Numerical Problems on Rankine’s formula.

 

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,

L_e=l / 2

L_{e = 3000/ 2}

L_{e = 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=  (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 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,

L_e=l / 2

L_{e =4000/ 2}

L_{e =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=  (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 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

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

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 the radius of the 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 ends.  Also, compute

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 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,

L_e=l

L_{e = 2500mm}

L_{e = 2500mm}

 Step 5:Calculation the radius of the 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 the 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 the gyration

K= √(I/A)

K = √(πd⁴/64 /πd²/4)

K = d/4

Step 4:Calculation of Effective length
condition=.One end is fixed and the 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 a circular section of 20mm diameter is hinged at 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 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

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 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

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,

L_e=l

L_{e = 7500mm}

L_{e = 7500mm}


 Step 5:Calculation the radius of the 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 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.

 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. 

Assume that the piston is firmly fixed to the piston and the crosshead. 

Take σc= 330N/mm2 and a=1/7500

Er. SP.ASWINPALANIAPPAN., M.E.,(Strut/.,)

Structural Engineer

http://civilbaselife.blogspot.com

LATERAL STRAIN, LONGITUDINAL STRAIN AND POISSON'S RATIO

 

LATERAL STRAIN, LONGITUDINAL STRAIN AND POISSON'S RATIO

Longitudinal strain: 

Whenever the bar is subjected to the axial load, there will be an increase in the length of the

 the bar along the direction of loading. Therefore the longitudinal strain is defined as the ratio

of increase in the length of the bar in the direction of the applied load to that of the original

 length (gauge length).

i.e, e = dL/L

where

e= longitudinal strain

dl= increase in length

L = gauge or original length

Lateral strain: Whenever the bar is subjected to the axial load, there will be a decrease

 in the dimensions of the bar in the perpendicular direction of loading. Therefore lateral 

strain is defined as the ratio of decrease in the length of the bar in the perpendicular direction 

of applied load to that of the original length (gauge length).

i.e, e = dB/B or dD/D

where

e= lateral strain

dd= decrease in depth

D= gauge or original depth

db= decrease in breadth

B = gauge or original breadth

Poisson’s ratio:  The ratio of lateral strain to that of the longitudinal strain is termed Poisson's ratio 

and it is represented by ϻ or 1/m.

i.e, ϻ or 1/m =  lateral strain/longitudinal strain

The value of the Poisson’s ratio for most materials lies between 0.25 and 0.33.

Er. SP.ASWINPALANIAPPAN., M.E.,(Strut/.,)

Structural Engineer

http://civilbaselife.blogspot.com

Pure Bending

 

Pure Bending

PURE BENDING

FIGURE  1

FIGURE  2

whenever the structure is applied with some magnitude of load, then there is a resultant of shear force 

and a couple simultaneously. Considering the above figures we can observe that some parts of the beam  

resulted in negligible or zero shear force but at the same part of the structure, the bending moment is

maximum or constant. Therefore, pure bending is the condition of stress where the bending moment is

maximum or constant simultaneously the shear force or the torsional force is zero or negligible.

Er. SP.ASWINPALANIAPPAN., M.E.,(Strut/.,)

Structural Engineer

http://civilbaselife.blogspot.com

Equilibrium of Forces

 

Equilibrium of Forces

Whenever a body (static or dynamic) is subjected to external force then it tends to remain in the same state as it was in the former case, so that the algebraic sum of horizontal, vertical forces and moments is equal to zero, as a result, the condition is termed to be in equilibrium.

Conditions of Equilibrium,

1) The algebraic sum of all the horizontal forces or forces in the X direction

 is equal to zero .i.e, ΣH=0 or ΣX=0

2) The algebraic sum of all the vertical  forces or forces in the Y direction

 is equal to zero .i.e, ΣV=0 or ΣY=0

3) The algebraic sum of all the moments for the system  is equal to zero 

.i.e, ΣM=0

Note:
ΣH or ΣV = 0 and ΣM_x=0 or ΣM_y=0 (In case of one dimensional system)
ΣH , ΣV = 0 and ΣM_x=0 , ΣM_y=0 (In case of two dimensional system)
ΣH , ΣV, ΣZ = 0 and ΣM_x=0 , ΣM_y=0 ,ΣM_z=0(In case of three dimensional system)

Er. SP.ASWINPALANIAPPAN., M.E.,(Strut/.,)

Structural Engineer

http://civilbaselife.blogspot.com

VOLUMETRIC STRAIN

 

VOLUMETRIC STRAIN

Volumetric strain: Volumetric strain of a deformed body is defined as the ratio of the

 change in volume of the body to the deformation to its original volume. If V is the 

original volume and dV is the change in volume that occurred due to the deformation, 

the volumetric strain EV induced is given by

                                                                      

  e_v=dV/V 

Consider a uniform rectangular bar of length l, breadth b and depth d as shown in the figure.

Its volume V is given by,

This means that the volumetric strain of a deformed body is the sum of the linear 

strains in three mutually perpendicular directions.

Er. SP.ASWINPALANIAPPAN., M.E.,(Strut/.,)

Structural Engineer

http://civilbaselife.blogspot.com