A force F = 1 kN acts at point C as shown in Figure 1.36. The strut AC and the tie-rod BC make angles 60° and 30° with the wall as shown. If length AC = 1 m and for both AC and BC, product of cross-sectional area (A) and Young’s modulus (E), that is, axial stiffness ‘AE’ is same, 10^6 N, then find

Question

A force F = 1 kN acts at point C as shown in Figure 1.36. The strut AC and the tie-rod BC make angles 60° and 30° with the wall as shown. If length [latex] \overline{A C}=1 m [/latex] and for both AC and BC, product of cross-sectional area (A) and Young’s modulus (E), that is, axial stiffness ‘AE’ is same, [latex] 10^6 [/latex] N, then find out horizontal and vertical components of displacement of point C.

Accepted Answer

To find out the axial force developed in AC and BC, let us consider the equilibrium of joint C [refer to Figure 1.37(a)].

From the triangle of forces, we can easily get [latex] F_{ AC }=1 kN ext { and } F_{ BC }=\sqrt{3} kN [/latex]. We can also find out the nature of the axial loads: [latex] F_{ AC } [/latex] is in compression and [latex] F_{ BC } [/latex] in tension. Now, contraction of AC is

[latex] \delta_{ AC }=\frac{1000 imes 1000}{10^6}=1 mm [/latex]

and elongation of BC is

[latex] \delta_{ BC }=\frac{\sqrt{3} imes 1000 imes 1000 \sqrt{3}}{10^6} \quad( ext { as } \overline{ BC }=\sqrt{3 AC }) [/latex]

As a result of these two changes in length, point C will move to a new point C′ as shown in Figure 1.37(b). How can this point C′ be located? This can be done with the help of a diagram which is known as ‘Williot diagram’. If [latex] \overline{ AC } ext { becomes } \overline{ AC ^{\prime \prime \prime}} ext { and } \overline{ BC } ext { becomes } \overline{ BC ^{\prime \prime}} [/latex], the perpendicular on AC at C′′′ and perpendicular on BC (extended) at C′′, as shown will meet at C′ the new location of ‘C’. Why is it so? It is because member AC will rotate about A and member BC about B, and C′ is the only point which lies on both the angular paths of these rotations.

This diagram can be physically constructed with any suitable scale so that the small changes in length of [latex] \overline{ AC } ext { and } \overline{ BC } [/latex] can be properly drawn. From that, the vertical and horizontal displacements of point C can be easily measured. Alternate way of doing it is to use the trigonometry of the Williot diagram. In Figure 1.37(c), the first method is followed which gives [latex] \Delta_n=4.73 [/latex] mm and [latex] \Delta_ν=6.20 [/latex] mm.
[Interested students can try to develop Williot diagram as per their convenient scale and repeat the process for better understanding.]

Question 1.15: A force F = 1 kN acts at point C as shown in Figure 1.36. Th...

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