Mechanics

Module II: Equilibrium

Lesson 10: Analysis of Frames - I

1 Frames

A frame is a structure with at least one multiforce member. A structural

element with more than two forces, or forces and couples is known as a multi

force member. In such members, we no longer have a simple configuration of

the unknown forces, as obtained in twoforce members. Machines and devices

with moving/driven parts that transfer and transform forces and torques are

typical examples frames.

Analysis of frames entails determination of internal forces and moments

carried by the members, including the ground forces if any. Frames may

be statically indeterminate, and all internal forces may not be determined

uniquely. In a large member of deterministic frames, identification of one (or

more) two-force member(s) that might be present can simplify the calcula-

tions.

Problem 1

Calculate the magnitude of the force acting on the pin D in the structure

shown in Fig. 1. Assume the contacts to be frictionless.

Figure 1:

Solution

The FBD of the members are shown in Fig. 2. Here, sin θ = 3/5 and cos θ =

4/5. In the triangular member ABC,

∑

MA = 0 ⇒ (0.2)FC − (0.14)100 = 0 ⇒ FC = 70 N

In the bar

∑

MD = 0 ⇒ (0.2)FE − 0.12FC sin θ = 0 ⇒ FE = 25.2 N

∑

Fx = 0 ⇒ Dx + FC sin θ − FE = 0 ⇒ Dx = 16.8 N

∑

Fy = 0 ⇒ Dy − FC cos θ = 0 ⇒ Dy = 56 N

Hence, FD =√

D2x + D2

y = 58.5 N.

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Figure 2:

Figure 3:

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Figure 4:

Problem 2

The clip shown in Fig. 3 requires a force F = 15 N for the jaws AB to release.

What is the normal reaction at the jaws when F = 0.

Solution

FBD of upper part (assuming internal horizontal force) is shown in Fig. 4.

When F = 15 N, RA = 0. In this situation,

∑

ME = 0 ⇒ 0.021RC − (0.032)15 = 0

⇒ RC = 22.9 N

When F = 0, we assume that the spring force RC remains the same as

above. Hence,

∑

ME = 0 ⇒ −0.039RA + (0.021)RC = 0

⇒ RA = 12.3 N

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Figure 5:

Problem 3

For the A-frame shown in Fig. 5, calculate the (shear) forces supported by

the pins B and D.

Solution

From the FBD in Fig. 6(a),

∑

Fx = 0 ⇒ Ax = 0

∑

MA = 0 ⇒ −14 + 11Fy = 0 ⇒ Fy = 14/11 kN

∑

Fy = 0 ⇒ Ay = −14/11 kN

From the FBD in Fig. 6(b),

∑

MB = 0 ⇒ 6Dy + 8Fy = 0 ⇒ Dy = −56/33 kN

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Figure 6:

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∑

Fy = 0 ⇒ By = 14/33 kN

∑

Fx = 0 ⇒ Bx + Dx = 0

From the FBD in Fig. 6(c),

∑

MC = 0 ⇒ −3Dx − 3Dy − 14 = 0 ⇒ Dx = 98/33 kN

Hence Bx = −Dx = −98/33 kN.

Finally, FB =√

B2x + B2

y = 3 kN, and FD =√

D2x + D2

y = 3.42 kN.

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