64}\times 0.5904\times D^4=0.009225\times \pi \times D^4$

$k=\sqrt{\frac{I}{A}}=\sqrt{\frac{0.009225\times \pi D^4}{\pi \times 0.09\times D^2}}=0.32D$

Crippling Load by Rankine Formula:

$P=\frac{{\sigma }_c.A}{1+a{\left(\frac{L_e}{k}\right)}^2}$

Or $1250000=\frac{550\times \pi \times 0.09D^2}{1+\frac{1}{1600}\times {\left(\frac{2000}{0.32D}\right)}^2}$ 

$\frac{1250000}{550\times \pi \times 0.09}=\frac{D^2}{1+\frac{24414}{D^2}}or8038=\frac{D^2\times D^2}{D^2+24414}$

Or D⁴ – 8038 D² – 196239700 = 0

$\therefore D^2=18592.5\mathrm{m}{\mathrm{m}}^2$

$\therefore D=\sqrt{18592.5}=136.3mm$

∴ External diameter = 136.3 mm

Rankine’s crippling load,

$P_R=\frac{f_C\times A}{1+a{(l/r)}^2}$

$A=\frac{\pi }{4}({150}^2-{100}^2)=9817.47m{m}^2$

r = radius of gyration = $\sqrt{\frac{I}{A}}=\sqrt{\frac{19941750}{9817.47}}=45mm$

$P_R=\frac{320\times 9817.47}{1+(1/1600)\times {(6000/45)}^2}=259.397kN$

Euler’s load,
$I=\frac{\pi }{64}({150}^4-{100}^4)=19941750m{m}^4$

$\begin{array}{rl}& P=\frac{{\pi }^{2}EI}{l^2}=\frac{\pi \times 2\times {10}^5\times 19941750}{{6000}^2}\\ & =1093.429kN\end{array}$