Solution for MECH 4450 Homework #3
1.
(a) The connectivity matrix EC
EC=[1 2 2 3;2 3 4 4]
EC =
1 2 2 3
2 3 4 4
(b) The MATLAB code to generate the property matrix :
E=3*10^7*ones(4,1);
A=1.5*ones(4,1);
L=[30;40;50;30];
Prop=[E A L]
The running result:
Prop =
1.0e+007 *
3.000000000000000
3.000000000000000
3.000000000000000
3.000000000000000
0.000000150000000
0.000000150000000
0.000000150000000
0.000000150000000
0.000003000000000
0.000004000000000
0.000005000000000
0.000003000000000
(c) The function to generate stiffness matrix of element is:
function K=local_stiff(E,A,L,th)
k=[cos(th).^2
sin(th).*cos(th);
sin(th).*cos(th) sin(th).^2];
K=E*A/L*[k -k;-k k];
Calling this function:
Angle=[0;pi/2;pi/2+atan(3/4);pi]
for i=1:4
E(i)=Prop(i,1);
A(i)=Prop(i,2);
L(i)=Prop(i,3);
th=Angle(i);
local_stiff(E(i),A(i),L(i),th)
end
Running result:
Angle =
0
1.5708
2.2143
3.1416
K=
1500000
0
-1500000
0
0 -1500000
0
0
0 1500000
0
0
0
0
0
0
K=
1.0e+006 *
0.0000
0.0000
-0.0000
-0.0000
0.0000 -0.0000 -0.0000
1.1250 -0.0000 -1.1250
-0.0000 0.0000 0.0000
-1.1250 0.0000 1.1250
K=
1.0e+005 *
3.2400 -4.3200 -3.2400 4.3200
-4.3200 5.7600 4.3200 -5.7600
-3.2400 4.3200 3.2400 -4.3200
4.3200 -5.7600 -4.3200 5.7600
K=
1.0e+006 *
1.5000 -0.0000 -1.5000 0.0000
-0.0000 0.0000 0.0000 -0.0000
-1.5000 0.0000 1.5000 -0.0000
0.0000 -0.0000 -0.0000 0.0000
(d) The function to generate the global matrix:
function Gmatrix=global_stiff(n,Prop,angle,EC)
globalstiffM=zeros(2*n);
for m=1:n
E(m)=Prop(m,1);
A(m)=Prop(m,2);
L(m)=Prop(m,3);
th=angle(m);
localM=local_stiff(E(m),A(m),L(m),th);
if m==3
%refer to element 3 in the EC table
I=2*(EC(1,m)-1);
J=2*(EC(2,m)-2);
globalstiffM(1+I:2+I,1+I:2+I)=globalstiffM(1+I:2+I,1+I:2+I)+
localM(1:2,1:2);
globalstiffM(1+I:2+I,3+J:4+J)=globalstiffM(1+I:2+I,3+J:4+J)+
localM(1:2,3:4);
globalstiffM(3+J:4+J,1+I:2+I)=globalstiffM(3+J:4+J,1+I:2+I)+
localM(3:4,1:2);
globalstiffM(3+J:4+J,3+J:4+J)=globalstiffM(3+J:4+J,3+J:4+J)+
localM(3:4,3:4);
else
for i=1:4
for j=1:4
I=i+2*(EC(1,m)-1);
J=j+2*(EC(1,m)-1);
globalstiffM(I,J)=globalstiffM(I,J)+localM(i,j);
end
end
end
end
Gmatrix=globalstiffM;
Calling this function:
EC=[1 2 2 3;2 3 4 4];
E=3*10^7*ones(4,1);
A=1.5*ones(4,1);
L=[30;40;50;30];
n=4;
Prop=[E A L];
Angle=[0;pi/2;pi/2+atan(3/4);pi]
global_stiff(n,Prop,Angle,EC);
Running result:
(e) Considering the boundary conditions,
 u1  0 
 P1  ? 
v  0
 Q ? 
1
 1



u 2  ? 
 P2  4000lb 


 Q ? 
2


v2  0 
=
Gmatrix

 P 0 


3


u3  ? 
 Q3  0 
v3  ? 




