# Chapter 16 : Columns¶

## Example 16.1 Page No : 417¶

In [1]:
import math

# Variables
E = 10.*10**6 			#psi
ys = 6000. 			#psi

# Calculations
lbyr = math.sqrt(math.pi**2*E/ys)

# Results
print  'Slenderness Ratio = %.f '%(lbyr)
Slenderness Ratio = 128

## Example 16.2 Page No : 417¶

In [3]:
import math

# Variables
fs = 3.
W = 50. 			#kips
l = 20. 			#ft
E = 30.*10**6 			#psi

# Calculations
Pcr = fs*W
I = Pcr*10**3*(l*12)**2/(math.pi**2*E)
r = 2.01
lbyr = l*12/r

# Results
print  'Required I  = %.1f in**4'%(I)
print  'slenderness ratio =%.f '%(lbyr)
Required I  = 29.2 in**4
slenderness ratio =119

## Example 16.3 Page No : 418¶

In [5]:
import math

# Variables
L1 = 18. 			#ft
L2 = 9. 			#ft
I1 = 12.1 			#in**4
I2 = 1.2 			#in**4
E = 30.*10**6 			#psi

# Calculations
r1 = 2.05
lbyr = L1*12/r1
r2 = 0.65
lbyr2 = L2*12/r2
Pcr1 = math.pi**2*E*I1/(L1*12)**2
Pcr2 = math.pi**2*E*I2/(L2*12)**2
P = Pcr1/2.5
P2 = Pcr2/2.5

# Results
print  'Design load of 1 = %.2f lb'%(round(P,-2))
print  'Design load of 2 =%.2f lb'%(round(P2,-2))
Design load of 1 = 30700.00 lb
Design load of 2 =12200.00 lb

## Example 16.4 Page No : 419¶

In [1]:
import math

# Variables
E = 30.*10**6
syp = 30000. 			#psi
I = 143.5 			#in**4
A = 7.32 			#in

# Calculations
I1 = 2*I
A1 = 2*A
L = math.sqrt(2*math.pi**2*E*I1/(syp*A1))

# Results
print  'Critical length of the column = %.0f in'%(L)
Critical length of the column = 622 in

## Example 16.5 Page No : 421¶

In [6]:
import math

# Variables
x = 30. 			#in
x1 = 10. 			#in
E = 30.*10**6
d = 0.5 			#in
syp = 60000. 			#psi
y1 = 8. 			#in
y2 = 2. 			#in

# Calculations
ratio = 0.8
l = x+x1
lr = ratio*l
I = (math.pi*(d)**4)/64
Pcr = math.pi**2*E*I/lr**2
scr = Pcr/(math.pi*(d/2)**2)
F = Pcr*y2/(y1+y2)

# Results
print  'Stress in the critical load = %.0f psi'%(round(scr,-1))
print  'Critical force F =%.0f lb'%(F)
Stress in the critical load = 4520 psi
Critical force F =177 lb

## Example 16.6 Page No : 423¶

In [10]:
import math

# Variables
l = 10. 			#ft
Ys = 33000. 			#psi
E = 30.*10**6
A = 13.24 			#in**4

# Calculations
r = 2.
lbyr = l*12/r
Cc = math.sqrt(2*math.pi**2*E/Ys)
fs = 5./3+3*(lbyr)/(8*Cc)+(lbyr)**3/(5*Cc**3)
Sa =((1-((lbyr)**2/(2*Cc**2)))*(Ys))/fs
Pa = Sa*A

# Results
print  'Premissible load = %.f kips'%(Pa/1000)

# note : rounding off error.
Premissible load = 212 kips

## Example 16.7 Page No : 425¶

In [1]:
import math
from sympy import Symbol,solve

# Variables
L = 12. 			#ft
Po = 100. 			#kips
e = 2. 			#ft
ys = 42000. 			#psi
A = 11.77 			#in**2
rmin = 195. 			#in
Zmin = 11.0 			#in**3
lbyr = 74.2
stress = 18. 			#ksi

# Calculations
#P = (stress-(Po/A)/((1/A)+((e*12)/Zmin)))
P = Symbol("P")
eq = (((Po+P)/A) + (P*(e*L)/11))/stress - 1
P = solve(eq,P)[0]

# Results
print  'Additional Load = %.2f kips '%(P)

## Example 16.8 Page No : 425¶

In [12]:

# Variables
l = 15. 			#ft
Al = 80. 			#kips
El = 60. 			#kips
Ys = 33. 			#ksi
e = 4. 			#in

# Calculations
A = 14.4
rmin = 2.54
Zxx = 54.6
lbyr = l*12/rmin
Smax = ((Al+El)/A)+El*e/Zxx

# Results
print  'Maximum stress %.1f ksi'%(Smax)
print ("10 WF 49 is the suitable one")
Maximum stress 14.1 ksi
10 WF 49 is the suitable one