SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Question : 192 of 200

Marks: +1, -0

The force in member AB is ______. (take angle BAC as 60° and angle BCA as 30°)

Solution:

Explanation:

Given:
∠BAC = 60°
∠BCA = 30°
Let

\F_\AB

= compressive

\F_\BC

= compressive

\F_\AB

= ?
Considering joint B,

∑\F_{\H}=0

{\F}_{{\AB}} \cos 60°^{}={\F}_{{\BC}} \cos 30° ^{}

{\F}_{{\AB}}=√{3} {\F}_{{\BC}}

∑\F_{\V}=0

{\F}_{{\AB}} \sin 60°^{}+{\F}_{{\BC}} \sin 30°^{}=10

Multiplying both sides by

√3

3 {\F}_{{\AB}}+{\F}_{{\AB}}=20 √{3}

{4\F}_{{\AB}}=20 √{3}

{\F}_{{\AB}}=5 √{3}

Compressive (As direction assumed comp)

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