 P4  ? 
u 4  0
 Q4  ? 
v 4  0 
Referring to the Gmatrix in part (d), we can get the condensed matrix:
P2  4000lb 
1.824 0
0 u 2 


 
6
1.5
0 ] u 3 
P3  0
  10 [ 0
Q  0

0
0 1.125 v 
 3

 3
u 2 
 
Once the values of u 3  are determined, loading matrix [P] can be directly
v 
 3
computed.
(f) Solving by MATLAB:
A=10^6*[1.824 0 0;0 1.5 0;0 0 1.125];
b=[4000;0;0];
u=inv(A)*b
Result:
u=
0.0022
0
0
Thus, u2  0.0022 in, u3  0 , v3  0
Substitute these values, we obtain that
 P1 
Q 
 1
 P2 
Q 
 2
 P  =1000*
 3
Q3 
 
 P4 
Q4 
(g) Displacement field of each member:
u1 
0
v 
 1
0
Element1:   = {
}
0.0022
u 2 
0
v 2 
u 2 
0.0022
v 
 2
0
Element2:   = {
}
0
u 3 
0
v3 
u 3 
0
v 
 3
0
Element4:   = { }
0
u 4 
0
v 4 
u 2 
0.0022
v 
 2
0
Element3:   = {
}
0
u 4 
0
v 4 
The forces in each element can be computed:
 P1 II 
 P1 I 
0
−3289.5
 II 
 I
Q
Q
 1 
 1 
0
0
 I  = { 3289.5 }
 II  = {0}
 P2 
 P2 II 
0
0
I
Q 
Q 
 2 
 2 
 P1 III 
 P1 IV 
0
709.56
III


 IV 
 Q1 
 Q1 
0
−946.08
 III  = {−709.56}
 IV  = {0}
 P2 III 
 P2 
0
946.08
Q 
Q IV 
 2 
 2 
The stress and strain in each element:
3289.5
𝜎
𝜎1 =
= 2193 𝑝𝑠𝑖, 𝜀1 = 1 = 7.31 × 10−5
1.5
𝐸
𝜎2 = 0, 𝜀2 = 0
√(709.562 +946.082 )
𝜎3 =
1.5
𝜎4 = 0, 𝜀4 = 0
= 788.4 𝑝𝑠𝑖, 𝜀3 =
𝜎3
𝐸
= 2.63 × 10−5
(h) From the calculations above, we can see that the reaction force at node 2 in the y
direction is Q2  947.4 lb
2.
(a)
The beam is divided into two elements as shown in the above figure. The local
element equation of each element is expressed as:
Element I
Q1 I   q1 I 
3L
6
3L   v1 
6
 I  I
 3L 2 L2  3L L2   
Q2  q 2  2 EI 
  1 
 I  I   3
6
 3L  v 2 
Q3  q3  L  6  3L


2
I
I
Q  q 
 3L 2 L2   2 
 3L L
 4   4 
Element II
Q1 II 
3L
6
3L  v 2 
6
 II 
 3L 2 L2  3L L2   
Q2  2 EI 
  2 
 II   3
6
 3 L   v3 
Q3  L  6  3L


2
II
Q 
 3L 2 L2   3 
 3L L
 4 
By assembling the systems, the global system is expressed as
 P1   q1 I 
M   I 
 1  q 2 
 P2  q 3 I  2 EI
  I  3
L
 M 2  q 4 
 P3   0 
   
 M 3   0 
3L
6
3L
0
0   v1 
6
 3L 2 L2  3L L2
0
0   1 

 6  3L 12
0
6
3L  v 2 

 
2
0
4 L2  3L L2   2 
 3L L
0
0
 6  3L
6
 3L   v 3 

 
2
0
3L
L
 3L 2 L2   3 
 0
(b)
The known boundary conditions are
v1  v2  0m
 1 0rad
The known loading conditions are
P3  2500  k  (v3 ) , M 2  1250N  m , M 3  0 N  m
 q1 I 
 qL / 2    2500 N 
 I
 qL2 / 12   2083 N  m
q 2  L

 

 I   0 i q  dx  


 q3 
 qL / 2    2500 N 
q I 
 qL2 / 12  2083 N  m 
 4 
(c)
By imposing boundary conditions and loading conditions, the condensed system is
expressed as:
 4 L2
 3L
L2   2 
 1250  2083
3
 
kL

 2 EI 
 3L   v 3 
  2500   3  3L 6 
2 EI
 

 L  2
0


 3L
2 L2   3 
 L
(d)
By using MATLAB to solve the equations, the unknown nodal displacements are
found as
 2   0.0000724rad 
  

v3     0.000879m 
   0.0002275rad 
 3 

Then unknown reactions can be found as
I
0   2   q1    974 N 
 P1 
3L 0
  2 EI  2
   I 

