Numerical Problems on Rankine’s formula.

1. The external and internal diameters of the hollow cast iron column are 5cm and 4cm respectively. If the length of this column is 3m and both of its ends are fixed, determine the crippling load using Rankine’s formula. Take σc= 550N/mm2 and a=1/1600.

Step 1: Data:
D = external diameter = 50mm
d= internal diameter = 40mm
Rankine's constant=a=1/1600.
length of column= 3m = 3000mm
condition=both of its ends are fixed
crippling load =??
_{critical stress =}  σc= 550N/mm2

Step 2: Calculation of the area of the column
A= π/4(D²-d²)
A= π/4(50²-40²)
A = 706.85 mm²

Step 3 :Calculation of the moment of inertia  of the column
I = π/64(D⁴-d⁴)
I = π/64(50⁴-40⁴)
I = 181.04X10³ mm⁴

Step 4:Calculation of Effective length
condition=both of its ends are fixed
therefore,

L_e=l / 2

L_{e = 3000/ 2}

L_{e = 1500mm}

Step 5:Calculation of radius of gyration

K= √(I/A)

K = √(181.04X10³ /706.85)

K = 16.00 mm

Step 6:Calculation of crippling load

P=  (σ_c A)/(1+a(L_e/k)²)

P = ((550) ₍706.85))/(1+(1/1600)(1500/16)²)

P = 60.33KN

2. The hollow cylindrical cast iron column is 4m long with both ends fixed. Determine the minimum diameter of the column if it has to carry a safe load of 250KN with a FOS of 5 take internal diameter as 0.8 times the external diameter. Take σc= 550N/mm2 and a=1/1600.

Step 1: Data:
D = external diameter
d= internal diameter = 0.8 D
Rankine's constant=a=1/1600.
length of column=4m = 4000mm
condition=both of its ends are fixed
_{critical stress =}  σc= 550N/mm2
safe load =  250KN
Factor of safety = 5

Step 2: Calculation of the area of the column
A= π/4(D²-d²)
A= π/4(D²-(0.8D)²)
A = 0.2826 D² mm²

Step 3 :Calculation of the moment of inertia  of the column
I = π/64(D⁴-d⁴)
I = π/64(D⁴-(0.8D)⁴)
I = 0.029 D⁴ mm⁴

Step 4:Calculation of Effective length
condition=both of its ends are fixed
therefore,

L_e=l / 2

L_{e =4000/ 2}

L_{e =2000mm}

Step 5:Calculation of radius of gyration

K= √(I/A)

K = √(0.029 D⁴/0.2826 D²)

K = 0.3203 D mm

Step 6: Calculation of crippling load 

FOS = Crippling load / safe load

5 = Crippling load / 250

Crippling load= 250X5 = 1250KN

Step 7:Calculation of diameter

P=  (σ_c A)/(1+a(L_e/k)²)

1250 = ((550) ₍0.2826 D²))/(1+(1/1600)(2000/0.3203 D)²)

D = 136.33 mm
d = 0.8 D = 109.06 mm

3. A hollow cast iron column of external diameter 250mm and internal diameter 200mm is 10m long with both ends fixed. Find the safe axial load with FOS of 4.Take σc= 550N/mm2 and a=1/1600.

Step 1: Data:
D = external diameter = 250mm
d= internal diameter = 200mm
Rankine's constant=a=1/1600.
length of column= 10m = 10000mm
condition=both of its ends are fixed
safe load =??
_{critical stress =}  σc= 550N/mm2

Factor of safety = 4

Step 2: Calculation of the area of the column
A= π/4(D²-d²)
A= π/4(250²-200²)
A = 17671.45 mm²

Step 3 :Calculation of the moment of inertia  of the column
I = π/64(D⁴-d⁴)
I = π/64(250⁴-200⁴)
I = 113.20X10⁶ mm⁴

Step 4:Calculation of Effective length
condition=both of its ends are fixed
therefore,

L_e=l / 2

L_{e =10000/ 2}

L_{e =5000mm}

Step 5:Calculation of radius of gyration

K= √(I/A)

K = √(113.20X10⁶/17671.45)

K = 80.03 mm

Step 6:Calculation of crippling load

P=  (σ_c A)/(1+a(L_e/k)²)

P = ((550) ₍17671.45))/(1+(1/1600)(5000/80)²)

P = 33.44MN

Step 7: Calculation of crippling load 

FOS = Crippling load / safe load

4= 33.44 / safe load

safe load= 8.36MN

4. Find the Euler’s crippling load for a hollow cylindrical steel column of 40mm external diameter and 4mm thick. The length of the column is 2.5m and is hinged at both the ends.  Also compute the Rankine’s crippling load using constants 350MPa and 1/7500.Take E=205GPa.

Step 1: Data:
D = external diameter = 40mm
Thickness = 4mm
d= internal diameter = 40-2(4)=32mm
Rankine's constant=a=1/7500.
length of column= 2.5m = 2500mm
condition=hinged at both the ends
crippling load =??
_{critical stress =}  σc=350N/mm2

E=205GPa.

Step 2: Calculation of the area of the column
A= π/4(D²-d²)
A= π/4(40²-32²)
A =452.38mm²

Step 3 :Calculation of the moment of inertia  of the column
I = π/64(D⁴-d⁴)
I = π/64(40⁴-32⁴)
I = 0.074X10⁶ mm⁴

Step 4:Calculation of Effective length
condition=both of its ends are fixed
therefore,

L_e=l

L_{e = 2500mm}

L_{e = 2500mm}

 Step 5:Calculation of radius of gyration

K= √(I/A)

K = √(0.074X10⁶ /452.38)

K = 12.789 mm

Step 6:Calculation of crippling load by Rankine's formula

P=  (σ_c A)/(1+a(L_e/k)²)

P = ((350) ₍452.38))/(1+(1/7500)(2500/12.789)²)
p= 965.04KN 

Step 7: Calculation of crippling load by Euler's formula
P = Π ²E I/ L^{2
P = Π 2 (2.05X105 )0.074X106 / (2500)2}
p=23.95KN

5. Design the section of circular cast iron column that can safely carry a load of 1000KN. The length of the column is 6m. Rankine’s constant is 1/1600, FOS 3.One end is fixed and other is free. Critical stress is 560Mpa.

