Numericals on Columns and struts_-_solved

Strut and Column Problems with their solution

1. 80
CHAPTER 9
Columns and Struts
Problem 1. Compare the ratio of the strength of solid steel column to that of the hollow steel column of the
same cross-sectional area. The internal diameter of the hollow column is 3/4th of the external diameter. The
columns have the same length and are pinned at both ends. Use Euler’s theory.
Solution: Let ‘d’ be the diameter of the solid strut and D be the outer diameter of the hollow strut.
As the cross-sectional areas are same,

2
4
d
π
=
2
23
4 4
D
D
 
π  
−  
  

d
2
=
2 2 29 7
16 16
D D D− =
Let P
crh
= buckling load of the hollow column
P
crs
= buckling load of solid column
I
h
= least moment of inertia of hollow column
=
4
4 43 175
64 4 64 256
D
D D
 
π π  
− = ×  
    

I
s
= Least moment of the solid column
=
4
64
d
π

Now P
crh
=
2
2
h
EI
L
π
and P
crs
=
2
2
s
EI
L
π

∴
crh
crs
P
P
=
4
4
175
64 256
64
h
s
D
I
I
d
π
×
=
π
×

=
4 4
4 2
2
175 175
256 7
256
16
D D
d
D
=
 
 
 
=
25
7

Thus
crh
crs
P
P
= 3.571. Ans.
Problem 2. A solid round bar of 60 mm diameter and 2.5 m long is used as a strut. Find the safe
compressive load for the strut using Euler’s formula if (a) both ends are hinged (b) both ends are fixed.
Take E = 2 × 10
5
N/mm
2
and factor of safety = 3.
Solution: l = 2.5 m = 2500 mm
d = 60 mm, E = 2 × 10
5
N/mm
2

I =
4 4
60
64 64
d
π π
× = × = 636172.51 mm
4

2. 81
Factor of safety = 3.
(a) Both ends are hinged:
P
cr
=
2 2 5
2 2
2 10 636172.51
(2500)
EI
l
π π × × ×
=
= 200920 N
= 200.92 kN
∴ Safe load = 200.92
Factor of safety 3
crP
= = 66.97 kN.
(b) Both ends are fixed:
P
cr
=
2 2 5
2 2
4 4 2 10 636172.51
(2500)
EI
l
π × π × × ×
=

= 803682 N = 803.682 kN
∴ Safe load = 803.682
Factor of safety 3
c r
P
=
= 262.89 kN. Ans.
Problem 3. What is the ratio of the strength of a solid steel column of 150 mm diameter to that of a hollow
circular steel column of the same cross-sectional area and a wall thickness of 15 mm? The two columns
have the same length and similar end conditions.
Solution: Diameter of circular column d = 150 mm
∴ C.S. Area =
2
150
4
π
×
Let the thickness of circular hollow column be t = 15 mm
Let external diameter of hollow circular column be D = mm
∴ Its internal diameter = D – 2t = D – 2 × 15 = (D – 30) mm
∴ C.S. area =
4
π
{D
2
– (D – 30)
2
}
This area is same as that of solid column
∴
2 2
{ ( 30) }
4
D D
π
− − =
2
150
4
π
×
D
2
– {D
2
– 60D + 900} = 150
2

60 D = 22500 + 900 = 23400
∴ D = 390 mm
∴ Internal diameter of hollow column = 390 – 30 = 360 mm.
Least moment of inertia:
I
s
=
4
150
64
π
× = 24850488.7 mm
4

