Question 1.18: Determine the central transverse load W which will produce c...
Determine the central transverse load W which will produce central deflection \delta in the initially straight bar AB of area A and length 2L hinged at both ends as shown in Fig. 1.28.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.
Learn more on how we answer questions.
Solution Equilibrium condition: The forces in section AC and CB will be equal due to symmetry. Therefore, from free body diagram,
W = 2F \sin \alpha= 2F \times \frac{\delta }{L^\prime} (i)
Compatibility condition: Elongation of section AC is given by,
\Delta L = L^\prime — L=\frac{FL}{AE} (ii)
From Eqs (i) and (ii),
W =\frac{2\left(L^\prime-L\right) A E}{L} \times \frac{\delta}{L^\prime}=\frac{2 \delta A E}{L}\left(1-\frac{L}{L^\prime}\right)=\frac{2 \delta A E}{L}\left(1-\frac{1}{\sqrt{L^2+\delta^2}}\right)
Related Answered Questions
Question: 1.19
Verified Answer:
Equilibrium condition: For static eq-uilibrium of ...
Question: 1.17
Verified Answer:
Area of steel bars,
A_s = 4 \times \left(\p...
Question: 1.16
Verified Answer:
First consider the external load alone. Equilibriu...
Question: 1.30
Verified Answer:
Static deflection of the beam,
\sigma_{s t}...
Question: 1.28
Verified Answer:
Using Eq. (1.28)
\sigma _{\text{max}}^2 -2\...
Question: 1.29
Verified Answer:
The deflection at midspan due to statically applie...
Question: 1.27
Verified Answer:
Using Eq. 1.28,
\sigma _{\text{max}}^2 -2\f...
Question: 1.26
Verified Answer:
An element ABCD shown in Fig. 1.35, deforms to [la...
Question: 1.25
Verified Answer:
Figure E. 1.34 shows an element under radial stres...
Question: 1.24
Verified Answer:
\sigma_x, \sigma_y and \sigma_z a...