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 the area of the column
A= π/4(D2-d2)
A= π/4(502-402)
A = 706.85 mm2
Step 3: Calculation 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,
Step 4:Calculation of Effective length
condition=both of its ends are fixed
therefore,
4. Find the Euler’s crippling load for a hollow cylindrical steel column of 40mm external diameter
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 the area of the column
A= π/4(D2-d2)
A= π/4(502-402)
A = 706.85 mm2
Step 3: Calculation 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,
Le=l / 2
Le = 3000/ 2
Le = 1500mm
Step 5:Calculation of the radius of gyration
K= √(I/A)
K = √(181.04X103 /706.85)
K = 16.00 mm
Step 6:Calculation of crippling load
P= (σc A)/(1+a(Le/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 and take the 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 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 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,
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 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 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,
Le=l / 2
Le =4000/ 2
Le =2000mm
Step 5:Calculation of the radius of the 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= (σc A)/(1+a(Le/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
d = 0.8 D = 109.06 mm
3. A hollow cast iron column with an external diameter of 250mm and an internal diameter of 200mm is 10m
long with both ends fixed. Find the safe axial load with a 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
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 the area of the column
A= π/4(D2-d2)
A= π/4(2502-2002)
A = 17671.45 mm2
Step 3: Calculation the moment of inertia of the column
I = π/64(D4-d4)
I = π/64(2504-2004)
I = 113.20X106 mm4
A= π/4(D2-d2)
A= π/4(2502-2002)
A = 17671.45 mm2
Step 3: Calculation 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,
Le=l / 2
Le =10000/ 2
Le =5000mm
Step 5:Calculation the radius of the gyration
Step 7: Calculation of crippling load
FOS = Crippling load / safe load
4= 33.44 / safe load
safe load= 8.36MN
Step 5:Calculation the radius of the gyration
K= √(I/A)
K = √(113.20X106/17671.45)
K = 80.03 mmStep 6:Calculation of crippling load
P= (σc A)/(1+a(Le/k)2)
P = ((550) (17671.45))/(1+(1/1600)(5000/80)2)
P = 33.44MN
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 ends. Also, compute
Rankine’s crippling load using constants 350MPa and 1/7500. Take E=205GPa.
5. Design the section of circular cast iron column that can safely carry a load of 1000KN.
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
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 the area of the column
A= π/4(D2-d2)
A= π/4(402-322)
A =452.38mm2
Step 3: Calculation 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,
Step 2: Calculation the area of the column
A= π/4(D2-d2)
A= π/4(402-322)
A =452.38mm2
Step 3: Calculation 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,
Le=l
Le = 2500mm
Le = 2500mm
Step 5:Calculation the radius of the gyration
Step 5:Calculation the radius of the gyration
K= √(I/A)
K = √(0.074X106 /452.38)
K = 12.789 mmStep 6:Calculation of crippling load by Rankine's formula
P= (σc A)/(1+a(Le/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/ L2
P = Π 2 (2.05X105 )0.074X106 / (2500)2
p=23.95KN
p= 965.04KN
Step 7: Calculation of crippling load by Euler's formula
P = Π 2E I/ L2
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 the other is free.
Critical stress is 560Mpa.
Step 1: Data:
safe load= 1000KN
length= 6m= 6000mm
Rankine's constant =a = 1/1600
σc = 560Mpa
FOS = 3
FOS = Crippling load / safe load
3= Crippling load / 1000
Crippling load = 3000KN
Step 4:Calculation of Effective length
condition=.One end is fixed and the other is free
therefore,
6. A column with a circular section of 20mm diameter is hinged at both ends.
Step 1: Data:
safe load= 1000KN
length= 6m= 6000mm
Rankine's constant =a = 1/1600
σc = 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 the gyration
K= √(I/A)
K = √(πd4/64 /πd2/4)
K = d/4Step 4:Calculation of Effective length
condition=.One end is fixed and the other is free
therefore,
Le=l / √2
Le =6000/√ 2
Le =4242.64 mm
Step 5:Calculation of diameter by Rankine's formula
P= (σc A)/(1+a(Le/k)2)
3000X1000 = ((560) (πd2/4))/(1+(1/1600)(4242.64/(d/4))2)
solving the above
d = 89.69mm
solving the above
d = 89.69mm
The column with two different lengths is tested under buckling load resulting
Determine the Rankine’s constant and crushing stress for the material of the column.
Step 1: Calculation of Area of the cross-section
7. A hollow cast-iron circular section column is 7.5mm long and pinned at both ends.
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 the 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 the radius of the gyration
K= √(I/A)
K = √(7853.98 /314)
K = 5 mm
Step 6:Calculation of crippling load by Rankine's formula
case1
case1
P= (σc A)/(1+a(Le/k)2)
60X1000= ((σc) (314))/(1+(a)(300/5)2)
60000+ (216X106 ) a = 314 σc
Case 2 :
P= (σc A)/(1+a(Le/k)2)
47000 = ((σc) (314))/(1+(a)(300/5)2)
47000+ (300.8X106 ) a = 314 σc
a= 1/5000=0.0002
σc = 286.538 N/mm 2
7. A hollow cast-iron circular section column is 7.5mm long and pinned at both ends.
The inner diameter of the column is 160mm and the thickness of the wall is 20mm.
Find the safe load by Rankine’s formula, using a 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
8. A hollow circular section 2.8m long is fixed at one end and hinged at the other ends.
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 the area of the column
A= π/4(D2-d2)
A= π/4(2002-1602)
A =11309.73mm2
Step 3: Calculation 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,
Step 2: Calculation the area of the column
A= π/4(D2-d2)
A= π/4(2002-1602)
A =11309.73mm2
Step 3: Calculation 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,
Le=l
Le = 7500mm
Le = 7500mm
Step 5:Calculation the radius of the gyration
Step 5:Calculation the radius of the gyration
K= √(I/A)
K = √( 46.36X106 /11309.73)
K = 64.03 mmStep 6:Calculation of crippling load by Rankine's formula
P= (σc A)/(1+a(Le/k)2)
P = ((550) (11309.73))/(1+(1/1600)(7500/64.03)2)
pr = 649 KN
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/ L2
P = Π 2 (8X104 )46.36X106/ (7500)2
pe=650.085KN
step10: Calculation of the ratio of Euler's load to bucking load
Pe/Pr = 650.085/649
= 1.0017
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/ L2
P = Π 2 (8X104 )46.36X106/ (7500)2
pe=650.085KN
step10: Calculation of the 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 the other ends.
The external diameter is 150mm and the thickness of the wall is 15mm. Rankine’s constant=1/1600
and σc= 550Mpa.Compare the buckling loads obtained by using the Euler formula and Rankine’s formula.
Also, find the length of the column for which both formulas give 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.
9. A column as shown in the figure below consists of three plates, each of thickness “t” welded together.
It carries an 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.
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 crosshead.
Take σc= 330N/mm2 and a=1/7500
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