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## Q. 3.3

Determine the tension in the member 13 that connects the cantilever to the bracket 234 as shown in Figure 3.8. The cross-sectional areas, second moment of areas, and modulus of elasticity of the members are $A_{13}=1 in^{2}, A_{34}=A_{23}=2 in^{2}$, $I_{12}=1440 in^{4}, E_{12}=10,000 \text{kips} / in^{2}$, and $E_{13}=E_{23}=E_{34}=30,000 \text{kips} / in^{2}$. ## Verified Solution

The force in member 13 is considered as the redundant, and the actual structure is replaced by system (i) plus R × system (ii).

In system (i), member 12 is subjected to a bending moment, and there are no axial forces in the members. Then, over the cantilever from x = 0 to x = 8.66 ft :

M = -15 x

and:

m = x

and the relative inward movement of 1 and 3 is:

\begin{aligned}\delta_{13}^{\prime} &=\int_{1}^{2} Mm dx/EI \\&=-15 \int_{0}^{8.66} x^{2} dx/EI \\&=-15(8.66)^{3} \times(12)^{2} /(3 \times 10,000 \times 1440) \\&=-32.47 \times 10^{-3} \text {ft} \end{aligned}

In system (ii), member 12 is subjected to a bending moment, and members 23 and 34 are subjected to axial forces, and the relative inward movement of 1 and 3 is:

\begin{aligned}\delta_{13}^{\prime \prime} &=\int_{1}^{2} m^{2} dx/EI +\sum u^{2} l / A E \\&=\int_{0}^{8.66} x^{2} dx/EI +2 \times(1)^{2} \times 10 \times 12^{2} / AE \\&=(8.66)^{3} \times(12)^{2} /(3 \times 10,000 \times 1440)+2 \times 10 /(2 \times 30,000) \\&=(2.17+0.33) \times 10^{-3} \\&=2.50 \times 10^{-3} \text {ft}\end{aligned}

The extension in member 13 in the actual structure is:

\begin{aligned}\delta_{13} &=-R \times 5(1 \times 30,000) \\&=-R \times 0.167 \times 10^{-3} \text {ft}\end{aligned}

Thus:

-0.167 R = -32.47 + 2.50 R

and:

R = 12.18 kips