AE 464
FALL 2009
MIDTERM EXAMINATION
SOLUTIONS
1a) READ(1,*) NUMFOR
read the number of nodes to which an external load
is applied
DO NNFORCE=1,NUMFOR
READ(5,*)NODNUM,PX,PY
P(NODNUM*3-2)=PX
P(NODNUM*3-1)=PY
P(NODNUM*3)=PZ
END DO
Data file unit=1:
3
(gives the number of nodes to which an external load
is applied- NUMFOR)
Data file unit=5:
41 100 100
343 150 0
819 120 120
0
180
100
b) P(I): External load vector
ICOUNT=0
DO J=1,NUMNP (loop over total number of nodes)
DO I=1,3 ( each node has 3 dof)
ICOUNT=ICOUNT+1
IKL=ID(I,J)
IF(IKL.GT.O)THEN
PEXT(IKL)=P(ICOUNT)
ENDIF
END DO
END DO
2- a)
6 L  12 6 L
0
0   w1 
 12
 6 L 4 L2  6 L 2 L2
0
0   1 

 12  6 L 24
0
 12 6 L  w2
EI/L3 
 
2
2
0
8L
 6 L 2 L2   2 
 6L 2L
 0
0
 12  6 L 12  6 L   w3

 
0
6 L 2 L2  6 L 4 L2   3 
 0
0 1 0 
b) 

0 2 0 
24 0 
c) EI/L3 
2
 0 8L 
 qL / 2
 0 


  qL 
d) 
 + reaction forces and moments at both both ends
 0 
 qL / 2


 0 
 qL 
e) 

 0 
f) w2=-qL4/24EI
2 =0
 1  1 0
AE 
3- k1 
 1 1 0
2L 
 0 0 0
0 0 0 
AE 
k2 
0 1  1
L 
0  1 1 
 0.5  0.5 0 
AE 
a) K  k1  k 2 
 0.5 1.5  1
L 
 0
1
1 
 EAT   R1

  
b) R1   EAT    0 
 0  0

  
 0  0
R2    EAT    0 
 EAT   R3

  
 0.5  0.5 0   u1  R1  EAT 
AE 
  

c)
 0.5 1.5  1 u 2  
0


L



 0
1
1  u3 R3  EAT 

0
 1.5  0.5  1 u 2 

AE 
  


 0.5 0.5
0   u1   R1  EAT 
L 
 

  1
0
1  
u3 R3  EAT 
d) u2=0
e) Since u1=u2=u3=0
R1  EAT =0
R1  EAT
R3  EAT =0
R3  EAT
f) Free body of element 1 gives (assume axial stress in element 1 to be in + x direction):
EAT  1 A  0
1  ET
(compression, as expected)
Free body of element 2 gives (assume axial stress in element 2 to be in + x direction):
 EAT   2 A  0
 2  ET
(compression, as expected)