Chapter 2

Q. 2.5

A strut and cable assembly ABC supports a vertical load as shown in Figure 2.18(a). The cable has an effective area of cross section 120 mm², and the strut has an area of 250 mm². Calculate the normal stresses in cable and strut and indicate the nature of stresses. If the cable elongates 1.3 mm, what is the strain? If the strut shortens 0.62 mm, what is the strain?



Verified Solution

The assembly is in equilibrium. The free-body diagram of strut is shown in Figure 2.18(b), and Figure 2.18(c) shows free-body diagram of joint B. First, we shall find force in the cable (cable can take only tension).

From Figure 2.18(a)

\begin{aligned} \tan \theta & =\left\lgroup \frac{1.5}{2} \right\rgroup \\ \theta & =\tan ^{-1}\left\lgroup \frac{1.5}{2} \right\rgroup =36.87^{\circ} \end{aligned}

Also,        \alpha=\tan ^{-1}\left\lgroup\frac{2}{1.5}\right\rgroup=53.13^{\circ}

Taking moments about C and applying \sum M_c=0

\begin{aligned} & P \times 2.0-T_{ BA } \sin 73.73^{\circ} \times \sqrt{2^2+1.5^2}=0 \\ & T_{ BA }=\frac{15 \times 1000 \times 2.0}{0.96 \times 2.5}=12500  N =12.5  kN \end{aligned}

Now, we shall consider joint B (the nature of the force in the strut is compressive).
Resolving all the forces parallel to the strut and equating to zero

\begin{gathered} -F_{ BC }+12.5 \cos 73.73^{\circ}+15 \cos 53.13^{\circ}=0 \\ F_{ BC }=12.5  kN \end{gathered}


\begin{aligned} & \text { Tensile stress in cable }=\sigma_t=\frac{12.5 \times 1000}{120}=104.16  MPa \\ & \text { Compressive stress in Strut }=\sigma_c=\frac{12.5 \times 1000}{250}=50  MPa \end{aligned}

The corresponding strains can be calculated using the known values of change in lengths.

\text { Length of cable }=\text { length of strut }=\sqrt{1.5^2+2.0^2}=2.5  m

\text { Strain in cable }=\varepsilon_{ AB }=\frac{d l}{l}=\frac{1.3}{2.5 \times 1000}=5.2 \times 10^{-4}

\text { Strain in strut }=\varepsilon_{ BC }=\frac{d l}{l}=\frac{0.62}{2500}=2.48 \times 10^{-4}