Transcript LECTURE # 13 - KFUPM Open Courseware :: Homepage

Chapter 5
Deflection and Stiffness
A. Aziz Bazoune
7/18/2015
Chapter 5
1
Outline
1.
2.
3.
4.
5.
6.
7.
Spring Rates
Deflection in Tension, Compression & Torsion
Deflection due to Bending
Strain Energy
Castigliano’s Theorem
Statically Indeterminate Problems
Compression Members
Long Columns with Central Loading
Intermediate Length Columns with Central Loading
7/18/2015
Chapter 5
2
Lec. 14
 Solved Problems
7/18/2015
Chapter 5
3
7/18/2015
Chapter 5
4
7/18/2015
Chapter 5
5
7/18/2015
Chapter 5
6
7/18/2015
Chapter 5
7
Problem
A column is pinned at both ends has a length of 4.50 ft and a solid circular
cross-section of 1.25 in diameter. If it is made of steel AISI 1020 CD what
would be a safe column loading?
Solution
 Material AISI 1020 CD Steel =  Sy = 57 ksi (Table A-20)
E = 30 Mpsi (Table A-5)
 Ends: Pinned = Rounded-rounded case = C=1 (Table 5-2)
 Eq. (5-48)
1/2
 l   2 CE 

   
 k 1  S y 
2
1/2
 2    1  30  10 


3
57

10


2
6
 101.9
l
 Calculate the slenderness ratio  
k
7/18/2015
Chapter 5
8
Problem
 Calculate the slenderness ratio  l k 
l k  
l
 I A

 d
l
4
64   d 2 4 
4.5 ft  12in



 172.8
2
d 16 d 4 1.25 in / 4
l
 Since  l k    l k 1
l
== Euler’s Column
 Pcr  C E 1    30  10

 9906


2
2
(172.8)
 A  (L / k )
2
2
6
 Therefore
D2
1.252
Pcr  9906  A  9906   
 9906   
 12150 lb
4
4
7/18/2015
Chapter 5
9
Problem
Determine the critical load on a steel column having a rectangular crosssection, 12 mm X 18 mm and a length of 280 mm? It is proposed to use AISI
1040 HR Steel. The lower end of the column is inserted into a close-fitting
socket and welded securely. The upper end is pinned.
Solution
 Material AISI 1020 HR Steel 
Sy = 290 MPa (Table A-20)
E = 207 GPa (Table A-5)
 Fixed-rounded end Condition 
C=1.2 (Table 5-2)
 Radii of gyration
I xx
kxx 

A
7/18/2015
1 12  b  h3
b h
h
18


 5.196 mm
12
12
Chapter 5
10
Problem
k yy 
I yy
A

It is clear that k yy
1 12  h  b 3
bh

b
12

 3.464 mm
12
12
 kxx and  l / k    l / k yy   280 / 3.464  80.83
On the other hand we have from Eq. 5-48
1/2
 2 CE 
 l / k 1  

 Sy 
2
Since
 l / k    l / k 1
1/2
 2    1.2  207  10 


6
290  10


2
9
 130
 Johnson's Column
 Apply the formula for Johnson’s column to find the critical load on the
steel column.
7/18/2015
Chapter 5
11
7/18/2015
Chapter 5
12
Problem
2
 Sy L  1
Pcr
 Sy  

A
 2 k  CE
2
 290  10

1
6
2
 290  10  
80.8 

234

10
N/m
9
2

1
.2

2
07

1
0


6
6
 Therefore
Pcr  234  106  A  234  106  18  12  106   50,544 N
7/18/2015
Chapter 5
13
Problem
Design a rectangular section steel column having the following
specifications:
 Width of the column (w) is three times its thickness (t)
 Maximum load that could be carried by the column is 100 kN
 Material has a yield strength (Sy= 500 MPa) and a modulus of elasticity
(E=207 GPa)
 Assume end conditions constant C=1 in the weakest direction.
 Column length l=150 mm and 400 mm)
Solution
 Cross-section area is
A  w  t  3t  t  3t 2
or
t A3
7/18/2015
Chapter 5
14
Problem
 Ixx
 Iyy
I xx  1 12  bh3  1 12  3t  t 3  t 4 4  0.25t 4
I yy  1 12  hb3  1 12  t   3t 3   27 t 4 12  2.25t 4
Solution
or
A  w  t  3t  t  3t 2
t A3
7/18/2015
Chapter 5
15
Problem
7/18/2015
Chapter 5
16
http://en.wikipedia.org/wiki/Leonhard_Euler
Leonhard Euler
Portrait by Johann Georg Brucker
Born
Died
April 15, 1707(1707-04-15)
Basel, Switzerland
September 18 [O.S. September 7] 1783
St Petersburg, Russia
Residence
Prussia
Russia
Switzerland
Nationality
Swiss
Field
Mathematics and physics
Institutions
Imperial Russian Academy of Sciences
Berlin Academy
Alma mater
University of Basel
Religion
Calvinist[1]
7/18/2015
Leonhard Paul Euler (pronounced Oiler; IPA [ˈɔʏlɐ]) (April
15, 1707 – September 18 [O.S. September 7] 1783) was a
pioneering Swiss mathematician and physicist, who spent
most of his life in Russia and
Germany. He published more papers than any other
mathematician of his time.[2]
Euler made important discoveries in fields as diverse as
calculus and graph theory. He also introduced much of the
modern mathematical terminology and notation, particularly
for mathematical analysis, s
uch as the notion of a mathematical function.[3] He is also
renowned for his work in mechanics, optics, and astronomy.
Euler is considered to be the preeminent mathematician of
the 18th century and one of the greatest of all time. He is
also
one of the most prolific; his collected works fill 60–80 quarto
volumes.[4] A statement attributed to Pierre-Simon Laplace
expresses Euler's influence on mathematics: "Read Euler,
read Euler, he is the teacher (master) of us all".[5]
Euler was featured on the sixth series of the Swiss 10-franc
banknote[6] and on numerous Swiss, German, and Russian
postage stamps. The asteroid 2002 Euler was named in his
honor. He is also commemorated by the Lutheran Church on
their Calendar of Saints on May 24.
Chapter 5
17
QUESTIONS ?
7/18/2015
Chapter 5
18