Landing Gear Repair

Q1.
Landing gear repair.
the figure below shows various configurations of landing gear assemblies. some are
repairable while others are non-repairable.
the landing gear design is very robust. Damaged parts are either welded or
replaced. The configurations A, B and C are repairable by welding methods while D
and E are non-repairable. A, B and C are made from tubing of steel and welding can
only be done if the steels were heat treated. Reheating is also done even after
welding.
Type of welding.
TIG. Tungsten Inert Gas.

Process of welding.
the steel parts are cleaned with oil and
a stainless steel brush first. the
cleaning is done to prevent hydrogen
embrittlement and the porosity which
may occur during welding. the
cleansing oil is then removed.
High frequency AC is used to start the
arcs. preheating in this case is
necessary since the thickness of the
tube is 0.12 inches. in other
applications which have a less
thickness of the tube, thinner than
0.12 inches, preheating is not
required.
to prevent brittleness from happening, stress relieving (post-weld tempering) is
required.
4130 filter rods are used. the work is preheated and post-heated after welding to
prevent cracking. For this particular application, a low speed is used. This causes a
fillet that is sufficiently large and a little bit concave or convex.
the weld is allowed to cool to room temperatures. then the process of heating
starts. the parts are heated between 1100o F and 1200O F using a neutral flame of
the oxy-acetylene for 45 minutes. In ambient lighting, this flame is dull red.
Advantages and disadvantages on structural members.
TIG is a method which produces quality welds on thin sections of metals. strong
welding and joints of high quality is achieved. The joints are more appealing and
most of them do not require finishing after welding. A wide variety of metals can be
welded by this method. Even different metals can be welded together. there is less
distortion of jointed parts as a result of small heat zone occurrences and the joints
are ductile and also non-corrosive.
Very thick sections become a problem when TIG method is used due to insufficient
penetration. the Welding process is complicated thus calling for an experienced
welder. one has to wait for 45 minutes when post-heating. this makes the method
to be time consuming.
Q2.
Jackscrew is a type of jack which is operated by a lead screw which is turned by a
handle, it is used to lift heavy loads such as vehicles and other parts of machine
tables. the jack raises whatever that is on top of the lead cap.
the design includes a nut and a screw. the nut is enclosed in a cast iron body. the
nut has setscrews which are used to lock the nut and by thus preventing its

rotation. The screw has a hole through its head. The hole is where the handle
passes through. The nut is restricted from moving out of the nut by a plain washer.
this washer has a big outer diameter than the outer diameter of the screw.
the parts are
1. Body
2. Nut collar
3. Spindle screw
4. lever
5. pin
6. cup
7. locknut and washer
Jackscrews are self-locking. However,
when used to lift a very large load,
slipping or tipping may occur.
The ratio of load to the amount of force required to lift the load, Mechanical
Advantage, is usually between 30% to 50%.
M.A = F load
F effort
where F load is the force appied to the load by the jackscrew
F effort is the force applied by the operator.
Jackscrews come in various designs such as the bottle jack, the scaffolds, the
floorjack (which was used to lift small houses) and the scissor jacks.
the figure below shows a power screw jack and a scissor jack
Fig: power-screw jack,
Fig: Scissor jack

Q3.
We resolve all the forces to x and y components.
 Resolving for W
W = (25000 × cos270)x + (25000 × sin270)y = (-25000y) N
 Resolving for force D
D = (3000 × cos350)x + (3000 × sin 350)y = (2954.4233x + -520.9445y) N
 Resolving for Force L
L = (24000 × cos80)x + (24000 × sin80)y = (4167.5563x + 23635.386y) N
 Resolving for force T
T = (10000 × cos170)x + (10000 × sin170)y = (-9848.07753x +1.7365y) N
Resultant force is given as:
FRes = W + FD + FL + FT
= (2954.4233 + 4167.5563 – 9848.0775)x +(-25000 – 520.9445 + 23635.386 +
1.7365)y
= (-2726.09793x – 149.0767y) N
In polar form, FRes = 2370.1710 N, -176.869o
%
Q4.
The forces acting on the helicopter can be illustrated below