# Calculating member stress in a truss

I have the following question:

Here is my attempt:

Is this correct ?

You made a mistake in solving the reaction at joint "A". See calc below:

$$\sum M_G = 0$$

$$R_A = \dfrac{22.31*8}{12} = 14.873$$ kN

Solve internal member force using the method of section:

Since there is only one unknown in the vertical direction, so we can solve the member force $$F_{BC}$$ directly by $$\sum F_X = 0$$

$$\sum F_X = 0$$

$$-F_{BC}cos 30^{o}$$ + R_A = 0

$$F_{BC} = \dfrac{R_A}{cos 30^{o}} = \dfrac{14.873}{0.866} = 17.17$$ kN (Tension. Direction as assumed - away from the joint B)

$$\sigma_{BC}$$ = $$\dfrac{17.17}{0.08} = 214.6 kN/m^2 = 214.6 kPa = 214,600 Pa$$

• Don't mean to criticize, but note, you need to improve the writing of numbers, people must not need to GUESS what it is. Sloppy writing can lead to confusion or costly mistakes, a big no-no for structural engineers.
– r13
Oct 20, 2021 at 18:21
• Yes that’s true sorry about that Oct 20, 2021 at 21:55