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

Questions and Numerical problems on Columns and Struts

Questions and Numerical problems on Columns and Struts

1. Define column and strut with examples.
2. Differentiate between column and strut.
3. Differentiate between long column and short column.
4. Mention some of the assumptions made by Euler’s theory for long columns.
5. Define slenderness ratio.
6. List the limitations of Euler’s theory
7. Derive the expression for columns with one end fixed and other end free.
8. Derive the expression for columns with both ends hinged.
9. Derive the expression for columns with both ends fixed.
10. Derive the expression for columns with one end fixed and other end hinged
11. Derive the  expression for Rankine’s theory

Numerical problems on columns and struts

1. A solid round bar 3 m long and  5cm in diameter is used as a strut with both the ends hinged. Determine the crippling load. Take E=2x10⁵ N/mm².

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) (0.306X10^{6 ) /}  (3000)²

P = 67.11 KN

2.      A mild steel tube 4m long, 30 mm internal diameter and 4mm thick is used as a strut with both ends hinged. Find the collapsing load. Take E=2.1x10⁵N/mm².

Step 1: Data

Length of the column = 4000mm

Internal diameter of the column = 30mm

Thickness = 4mm

Condition = Both ends hinged

Crippling load =??

E=2.1x10⁵N/mm²

Step 2 : Calculation of external diameter

External diameter D= (d+2t)

D = 30 + 2(4)

D = 38mm

Step 3: Calculation of moment of inertia

I =Π (D⁴-d⁴) / 64

I= Π ((38)⁴-(30⁴)) / 64

I = 0.0625X10⁶mm⁴

Step 4: Calculation of  crippling load

Condition = Both ends hinged

P = Π ²E I/ L²

P = Π ² (2.1x10⁵) (0.0625 X 10 ⁶  ) /  (4000)²

P = 8.096 KN

3. A strut 2.5m long is 60mm in diameter. One end of the strut is fixed while its other end is hinged. Find the safe compressive load with FOS=3.5. Take E=2.1x10⁵N/mm²

Step 1: Data

Length of the column = 2500mm

Diameter of the column = 60mm

Condition = one end hinged and other end fixed

Crippling load =??

FOS=3.5

E=2.1x10⁵N/mm²

Step 2: Calculation of moment of inertia

I = π d ⁴ / 64

I= π (60)⁴ / 64

I = 0.636X10⁶mm⁴

Step 3: Calculation of  crippling load

Condition = one end hinged and other end fixed

P = 2 Π ²E I/ L²

P = 2 Π ²⁽2.1x10⁵) (0.636X10⁶) / (2500)²

P = 421.81 KN

Step 4: Calculation of safe load

Safe load = crippling load /FOS

Safe load = 421.81/3.5

Safe load = 120.52KN

4. A column of timber section 15cmx 20cm is 6m long both ends being fixed .E for timber is 17.5KN/mm², Determine

a) Crippling load

b) Safe load for the column if factor of safety=3

Step 1: Data

Length of the column = 6000mm

Width of the column = 150mm

Depth of the column = 200mm

Condition = Both ends being fixed

Crippling load =??

FOS=3

E=17.5 KN/mm²

Step 2: Calculation of moment of inertia

I _xx = bd ³ / 12

I _xx = 150 (200)³ / 12

I _xx = 100X10⁶mm⁴

I _yy = db ³ / 12

I _yy = 200 (150)³ / 12

I _yy= 56.25X10⁶mm⁴

Choose whichever is least

Therefore, I = 56.25X10⁶mm⁴

Step 3: Calculation of crippling load

Condition = Both ends being fixed

P = 4Π ²E I/ L²

P = 4Π ²⁽17.5 x1000) (56.25 X 10⁶ ) /  (6000)²

P = 1079.48 KN

Step 4: Calculation of safe load

Safe load = crippling load /FOS

Safe load = 1079.48/3

Safe load = 359.82KN

5. A solid round bar 3m long and 5cm in diameter is used as a strut, determine the crippling load. Take E=2x10⁵N/mm².

a) One end hinged and other end fixed

b) One end is fixed and other end is free.

c) Both the ends are fixed.

Step 1: Data

Length of the column = 3000mm

Diameter of the column = 50mm

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 = One end hinged and other end fixed

P = 2Π ²E I/ L²

P = 2Π ²⁽2x10⁵) (0.306X10⁶ ) /  (3000)²

P = 134.22 KN

Step 4: Calculation of crippling load

Condition = One end is fixed and other end is free.

