MechanicalStaticseasyFE ExamCoursework

Two-Bar Bracket Member Force

A pin-jointed bracket carries a vertical load $P = 10\ \text{kN}$ hanging at joint $C$ .

Member $AC$ is horizontal. Member $BC$ runs up to the wall at $30^\circ$ above the horizontal. Both members are two-force members.

Find the force in member $BC$ , and state whether it is in tension or compression.

Given

P	10 kN	Applied vertical load at C
θ	30 °	Angle of BC above horizontal

Hint 1

Only two members meet at joint C, so joint equilibrium alone determines both forces.

Hint 2

Resolve into vertical and horizontal components. Which member has a vertical component?

Hint 3

Vertically: F_BC·sin30° must carry the entire 10 kN load.

Worked solution — try the problem first

Isolate joint $C$ . Only two members meet there, so the method of joints gives the answer directly.

Vertical equilibrium at C. Only $BC$ has a vertical component:

$\sum F_y = 0: \quad F_{BC}\sin 30^\circ - P = 0$

$F_{BC} = \frac{P}{\sin 30^\circ} = \frac{10}{0.5} = 20\ \text{kN}$

Sense. The load pulls joint $C$ down, so $BC$ must pull it up — the member pulls away from the joint, which is tension.

Check with horizontal equilibrium:

$\sum F_x = 0: \quad F_{AC} + F_{BC}\cos 30^\circ = 0$ $F_{AC} = -20(0.866) = -17.3\ \text{kN} \quad \text{(compression)}$

That $AC$ comes out in compression is the expected result: it is the strut holding $C$ away from the wall.

$\boxed{F_{BC} = 20.0\ \text{kN (tension)}}$

Concepts:Method of jointsTwo-force membersStatic equilibrium

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