Plan the Solution Since this is a statically determinate structure. Part (a) is very straightforward. We simply use equilibrium to determine an expression for the force in the wire and divide that force by the cross-sectional area of the wire to get the axial stress in the wire. In Part (b), getting the correct geometrical relationship between the elongation of the wire and the tip displacement δ_D will be the challenge! A deformation diagram will help us visualize the geometry of deformation.

(a) Axial stress in wire AD.

Equilibrium: Using the free-body diagram in Fig. 2, we can write an equilibrium equation that relates the tensile force F in the wire to the external load P.

+\circlearrowleft\left(\sum{M} \right) _B = 0 :      -P\left(\frac{2L}{5}\right) + \left(\frac{3}{5}F\right)\left(\frac{4L}{5}\right) = 0

or

F = \frac{5}{6}P     Equilibrium       (1)

Therefore, the axial stress in the wire is

σ = \frac{F}{A} = \frac{5}{6}\frac{P}{A}            Ans. (a)

(b) Tip displacement of the beam. To determine the tip displacement of the beam, we need to determine the elongation of the support wire (force-deformation behavior) and then determine how far this elongation lets the beam rotate (geometry of deformation).

Element Force-Deformation Behavior: From the element force-deformation equation. Eq. 3.13, we have

e = \frac{FL}{AE}    Element Force-Deformation      (2)

Geometry of Deformation; Compatibility Equation: Figure 3 is a deformation diagram that enables us to relate the elongation e of the loaded support wire to the vertical displacement of D. The beam BD is rigid, so end D would actually follow a circular arc about end B when load P is applied to the beam, as indicated by the dashed circular arc in Fig. 3a. However, in Fig. 3b we have employed the simplifying assumption that point D moves vertically downward to D*, which differs little from moving along the circular arc so long as δ_D is small (i.e., δ_D \ll L)^4.
Figure 3b is an expanded view of the vicinity of D showing the deformation triangle DD′D*, with side DD′ drawn perpendicular to the wire in its final position AD*. Side DD* is the (approximated) vertical displacement of D, while D′ D* is the elongation of the wire, e = L* – L. Angle α in Figs. 3a,b is a vertex of the 3-4-5 triangle ADB. The two angles labeled α* in Fig. 3b are equal to each other and clearly, if δ_D is very small, α* ≈ α.Therefore, we can identify triangle D′DD* as a 3-4-5 triangle, so

e = \frac{3}{4}δ_D  Geometry of Deformation      (3)

This equation may be called a compatibility equation, since it enforces the condition that the end of the wire and the end of the beam undergo the same displacement from D to D*.
Solution of the Equations: Since this is a statically determinate problem, we were able to determine the element force in Eq. (1) from equilibrium alone. This force can be substituted into the element force-deformation equation, Eq. (2), giving

e = \frac{5}{6}\left(\frac{PL}{AE}\right)                (4)

Finally, this result can be substituted into the compatibility equation, Eq. (3), to give the tip displacement δ_D.

δ_D = \frac{25}{18}\left(\frac{PL}{AE}\right)  Ans. (b)

Review the Solution The right-hand side of the answer has the correct dimension of length. A larger force should produce a larger tip displacement, which it does. Likewise, a larger area or a larger value of E should reduce the tip displacement, as is the case. Therefore, the answer we have obtained seems reasonable.