P = Π ²E I/4 L²

P = Π ²⁽2x10⁵) (0.306X10⁶) / 4(3000)²

P = 16.778 KN

Step 5: Calculation of crippling load

Condition = Both the ends are fixed

P = 4Π ²E I/ L²

P = 4Π ²⁽2x10⁵) (0.306X10⁶ ) /  (3000)²

P = 268.45 KN

6. A simply supported beam of length 4m is subjected to a uniformly distributed load of 30KN/m over the whole span and deflects 15mm at the centre. Determine the crippling load when the beam is used as a column for the following conditions.

a) One end is fixed and other end is hinged

b) Both the ends are pin jointed.

Step 1: Data

Length of the column = 4000mm

Udl = 30KN/m

Deflection = 15mm

Crippling load =??

Step 2: Calculation of flexural rigidity

Δ =   5wL^{4   /} 384EI

15 = 5(30) (4000)⁴/384 EI

EI ⁼ 6.667X10¹² N-mm²

Step 3: Calculation of crippling load

Condition = One end hinged and other end fixed

P = 2Π ²E I/ L²

P = 2Π ²(6.667X10¹²) / (4000)²

P = 822.5 KN        

Step 4: Calculation of crippling load

Condition = Both ends pinned

P = Π ²E I/ L²

P =Π ²(6.667X10¹²) / (4000)²

P = 411.20 KN        

 

7. A solid round bar 4m long and 5cm diameter was found to extend 4.6mm under the tensile load of 50KN.This bar is used as a strut with both ends hinged. Determine the buckling load for the bar and also safe load taking factor of safety as 4.

Step 1: Data

Length of the column = 4000mm

Diameter = 50mm

Extension = 4.6mm

Tensile load = 50KN

FOS = 4

Condition = Both ends hinged

Buckling load=??

Step 2: Calculation of strain

e = dl / l

e = 4.6/ 4000

e = 1.15X10^-3

Step 3: Calculation of Area of cross section 

A= π d²/4

   = π (50)²/4

 A = 1963mm²

Step 4: Calculation of stress

σ = P/A

σ = 50X10³ /1963

σ = 25.47 N/mm²

Step 5: Calculation of Young's modulus

E = σ / e

E = (25.47 ) / (1.15X10^-3)

E = 0.221X10⁵ N/mm²

Step 6: Calculation of Moment of inertia

I= π d⁴/64

 = π (50)⁴/64

I = 3.068X10⁵mm⁴

Step 7: Calculation of buckling load

Condition =  both  ends  hinged

P = Π ²E I/ L²

P =Π ²(0.221X10⁵) ⁽3.068X10⁵) /  (4000)²

P = 4.182 KN      

Step 8: Calculation of Safe load

Safe load = crippling load /FOS

Safe load = 4.182X1000/4

Safe load = 1.0455 KN

8. A hallow alloy tube 5m long with external and internal diameters 40mm and 25mm was found to extend 6.4mm under the tensile load of 60KN. Find the buckling load for the tube of column with both ends pinned. Also find the safe load for the tube, taking FOS=4.

Step 1: Data

Length of the column = 5000mm

External diameter = 40mm

Internal diameter =25mm

Extension = 6.4mm

Tensile load = 60KN

FOS = 4

Buckling load =??

Safe load =??

Condition = Both ends fixed

Step 2: Calculation of strain

e = dl / l

e = 6.4/ 5000

e = 1.28X10^-3 

Step 3: Calculation of Area of cross section 

A = π (D²- d²)/4

    = π ((40)²- (25)²) / 4

 A= 765.76mm²

Step 4: Calculation of stress

σ = P/A

σ = 60X10³/765.76

σ = 78.35 N/mm²

Step 5: Calculation of Young's modulus

E = σ /e

E = 78.35 / 1.28X10^-3 

E = 0.198X10⁵ N/mm²

Step 6: Calculation of  Moment of inertia

I= π (D⁴- d⁴⁾/64

  = π ((40)⁴- (25)⁴) /64

 I = 1.065X10⁵mm⁴

Step 7: Calculation of Buckling load

Condition =  both  ends  fixed

P = 4Π ²E I/ L²

P =4Π ²(0.198X10⁵) (1.065X10⁵)/ (5000)²

P = 4.182 KN   

9. Determine the ratio of the buckling strengths of two circular columns one hollow and other solid. Both the columns are made of same material and have same length, cross sectional area and end conditions. The internal diameter of the hollow column is half the external diameter.

