Solution to Problem 247 Statically Indeterminate

$R_2 = 240 - R_1$

$P_{al} = R_1$

$P_{st} = 150 - R_1$

$P_{br} = R_2 = 240 - R_1$
 

 

From the FBD of each material shown, δ_al is shortening, δ_st is lengthening, and δ_br is also lengthening. Thus,
$\delta_{al} = \delta_{st} + \delta_{br}$

$\left( \dfrac{PL}{AE} \right)_{al} = \left( \dfrac{PL}{AE} \right)_{st} + \left( \dfrac{PL}{AE} \right)_{br}$

$\dfrac{R_1 \, (500)}{900(70\,000)} = \dfrac{(150 - R_1 )(250)}{2000(200\,000)} + \dfrac{(240 - R_1 )(350)}{1200(83\,000)}$

$\dfrac{R_1}{126\,000} = \dfrac{150 - R_1}{1\,600\,000} + \dfrac{(240 - R_1 )7}{1\,992\,000}$

$\frac{1}{63}R_1 = \frac{1}{800}(150 - R_1) + \frac{7}{996}(240 - R_1)$

$( \frac{1}{63} + \frac{1}{800} + \frac{7}{996} ) R_1 = \frac{1}{800}(150) + \frac{7}{996}(240)$

$R_1 = 77.60 \, \text{ kN }$
 

$P_{al} = R_1 = 77.60 \, \text{kN}$

$P_{st} = 150 - 77.60 = 72.40 \, \text{kN}$

$P_{br} = 240 - 77.60 = 162.40 \, \text{kN}$
 

$\sigma = P/A$
 

$\sigma_{al} = 77.60(1000)/900$

$\sigma_{al} = 86.22 \, \text{ MPa}$           answer
 

$\sigma_{st} = 72.40(1000)/2000$

$\sigma_{st} = 36.20 \, \text{ MPa}$           answer
 

$\sigma_{br} = 162.40(1000)/1200$

$\sigma_{br} = 135.33 \, \text{ MPa}$           answer