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 0.0044232,u  9 0.0053281in.u 
y (in) exact defl. Two-element eight element
0 0 0 0
1.25 0.00049645 0.00049628
2.5 0.0010272 0.0010268
3.75 0.0015972 0.0015965
5.0 0.0022129 0.002197 0.0022119
6.25 0.0028822 0.0028808
7.5 0.0036154 0.0036134
8.75 0.0044259 0.0044232
10.0 0.0053319 0.005274 0.0053281

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

 
     

(3) 18.5 Psi C
 

Loading page 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
    
 
 
 

Loading page 5...

1.3
AE
K 
9
1 3
(0.08)(0.006)(68.9 10 )
0.025
K K 
 
9
1 3
N
1.32288 10 m
K K  
9
2
(0.02)(0.006)(68.9 10 )
0.1
K 

9
2
N
0.08268 10 m
K  
9
1.32288
10
1
1.32288
0
1.32288
1
0 2
3
4
0
0
1.32288 0.08268 0.08268 0
0.08268 0.08268 1.32288 1.32288 1800
1.32288 1.32288
u
u
u
u
                     
     
          
 
2
3
9
4
6 5 5
2 3 4
1.40556 0.08268 0 0
0.08268 1.40556 1.32288 0
0 1.32288 1.32288 1800 10
1.360668 10 m 2.313135 10 m 2.449202 10 m
u
u
u
u u u

  
                         
     
6
(1) 92 1 (1.360668 10 0)
(68.9 10 ) 3750000 3.75 MPa
0.025
u u
E


  
    
as a check: (1) 1800 N 3.75 MPa
(0.08)(0.006)
F
A
   
5 6
(2) 93 2 (2.313135 10 1.360668 10

Loading page 6...

Loading page 7...

1.5
1 2 1( ) 0R K u u  
1 2 1 2 3 2( ) ( ) 0K u u K u u   
2 3 2 3 4 3( ) ( ) 0K u u K u u   
` 3 4 3( ) 0K u u P  
In matrix form:
1 1 1
1 1 2 2 2
2 2 3 3 3
3 3 4
0
0
K K u R
K K K K u
K K K K u
K K u P
      
                
       
        

Loading page 8...

1.6
Size of the global matrix: 5 × 5
 1
[ ] 5 5
5 5
K   
  
① ②
①
②
 2
[ ] 8 8
8 8
K   
  
② ③
②
③
 3
[ ] 5 5
5 5
K   
  
② ③
②
③
 4
[ ] 20 20
20 20
K   
  
② ④
②
④
 5
[ ] 10 10
10 10
K   
  
③ ④
③
④
 6
[ ] 20 20
20 20
K   
  
④ ⑤
④
⑤
(6)
5 5
5 5 8 5 20 8 5 20
[ ] 8 5 8 5 10 10
20 10 20 10 20 20
20 20
K
 
        
      


Loading page 9...

Loading page 10...

1.7
5.88
150
1
150
5.88
0
5.88 5.88 2.27 2.27
2.27 2.27 10 10
10 10 0.581 0.581
0.581 0.581 0.781 0.781
0.781 0.781 2.22 2.22
2.22 2.22 1.47 1.47
1.47
  
  
  
  
  
  
 0
1.47
1
2
3
4
5
6
7
8
1
150
10
0
0
0
0
0
0
68
T
T
T
T
T
T
T
T
 
                                                                
 
  
10
12.04
17.31
18.51
{ } 39.12
54.46
59.
F
85
68
T
 
 
 
 
 
 
  
 
 


Loading page 11...

