Answer
(a) $σ_{BE}= - 281 psi$
(b)$σ_{BE}= 107.1 psi$
Work Step by Step
In the given problem we have to find the normal stress in member Be and Cf.
Cross-sectional areas of each member are to give to us.
We will first estimate the support reactions at point D and then consider the equilibrium of part DEF to estimate axial forces in members BE and CF.
Take the sum of moments about point A
The 480 lb force is at a distance of 75 in from point A.
Only one reaction ($F_{DX}$= will cause a bending moment at A.
Since joint is a hinged support sum of moments there will be zero. )
$∑ M_{A}=0$
$40 \times F_{DX}-( 45 +30 ) \times 480 =$0
$40 \times F_{DX}=(75) \times 480$
$F_{DX}=\frac{(75) \times 480}{40}$
$F_{DX}= 900 lb$
Now , Take sum moments about point F
$∑ M_{F}= 0$
$- 45 \times F_{DY} - (30) \times sin ( 53.13 ') \times F_{BE}=0$
$F_{BE}=- \frac{45(900)}{24}$
$F_{BE}= -2250 lb$
So, the member BE is in compression.
Take sume of moments about point E
$∑ M_{f}=0$
$- 15 \times F_{DY}+ (30) \times sin ( 53.13 ') \times F_{CE}=0$
$F_{CE}=\frac{15(900)}{24}$
$F_{CE}= 750 lb$
So, the memeber CF is tension .
Part (a) Stress in member BE
The cross-sectional area of BE
$A= 2 \times 4$
$A= 8 in^{2}$
Calculate the stress in the compression member BE using the formula
$σ_{BE}= \frac{F_{BE}}{A}$
$σ_{BE}=\frac{-2250}{8}$
$σ_{BE}= - 281 psi$
Part (b) Stress in member CF
The minimum cross-sectional area of CF occurs at the pin
$A_{min}= 2 ( 4- 0.5)$
$A_{min}= 7 in^{2}$
Calculate the stress in the compression member BE using the formula
$σ_{CF}= \frac{F_{Cf}}{A_{min}}$
$σ_{CF}=\frac{750}{7}$
$σ_{BE}= 107.1 psi$
