1. Never stop thinking with us
KITES TEAM

2.  To this lecture we have one way to deal with
our problems that :
Determine all
forces (support
reaction and
external forces)
Determine the
thermal strain(if
exist)
Know how
will be the
deflection/the
force effect

3.  Links BC and DE are
both made of steel
(E=200 GPA) and are
12mm wide and 6mm
thick . Determine (a)
the force in each link
when a 2.4 KN force P
is applied to the rigid
member AF shown (B)
the corresponding
deflection of point A
 Simple diagram
100 mm
P
50 mm
50 mm
A
C
D
E
B
F
100m
m 125 mm

4.  Use equilibrium relations (∑ FX=0 , MB =0)
 From ∑FX=0 so XB=XE+P (1) .
 From MB=0 50*XE=100*2.4 so XE=4.8 KN
 XB=7.2 KN
 So F CB=7.2 KN
 So F DE=4.8 KN
 Note that we consider XF =0

5.  A =W*t=.012*.006=7.2*(10^-5 ) (m^2)
 ∆DE=F DE*L DE/E*A =4.8*125/200*7.2*(10^4)
=0.4166*(10^-4) m
 ∆BC=F BC*L BC/E*A
 =7.2*100/200*7.2*(10^4) = 0.5*(10^-4) m
 Def of point A=(.05-.04166)=.008 mm to right.

6.  A brass link
(W=100mm
,αb=20.9*(10^-6)/C ,
Eb=105 G pa )and a
steel rod its center line
at the center of the link
as shown (Es=200 G pa ,
αs=11.7*(10^-6)/C ).
Intial temp =20 C , final
temp=45 C.
 Determine the final
length of steel rod.
 Simple diagram
 D steel rod=38mm.
 Length of steel rod=
 250mm
 C at distance=50mm
 From the right edge.
2KN
C
3KN
38 mm
T=7mm
A
A
Sec A-A
4mm