0
0   v3   q 2    3707 N  m
M 1   3  L
 P  L  0  6 3L    q I   8456 N 
 2

 3   3  

(e) The MATLAB code for shear force and moment:
P=[-974 8456];
q=-1000;
M1=-3707;
x1=linspace(0,5,20);
x2=linspace(5,10,20);
for i=1:length(x1)
y1(i)=P(1)+q*x1(i);
y3(i)=P(1)*x1(i)+0.5*q*x1(i)^2-M1;
end
for j=1:length(x2)
y2(j)=P(2)+y1(i);
y4(j)=(P(2)+y1(i))*(x2(j)-10);
end
subplot 211;
plot(x1,y1,'k',x2,y2,'k','linewidth',3);
ylabel('N','Fontsize',20);
legend('Shear force(N)');
grid on;
subplot 212;
plot(x1,y3,'r-o',x2,y4,'r-o','linewidth',3);
xlabel('x(m)','Fontsize',20);
ylabel('Nm','Fontsize',20);
legend('Moment')
grid on;
The diagram:
3.
a)
The MATLAB code for generating the local matrix:
function K=Q3local_stiff(E,A,I,L,th)
Q1=[cos(th)
sin(th) 0;
-sin(th) cos(th) 0;
0
0
1];
Q=[Q1 zeros(3);zeros(3) Q1];
k=[A*E/L 0 0, -A*E/L 0 0;
0 12*E*I/L^3 6*E*I/L^2, 0
0 6*E*I/L^2 4*E*I/L,
0
-A*E/L 0 0, A*E/L 0 0;
0 -12*E*I/L^3 -6*E*I/L^2,
0 6*E*I/L^2 2*E*I/L,
0
K=Q'*k*Q;
-12*E*I/L^3 6*E*I/L^2;
-6*E*I/L^2 2*E*I/L;
0 12*E*I/L^3 -6*E*I/L^2;
-6*E*I/L^2 4*E*I/L;];
Calling this function:
clc;
clear all;
format short;
EC=[1 2;2 3];
E=3*10^7*ones(2,1);
A=100*ones(2,1);
I=[1000,1000]';
L=[30*sqrt(2)*12,40*12]';
Prop=[E A I L];
Angle=[pi/4;0]
for i=1:2
th=Angle(i);
Q3local_stiff(E(i),A(i),I(i),L(i),th)
end
The result:
(b) The function to obtain the global stiffness matrix:
function Gmatrix=Q3global_stiff(Prop,angle,EC)
globalstiffM=zeros(9);
for m=1:2
E(m)=Prop(m,1);
A(m)=Prop(m,2);
I(m)=Prop(m,3);
L(m)=Prop(m,4);
th=angle(m);
localM=Q3local_stiff(E(m),A(m),I(m),L(m),th);
for i=1:6
for j=1:6
I=i+3*(EC(1,m)-1);
J=j+3*(EC(1,m)-1);
globalstiffM(I,J)=globalstiffM(I,J)+localM(i,j);
end
end
end
Gmatrix=globalstiffM;
Calling the function Q3global_stiff:
clc;
clear all;
format short;
EC=[1 2;2 3];
E=3*10^7*ones(2,1);
A=100*ones(2,1);
I=[1000,1000]';
L=[30*sqrt(2)*12,40*12]';
Prop=[E A I L];
Angle=[pi/4;0]
GlobalMatrix=Q3global_stiff(Prop,Angle,EC)
The result:
(c) Boundary conditions:
𝑢1 = 0, 𝑣1 = 0, 𝜃1 = 0;
𝑅2 = 0, 𝑆2 = 0, 𝑀2 = 0;
𝑢3 = 0, 𝑣3 = 0, 𝜃3 = 0;
 q1 II 
 qL / 2    2  10 4 lb 
 II 

 qL2 / 12  
6
L
q 2 

  1.6  10 lb  in 
 II   0 i q  dx  


4
q3 
 qL / 2    2  10 lb 
q II 
 qL2 / 12  1.6  10 6 lb  in 
 4 
By row operations, the condensed system can be obtained:
0
u 2 
0.092 0.0294 0.0049
4
 
8
{  210 } = 10 × [0.0294 0.0295 0.0029] × v 2 
 
0.0049 0.0029 4.8570
 1.6  10 6
 2
P1


−0.0295 −0.0294 −0.0049


S
1


−0.0294 −0.0295 0.0049


M1
0.0049 −0.0049 1.1785 ×
8

  10 × −0.0625
0
0
R
3


0
0
−0.0078
4
 S 3  2  10 
[
0
0.0078
1.25 ]


M 3  1.6  10 6 
u 2 
 
v 2 
 
 2