Solution to Problem 112 Normal Stress

$\Sigma F_V = 0$

$R_{AV} = 40 + 25 = 65^k$
 

$\Sigma M_A = 0$

$18R_D = 8(25) + 4(40)$

$R_D = 20^k$
 

$\Sigma F_H = 0$

$R_{AH} = RD = 20^k$
 

 

Check:
$\Sigma M_D = 0$

$12R_{AV} = 18(R_{AH}) + 4(25) + 8(40)$

$12(65) = 18(20) + 4(25) + 8(40)$

$780 \text{ ft} \cdot \text{kip} = 780 \text{ ft} \cdot \text{kip}$ (OK!)
 

For member AG (At joint A):
 

 

$\Sigma F_V = 0$

$\frac{3}{\sqrt{13}}AB = 65$

$AB = 78.12^k$
 

$\Sigma F_H = 0$

$AG + 20 = \frac{2}{\sqrt{13}} AB$

$AG = 23.33^k \, \text{Tension}$
 

$AG = \sigma_{\text{tension}} \, A_{AG}$

$23.33 = 20 A_{AG}$

$A_{AG} = 1.17 \, \text{in}^2$           answer
 

For member BC (At section through MN):
 

 

$\Sigma M_F = 0$

$6(\frac{2}{\sqrt{13}}BC) = 12(20)$

$BC = 72.11^k$ Compression
 

$BC = \sigma_{\text{compression}} \, A_{BC}$

$72.11 = 14 A_{BC}$

$A_{BC} = 5.15 \, \text{in}^2$           answer
 

For member CE (At joint D):
 

 

$\Sigma F_H = 0$

$\frac{2}{\sqrt{13}} CD = 20$

$CD = 36.06^k$
 

$\Sigma F_V = 0$

$DE = \frac{3}{\sqrt{13}} CD = \frac{3}{\sqrt{13}} (36.06) = 30^k$
 

At joint E:
 

 

$\Sigma F_V = 0$

$\frac{3}{\sqrt{13}} EF = 30$

$EF = 36.06^k$
 

$\Sigma F_H = 0$

$CE = \frac{2}{\sqrt{13}} EF = \frac{2}{\sqrt{13}} (36.06) = 20^k$ Compression
 

$CE = \sigma_{\text{compression}} A_{CE}$

$20 = 14 A_{CE}$

$A_{CE} = 1.43 \, \text{in}^2$           answer