Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition

Name: Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition
Brand: Academic Documents Platform
Price: 16.99 USD
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Author: Charlotte Garcia

Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition is the ultimate guide for understanding and solving textbook problems.

Charlotte Garcia

Contributor

4.9

7 months ago

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Chapter 1
1.1
(a) Using two elements
( ) 0.25 0.0125A y y 
2
1 0.25 inA  2
2 0.1875 inA 
2
3 0.125 inA 
1( )
2
i i
eq
A A E
K  

6
1
(0.25 0.1875)(10.4 10 ) lb
455,000
2(5) in
K  
 
6
2
(0.1875 0.125) (10.4 10 ) lb
325,000
2(5) in
K  
 
3
455
10
1
455
0
455
1
2
0
3
1 2 3
0
455 325 325 0
1000
325 325
0, 0.002197 in 0.005274 in
u
u
u
u u u
                
     
      
  
(b)
2
1 0.25 inA 
2
2 0.234375 inA 
2
3 0.21875 inA 
2
4 0.203125 inA 
2
5 0.1875 inA 
2
6 0.171875 inA 
2
7 0.15625 inA 
2
8 0.140625 inA 
2
9 0.125 inA 

1.1 Cont.
1
lb
2015000 in
K  2
lb
1885000 in
K  3
lb
1755000 in
K 
4
lb
1625000 in
K  5
lb
1495000 in
K  6
lb
1365000 in
K 
7
lb
1235000 in
K  8
lb
1105000 in
K 
2015000
1
2015000
0
2015000 0
2015000 1885000
1885000
1885000
1885000 1755000
1755000
1755000
1755000 1625000
1625000
1625000
1625000 1495000
1495000
1495000
1495000 1365000
1365000
1365000
1365000 1235000
1235000
1235000
1235000 11050


 

 

 

 

 

 
1
2
3
4
5
6
7
8
9
0
0
0
0
0
0
0
0
1000
1105000
00
1105000 1105000
u
u
u
u
u
u
u
u
u
 
 
 
 
 
 
     
     
     
     
     
     
                                          
 
 
 
 
 
  
1 0,u  2 0.00049628,u  3 0.0010268,u  4 0.0015965,u  5 0.0022119,u 
6 0.0028808,u  7 0.0036134,u  8

1.2
avg.A E
K 
6
1 5
10.5 12 (3)(3.27 10 )
2
6
K K
   
  
1 5
lb
18,393,750 in
K K 
6
2 4
10.5 9 (3)(3.27 10 ) lb2 15,941,250
6 in
K K
   
   
6
3
(9)(3)(3.27 10 ) lb
22, 072,500
4 in
K 
 
18393750
1
18393750
0
18393750 10
2
3
4
5
6
18393750 15941250
15941250
15941250
15941250 22072500
22072500
22072500
22072500 15941250
15941250
15941250
15941250 18393750
18393750
18393750 18393750
u
u
u
u
u
u
 
 
 
            
   
 
 
  
 
  
0
0
0
0
0
500
  
  
  
      
   
   
   
    
5
5
5
4
4
0
2.71831 10
5.85483 10
{ } in
8.12009 10
1.12566 10
1.39749 10
u





 
 
  
 
  
  
  
 
  
   
5 5 6
(3) ( 8.12009 10 5.85483) 10 (3.27 10 )
4

 
    

1.2 Cont.
as a check:
(3) 500 18.5 Psi
(9)(3)
F
A
   
1( )i iu u E
  

5 6
(1) ( 2.71831 10 0)(3.27 10 ) 14.8 Psi C
6


   
 
4 6
(5) ( 1.39749 1.12566) 10 (3.27 10 ) 14.8 Psi C
6


    
 
as a check:
(1) (5) 500 14.8 Psi C
10.5 12 (3)
2
F
A
    
 
 
 
    5 6
2 5.85483 2.71831 10 (3.27 10 ) 17.1 Psi C
6


    
 
4 5 6
(4) ( 1.12566 10 8.12009 10 ) (3.27 10 ) 17.1 Psi C
6

 
    
 
as a check:
(2) (4) 500 17.1 Psi
10.5 9 (3)
2
F
A
    
 
 
 

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Solution Manual for Finite Element Analysis: Theory and Application with ANSYS, 5th Edition

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