I
h
=
4 4
(390 360 )
64
π
− = 311128119.5 mm
4

P
crh
=
2
2
h
e
EI
l
π

P
crs
=
2
2
s
e
EI
l
π

3. 82
∴
crh
crs
P
P
=
311128119.5
24850488.7
h
sI
I
=
= 12.52. Ans.
Problem 4. Find the Euler’s crushing load for a hollow cylindrical cast iron column 120 mm external
diameter and 20 mm thick, if it is 4.2 m long and is hinged at both ends. Take E = 80 kN/mm
2
. Compare
this load with the crushing load as given by Rankine’s formula using constants f
c
= 550 N/mm
2
and
a = 1/1600. For what length of strut does the Euler’s formula cease to apply?
Solution: External diameter = 120 mm
Thickness = 20 mm
Internal diameter = 120 – 2 × 20 = 80 mm
Least moment of inertia =
4 4
(120 80 )
64
π
−
= 8168140.89 mm
4

Column is hinged at both ends.
∴ l
e
= 4.2 m = 4200 mm
Euler’s buckling load =
2 2 3
2 2
80 10 8168140.89
(4200)
EI
l
π π × × ×
=

= 365606.89 N . Ans.
A =
2 2
(120 80 )
4
π
− = 6283.18 mm
2

K
2
=
8168140.89
1300
6283.18
I
A
= =

∴ K
2
= 36.05 mm
∴ Rankine’s critical load P
R
=
2 2
550 6283.18
1 4200
1 1
1600 36.05
cf A
l
a
K
×
=
   
+ +
   
   
= 364415.16 N
∴
E
R
P
P
=
365606 89
364415 16
.
.
= 1.003. Ans.
Now, P
E
=
2
2
EI
l
π

Equating it to crushing load, we get

2
2
EI
l
π
= f
c
A

2
2
2
E
K
l
π
= f
c


2 3
2
80 10 1300
l
π × × ×
= 550
l
2
=
2
80 1000 1300
550
π × × ×

l = 1366.108 mm. Ans.

4. 83
Problem 5. An ISLB 300 section is provided with a flange plate 200 mm × 12 mm for each flange. The
composite member is used as a column with one end fixed and the other end hinged. Calculate the length of
the column for which, crippling loads given by Rankine’s formula and Euler’s formula will be the same.
Take E = 210 kN/mm
2
, f
c
= 330 N/mm
2
, a = 1/7500
Properties of ISLB 300 section are:
Overall width = 150 mm,
Overall depth = 300 mm,
Thickness of flange = 9.4 mm,
Thickness of web = 6.7 mm
I
xx
= 73.329 × 10
6
mm
4

I
yy
= 3.762 × 10
6
mm
4

A = 4808 mm
2
.
Solution: f
c
= 330 N/mm
2
, a =
1
,
7500
E = 210 × 10
3
N/mm
2

Area A = 4808 mm
2

I
xx
= 73.329 × 10
6
mm
4
, I
yy
= 3.762 × 10
6
mm
4

Sectional area of ISLB 300 column.
A = 4808 + 2 × (200 × 12) = 9608 mm
2

Moment of inertia about x – x axis.
I
xx
= 73.329 × 10
6
+ 2
3
2200 12
(200 12) 156
12
 ×
+ × 
  

= 190199406 mm
4

Moment of inertia about y-y axis.
I
yy
=
3
6 12 200
3.762 10 2
12
 ×
× +  
 
 
= 19762000 mm
4

Since I
yy
< I
xx
, the column buckles about y-y axis.
∴ I = I
min
= 19762000 mm
4

