Question

The ILD for member $U_2{L}_2$ in the shown below is [Take compression positive]

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

$From\Delta O{L}_1{U}_{1}and\Delta O{L}_4{U}_4$

$\frac{O{L}_1}{O{L}_4}=\frac{6}{12}$

$\frac{O{L}_1}{O{L}_1+24}=\frac{6}{12}$

$\Rightarrow O{L}_1=24m$

$\text{Now, Case I : Consider the load is between}$ $L_1{L}_2$

$\Sigma M_{L1}=0V_{L7}=\frac{x}{48}$

Consider structure to the right of section (a)-(a)

$\Sigma M_O=0$

$F_u{{{}_{{}_2}}_L}_{{}_2}\times (24+x)=V_{L_7}\times (24+48)$

$F_u{{{}_{{}_2}}_L}_{{}_2}=\frac{x}{48}\times \frac{72}{(24+x)}$

$Atx=8m,i.e.at{L}_2$

$F_u{{{}_{{}_2}}_L}_{{}_2}=\frac{8\times 72}{48\times 32}=0.375kN\left(Tensile\right)$

$\text{Now, Case II: Consider the load is between}$ $L_3{L}_7$

$V_{L_1}=(1-\frac{x}{48})$

Consider structure to the left of section (a)-(a)

$\Sigma M_0=0$

$V_{L_1}\times 24=F_u{{{}_{{}_2}}_L}_{{}_2}\times 32$

$F_u{{{}_{{}_2}}_L}_{{}_2}=0.75{V_L}_{{}_1}$

$=0.75(1-\frac{x}{48})$

$At{L}_{3}i.e.x=16$,

$F_u{{{}_{{}_2}}_L}_{{}_2}=0.5kN\left(Compressive\right)$