Mechanics
Module II: Equilibrium
Lesson 11: Analysis of Frames - II
In a number of frame problems, presence of a two-force member simplifies
the calculation since the force configuration in two-force members is fixed. In
the following, we consider a few such examples.
Figure 1:
Figure 2:
Problem 1
In the lifting tongs shown in Fig. 1, determine the force in link AB. Neglect
the weight of the members.
Solution
In this problem, there are three two-force members, namely AB, CE and
DE. The forces in these members are, therefore, along the lines joining the
connection points.
The FBDs of an arm and the box are shown in Fig. 2. From the FBD of
the box, using symmetry,
X
Fy = 0 ⇒ 2Fy = 400g N ⇒ Fy = 200g N
2
From the FBD of the tong
X
X
Fy = 0 ⇒ −200g + FED sin θ = 0 ⇒ FED = 3.532 kN
MF = 0 ⇒ −0.72FAB + 0.99FED sin θ + 1.32FED cos θ = 0
⇒ FAB = 80.09 kN
Figure 3:
Problem 2
An aircraft landing gear supports a vertical load of 24 kN, as shown in Fig. 3.
Determine the shear force supported by the pin A.
Solution
The member BC is a twoforce member. Hence, the force in this member is
along BC. The FBD of the member OB is shown in Fig. 4.
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Figure 4:
From geometry
0.7
0.25
=
⇒ θ = 18◦.
◦
sin 60
sin θ
Also
0.7
CA
=
⇒ CA = 0.791 m.
◦
sin 102
sin 60◦
From the FBD
X
X
MC = 0 ⇒ (CA)Ax − (0.25 cos 30◦)24 = 0 ⇒ Ax = 6.57 kN.
MB = 0 ⇒ (−0.25 cos 30◦)Ay + (0.25 sin 30◦)Ax − (0.5 cos 30◦)24 = 0
⇒ Ay = −44.2 kN.
q
Thus, the net shear force at pin A FA = A2x + A2y = 44.68 kN.
Problem 3
Determine the moment M required to hold the nosewheel assembly at the
4
Figure 5:
configuration shown in Fig. 5 with θ = 30◦. The combined mass of the wheel
and arm OA is 50 kg and the center of gravity is at G. neglect the mass of
the other members.
Solution
Here, the member CD is a two-force member. The FBDs of the wheel assembly and element MC are shown in Fig. 6.
For the member AO
X
MA = 0 ⇒ (0.8)FCD sin 60◦ − (cos 60◦)50g = 0 ⇒ FCD = 353.6 N
For the member BC
X
MB = 0 ⇒ −(0.5)FCD sin 60◦ + M = 0 ⇒ M = 153 Nm
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Figure 6:
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