1.8
5.88
150
1
150
5.88
0
5.88 5.88 1.23 1.23
1.23 1.23 0.76 0.76
0.76 0.76 0.053 0.053
0.053 0.053 2.22 2.22
2.22 2.22 1.47 1.47
1.47
  
  
  
  
  
 0
1.47
1
2
3
4
5
6
7
1
150
20
0
0
0
0
0
68
T
T
T
T
T
T
T
 
                                                       
  
20
20.37
22.12
{ } 24.95
65.57
66.
F
54
68
T
 
 
 
 
 
  
 
 
 
 



q UA T 
as an example:
Btu
(1.47)(150)(68 66.54) 320 hr
q   
or
in out
1 (150)(68 20) Btu
( ) 320
Resistance 0.17 0.81 1.32 19 0.45 0.68 hr
q A T T 
   
    

Loading page 12...

Loading page 13...

1.10
1.47
1000
1
1000
1.47
0
1.47 1.47 0.053 0.053
0.053 0.053 2.22 2.22
2.22 2.22 1.47 1.47
1.47
  
  
  
 0
1.47
1
2
3
4
5
1
1000
15
0
0
0
70
T
T
T
T
T
 
                    
     
     
     
       
 
  
15
16.81
{ } F66.99
68.19
70
T
 
 
  
  
 
 
  
q UA T 
as an example:
Btu
(1.47)(1000)(70 68.19) 2660 hr
q   
or
Btu
(1.47)(1000)(16.81 15) 2660 hr
q   

Loading page 14...

1.11
5.88
22.5
1
22.5
5.88
0
5.88 5.88 2.56 2.56
2.56 2.56 1.47 1.47
1.47
  
  
 0
1.47
1
2
3
4
1
22.5
20
0
0
70
T
T
T
T
 
               
     
     
        
  
20
26.85
{ } F
42.59
70
T
 
 
 
  
 
  
1
Btu
(5.88)(22.5)(26.85 20) 910 hr
q U A T    
or as another example:
2
Btu
(2.56)(22.5)(42.59 26.85) 910 hr
q U A T    
or
in out
1 1
( ) (22.5)(70 20)
Resistance (0.17 0.39 0.68)
q A T T   
 
Btu
910 hr
q 

Loading page 15...

Loading page 16...

Loading page 17...

Loading page 18...

1.14
(a) 0 0M 
( 6 lb)(5 in) (35 lb)(6 in) (8 in) 0sF   
30 lbsF 
30 lb 60 0.5 insF KX X X  
(b)
( )
1 1
0
n m
e i i
i i ii i
F u
u u u

 
  
   
  
 
21 3 5
35 (6) 0
2 4 8
KX X X
X X
        
               
3 5
(35) (6) 0
4 8
KX    
     
   
3 5
(35) (6)
4 8
60
X
   
   
   
0.5 inX 

Loading page 19...

Loading page 20...

1.16
(1) 1 11
[ ] 1 15
K  
   
(2) 1 11
[ ] 1 110
K  
   
(3) 1 11
[ ] ;
1 115
K  
    Apply 3 0V  as a boundary condition
1 1 0
5 5
1 1 1 1 1 1
5 5 10 15 10 15
1 1
0 10 15

    
 
0
1 1
10 15

1
2
3
1
1 2
0.01
0
0
0.11 0.06
V
V
V
V V
 
 
                        
 
 
  
 
1 2 0.05 VoltsV V 
2 3 0.06 VoltsV V 

Loading page 21...

Loading page 22...

Loading page 23...

Loading page 24...

Loading page 25...

Loading page 26...

Loading page 27...

Loading page 28...

Loading page 29...

Loading page 30...

Loading page 31...

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

Study Now!

Document Details

Related Documents

Biofluid Mechanics: An Introduction to Fluid Mechanics, Macrocirculation, and Microcirculation, 3rd Edition (2021)

Solution Manual For Modern Compressible Flow: With Historical Perspective, 3rd Edition

Solution Manual for Engineering Mechanics: Dynamics, 2nd Edition

Solution Manual for Introduction to Finite Elements in Engineering, 4th Edition

Solution Manual For Introduction To Engineering Experimentation, 3rd Edition

Solution Manual for Engineering Mechanics Statics, 13th Edition

Solution Manual for Mechanics of Materials SI, 9th Edition

Solution Manual for Engineering Mechanics Dynamics, 13th Edition