By symmetry

$R_A = R_{Ov} = \frac{1}{2}(20)(8) = 80 ~ \text{kN}$

$y = 9 \tan 30^\circ = 3\sqrt{3} ~ \text{m}$

$z = \frac{1}{2}\sqrt{9^2 + y^2} = \frac{1}{2}\sqrt{9^2 + \left( 3\sqrt{3} \right)^2} = 3\sqrt{3} ~ \text{m}$

$x = \dfrac{z}{\cos 30^\circ} = \dfrac{3\sqrt{3}}{\cos 30^\circ} = 6 ~ \text{m}$

$\Sigma M_E = 0$

$x(F_{FH} \sin 30^\circ) + 10x + 20(x - 2.25) + 20[ \, x - 2(2.25) \,] \\

~ ~ ~ ~ ~ = 20[ \, (9 - x) - 2.25 \, ] + 80x$

$6\left( \frac{1}{2} \right)F_{FH} + 10(6) + 20(6 - 2.25) + 20(6 - 4.5) \\

~ ~ ~ ~ ~ = 20[ \, (9 - 6) - 2.25 \, ] + 80(6)$

$3F_{FH} + 60 + 75 + 30 = 15 + 480$

$3F_{FH} = 330$

$F_{FH} = 110 ~ \text{kN compression}$ *answer*

$\Sigma M_A = 0$

$xF_{GH} \sin 60^\circ = 2.25(20) + 2(2.25)(20) + 3(2.25)(20)$

$6F_{GH} \sin 60^\circ = 270$

$F_{GH} = 30\sqrt{3} ~ \text{kN} = 51.96 ~ \text{kN tension}$ *answer*

$\Sigma M_H = 0$

$yF_{EK} + 9(10) + 3(2.25)(20) + 2(2.25)(20) + 2.25(20) = 9(80)$

$3\sqrt{3}F_{EK} + 90 + 135 + 90 + 45 = 720$

$3\sqrt{3}F_{EK} = 360$

$F_{EK} = 40\sqrt{3} ~ \text{kN} = 69.28 ~ \text{kN tension}$ *answer*

Your solution to

$\sum M_E = 0$

You have $9F_{FH}+120+150+60=3F_{FH}+30+960$

Then It should be $6F_{FH}=660$ and not $3F_{FH}$ right?

Yes po, tama ka, maraming salamat.

So 110 kN na po yung sa unang member? hehe

Also, yung 18 m po hehe

The solution has been updated and also the figure. Thank you Infinitesimal sir.

No problem, sir :)

:)

Bakit po ang x= z/cos 30?

hindi po ba dapat x= z cos30?

In triangle ADE, x is hypotenuse.

Bakit po sa Sum of Moments @ E ay Ffh sin 30 lang yung ininclude di po kasama yung x(?) component ng Ffh?

Nakuha na kasi ang perpendicular distance between F

_{FH}at point E which is 1/2 of x. Ang triangle CDE kasi ay equilateral.