Step 1: Data:
safe load= 1000KN
length= 6m= 6000mm
Rankine's constant =a = 1/1600
σ_{c =} 560Mpa
FOS =  3

Step 2: Calculation of crippling load 

FOS = Crippling load / safe load

3= Crippling load / 1000

Crippling load = 3000KN

Step 3:  Calculation of radius of gyration

K= √(I/A)

K = √(πd⁴/64 /πd²/4)

K = d/4

Step 4:Calculation of Effective length
condition=.One end is fixed and other is free
therefore,

L_e=l / √2

L_{e =6000/}√ ₂

L_{e =4242.64 mm}

Step 5:Calculation of diameter  by Rankine's formula

P=  (σ_c A)/(1+a(L_e/k)²)

3000X1000 = ((560) ₍πd²/4))/(1+(1/1600)(_4242.64/(d/4))²)
solving the above 
d = 89.69mm

6. A column with circular section of 20mm diameter is hinged at its both ends. The column with two different lengths is tested under buckling load resulting

Length in mm	Buckling load (KN)
300	60
400	47

Determine the Rankine’s constant and crushing stress for the material of the column.

Step 1: Calculation of Area of cross section 

A= πd²/4

   =  π(20)²/4

   = 314 mm²

Step 2: Calculation of  Moment of inertia

I= πd⁴/64

   =π(20)⁴/64

   = 7853.98 mm⁴

Step 5:Calculation of radius of gyration

K= √(I/A)

K = √(7853.98 /314)

K = 5 mm

Step 6:Calculation of crippling load by Rankine's formula
case1

P=  (σ_c A)/(1+a(L_e/k)²)

60X1000= ((σ_c) ₍314))/(1+(a)(300/5)²)

^{60000+ (216X106 ) a = 314} σ_c

_{Case 2 :}

P=  (σ_c A)/(1+a(L_e/k)²)

^{47000 =} ((σ_c) ₍314))/(1+(a)(300/5)²)

47000^{+ (300.8X106 ) a = 314} σ_c

a= 1/5000=0.0002

σ_{c = 286.538 N/mm 2}

7. A hollow cast iron circular section column is 7.5mm long and pinned at both the end. The inner diameter of the column is 160mm and thickness of the wall is 20mm. Find the safe load by Rankine’s formula, using FOS of 5.Also find the slenderness ratio and ratio of the Euler’s and Rankine’s critical loads. For cast iron take σc= 550N/mm2 and a=1/1600 and E=8X104N/mm2.

Step 1: Data:
Thickness = 20mm
d= internal diameter = 160mm
D = external diameter =D=d+2t=160+2(20)=200mm
Rankine's constant=a=1/1600
length of column= 7.5m = 7500mm
condition=hinged at both the ends
crippling load =??
FOS=5
_{critical stress =}  σc=550N/mm2

E=8X104N/mm2

Step 2: Calculation of the area of the column
A= π/4(D²-d²)
A= π/4(200²-160²)
A =11309.73mm²

Step 3 :Calculation of the moment of inertia  of the column
I = π/64(D⁴-d⁴)
I = π/64(200⁴-160⁴)
I = 46.36X10⁶ mm⁴


Step 4:Calculation of Effective length
condition=both of its ends are fixed
therefore,

L_e=l

L_{e = 7500mm}

L_{e = 7500mm}


 Step 5:Calculation of radius of gyration

K= √(I/A)

K = √( 46.36X10⁶ /11309.73)

K = 64.03 mm

Step 6:Calculation of crippling load by Rankine's formula

P=  (σ_c A)/(1+a(L_e/k)²)

P = ((550) ₍11309.73))/(1+(1/1600)(7500/64.03)²)
pr = 649 KN

Step 7: Calculation of crippling load 

FOS = Crippling load / safe load

5 = 649  / safe load

safe load=  129.8 KN


Step 8: Calculation of slenderness ratio
slenderness ratio = l/k
                            = 7500/ 64.03
                            = 117.10

Step 9: Calculation of crippling load by Euler's formula
P = Π ²E I/ L^{2
P = Π 2 (}8X104 ⁾46.36X10^6/ (7500)2
pe=650.085KN

step10 : Calculation of ratio of euler's load to bucking load

Pe/Pr = 650.085/649
          = 1.0017

8. A hollow circular section 2.8m long is fixed at one end and hinged at other end. External diameter is 150mm and thickness of the wall is 15mm.Rankine’s constant=1/1600 and σc= 550Mpa.Comparwe the buckling loads obtained by using Euler formula and Rankine’s formula. Also find the length of the column for which both formulas gives the same load. Take E=80GPa.

9. A column as shown in the figure below consists of three plates, each of thickness “t” welded together. It carries a axial load of 400 KN over an effective length of 4m.Taking σc= 320MPa, a=1/7500 and FOS= 2.5, determine the value of t.

10. The following particulars are given below
a) Diameter of the cylinder =400mm
b) Steam pressure in cylinder =0.6N/mm2
c) Distance between the piston and cross head = 1.25m.
Find the diameter of the piston rod allowing a FOS of 4.Assume that the piston is firmly fixed to the piston and the cross head. Take σc= 330N/mm2 and a=1/7500