Step 1: Data:

d= Diameter of the solid column

Di= Internal diameter of the hallow column

De= External diameter of the hallow column

Step 2:Calculation of Area of cross section of the hallow column

A_h= π (D_e²- D_i²)/4

A_h = π ((D_e)²- (D_e/2)²) /4      (Since D_i = D_e/2)

Step3: Calculation of Area of cross section of the solid column 

A_s= πd²/4

Step 4: Condition: Both the columns are made of same material and have same length, cross sectional area and end conditions

Therefore,

Ah=As

π ((D_e)²- (D_e/2)²) /4= πd²/4

d = 0.866 D_e

Step 5: General expression for the crippling load

P = Π ²E I/ L_e²

P α I

P_s/P_h=I_s/I_h

P_s/P_{h =}πd⁴/64_/ π (D_e⁴ - D_i⁴) / 64

P_s/P_{h =} 0.6

Therefore P_h/P_s = 1.67 

10. Determine the ratio of buckling strengths of one hollow and one solid column. Both are made of same material and have same length, cross sectional area and end conditions. The internal diameter of the hollow section is 2/3rd of its external diameter.

Step 1: Data:

d= Diameter of the solid column

Di= Internal diameter of the hallow column

De= External diameter of the hallow column

Step 2:Calculation of Area of cross section of the hallow column

A_h= π (D_e²- D_i²) / 4

     = π((D_e)²- ((2/3)D_e)²)/4     (Since D_i = (2/3)D_e)

Step3: Calculation of Area of cross section of the solid column 

A_s= πd²/4

Step 4: Condition: Both the columns are made of same material and have same length, cross sectional area and end conditions

Therefore,

Ah=As

= π ((D_e)²- ((2/3)D_e)²)/4  = πd²/4

=d = 0.745 D_e

Step 5: General expression for the crippling load

P = Π ²E I/ Le²

Pα I

P_s/P_h=I_s/I_h

P_s/P_{h =}πd⁴/64_/ π(D_e^{4 -} ((2/3)De)⁴) /64

P_s/P_{h =} 0.3846

Therefore Ph/Ps =2.6 

11. Determine the crippling load for the “T”section of length 5m when it is used as a strut with both ends hinged. E=2.0x105N/mm2.

Step 1: 

7. Calculate the critical load for a strut which is made of a bar circular in section and 5m long and which is pin jointed at both ends. The same bar when freely supported gives a mild span deflection of 10mm under a load of 80N at the center.

7. Find the shortest length L for a pin ended steel column having a cross section of 60mmx100mm for which Euler’s formula applies. Take E=2x105N/mm2 and critical proportionality limit is 250 N/mm2.
8. A hollow mild steel tube 6m long 4cm internal diameter and 6mm thick is used as a strut with both ends hinged. Find the crippling load and safe load taking factor of safety as 3 Take E=2x105N/mm2.




11. A hallow alloy tube 5m long with external and internal diameters 40mm and 25mm was found to extend 6.4mm under the tensile load of 60KN.Find the buckling load for the tube of column with both ends pinned. Also find the safe load for the tube, taking FOS=4.

12. Calculate the safe compressive load on a hollow cast iron with one end is built up and one end is hinged of 15mm external diameter, 10cm internal diameter and 10m in length. Take FOS of 5 and E=95KN/mm2.

13. A steel bar of rectangular section 30mmx40mm pinned at each end is subjected to axial compression. The bar is 1.75m long. Determine the buckling load and the corresponding axial stress using Euler’s formula. Determine the minimum length for which Euler’s equation may be used to determine the buckling load, if the proportionality limit of the material is 200N/mm2. Take E=2x105N/mm2.



15. Find the ratio of the strength of a solid circular column with that of a hollow circular column of equal area, whose internal diameter is two third, the external diameter. Both the columns are of same material, having same length and are hinged at their ends.

16. Determine the crippling load for an I section shown below with 5m long and built up at
both ends. E=2.1x105N/mm2

18. Calculate Euler’s critical load for a strut of T section the flange width being 10cm, overall depth 8cm and both stem and flange is 1cm thick. The strut is 3m long and is built in at both ends. Take E=2.0x105N/mm2.

20.

21. A solid round bar of 60mm diameter and 2.5m is used as a strut. Find the safe compressive load for the strut if
a) Both ends are pinned
b) Both ends are built in ,Take E=2.0x105N/mm2 and  FOS = 3

22. A column has an “I” section with equal flanges of 200mmX10mm and web 300mmX10mm.When the column is simply supported with a udl of 24KN/m ,maximum deflection at the mid span is 6mm.Determine the safe load the column can carry at its free end with its other end fixed. Take FOS as 2.5 and E=210 GPa.

23. A column of timber section is 200mmX300mm and 5m long. One end of the column is fixed and other end is free. If the young’s modulus of the timber is 17.5KN/mm2,determine
a) Crippling load
b) Safe load if FOS=2.5