Least radius of gyration = K =
19762000
9608
I
A
=

= 45.35 mm.
Let L = effective length

2
2
EI
L
π
=
2
2
1
c
f A
L
a
K
+


2 3
2 2
210 10 19762000 330 9608
1
1
7500 2056.82
L L
π × × × ×
=
+

12918229.65 =
2
2 8
1 6.4825 10
L
L
−
+ × ×

12918229.65 + 0.8374 L
2
= L
2

∴ L = 8914 mm

5. 84
Therefore, required actual length for one end hinged and other end fixed column for which critical load
by Rankine’s formula and Euler’s formula will be equal is
l = 2 2 8914L= × = 12606 mm
= 12.606 m. Ans.
Problem 6. A built up steel column, 8 m long and ends firmly fixed is having cross-section as shown in
Fig. 1. The properties of I-section are Area = 9300 mm
2
, I
xx
= 3 × 10
6
min
4
, I
yy
= 8.4 × 10
6
mm
4
. Determine:
(i) The safe axial load the column can carry with a factor of safety of 3.5 using
( a) Euler’s Formula, (b) Rankine’s Formula.
(ii) The length of the column for which both formulae give the same crippling load.
(iii) The length of the column for which the Euler’s formula ceases to apply.
Take E = 2 × 10
5
N/mm
2
, f
c
= 330 N/mm
2
, a = 1/7500

Fig. 1
Solution: (i) Length of the column = 8 m = 8000 mm
Factor of safety = 3.5, f
c
= 330 N/mm
2
,
a =
1
7500
, E = 2 × 10
5
N/mm
2

A = 2 (9300 + 350 × 25) = 36100 mm
2

Moment of inertia of column section about x-x axis:
I
xx
= 2 × 3 × 10
6
+ 2
3
2350 25
350 25 237.5
12
 ×
+ × × 
  

= 994020833.5 mm
4

Moment of inertia of the column section about y-y axis:
I
yy
= 2 × 8.4 × 10
6
+ 2
3
225 350
9300 100
12
 ×
+ × 
  

= 381445833.3 mm
4

I
yy
< I
xx

∴ I = I
min
= 381445833.3 mm
4

K =
381445833.3
36100
I
A
=

= 102.793

6. 85
Since column is fixed at both ends,
P
E
=
2 2 5
2 2
4 4 2 10 381445833.3
(8000)
EI
l
π × π × × ×
=

= 47058993.44 N
∴ Safe Load P = 47058993.44
3.5 3.5
E
P
=
= 13445.40 × 10
3
N
= 13445.4 kN. Ans.
(b) P
R
=
2
1
cf A
L
a
K
 
+
 
 
where L =
8000
2 2
l
=
= 4000 mm
=
2
330 36100
1 4000
1
7500 102.793
×
 
+
 
 

= 9911.806 × 10
3
N
= 9911.806 kN.
∴∴∴∴ Safe load P =
9911.806
3.5
= 2831.95 kN. Ans.
(ii) Let L
1
be the effective length

2
2
1
EI
L
π
=
2
1
2
1
c
f A
L
a
K
+


2 5
2
1
2 10 381445833.3
L
π × × ×
=
2
1
2
330 36100
1
1
7500(102.793)
L
×
+

or
2
1
0.797 63203550.35L+ =
2
1
L
or L
1
= 17670 mm
= 17.67 m. Ans.
(iii) Let ‘l’ be the length of column for which Euler’s formula ceases to apply. Then
P
E
=
2
2
4
EI
l
π

f
c
A =
2
2
4
EI
l
π

330 =
2 2 2 5 2
2 2
4 4 2 10 (102.793)EK
l l
π π × × ×
=

∴ l
2
= 252815021.4
l = 15900 mm
= 15.9 m. Ans.
Problem 7. Find the safe load carrying capacity of column given in Problem 6 by Indian Standard code
procedure.
Given f
c
= 250 N/mm
2
.

7. 86
Solution: I = 381445833.3 mm
4

A = 36100 mm
2

K =
4
I
= 102.793 mm
Slenderness ratio λ =
4000
102.793
l
k
=
= 38.914
From I.S. table, λ = 30 and f
c
= 250 N/mm
2

σ
ac
= 145 N/mm
2

and for λ = 40, f
c
= 250 N/mm
2

σ
ac
= 139 N/mm
2

Linearly interpolating between these two values of λ,
σ
ac
=
8.914
145 (145 139)
10
− −

= 139.65 N/mm
2

Therefore, safe load carrying capacity of the column is,
P = σ
ac
× A = 139.65 × 36100
= 5041444 N
P = 5041.444 kN. Ans.