Consider the beam ACDEB given in the figure. Which of the following statements is/are correct:

Let \( V_A \) and \( V_D \) be the vertical reactions at points A and D, respectively.
Using the moment equilibrium condition about point B:
\[ \sum M_B = 0 \]
We get:
\[ - V_D(4) + 4(2) + 2(6) = 0 \]
Solving for \( V_D \):
\[ V_D = 5 \, \text{kN} \]
Now, using the equilibrium of vertical forces:
\[ V_A + V_D = 5 \]
Substituting \( V_D = 5 \) kN:
\[ V_A = 0, \quad M_A = 0. \]
- At point A, the shear force is positive (+).
- At point B, the shear force is negative (−).
- The shear force remains constant in spans BC and CD.
- The bending moment at point A is zero.
- The bending moment is also zero in the spans AB and between C and D.
- The bending moment increases from C to D, then decreases towards point E.
Thus, the correct answers are (A) and (D).
A 2D thin plate (plane stress) has $E=1.0~\text{N/m}^2$ and Poisson’s ratio $\mu=0.5$. The displacement field is $u=Cx^2y$, $v=0$ (in m). Distances $x,y$ are in m. The stresses are $\sigma_{xx}=40xy~\text{N/m}^2$ and $\tau_{xy}=\alpha x^2~\text{N/m}^2$. Find $\alpha$ (in $\text{N/m}^4$, integer).
The infinitesimal element shown in the figure (not to scale) represents the state of stress at a point in a body. What is the magnitude of the maximum principal stress (in N/mm², in integer) at the point?

| Point | Staff Readings Back side | Staff Readings Fore side | Remarks |
|---|---|---|---|
| P | -2.050 | - | 200.000 |
| Q | 1.050 | 0.95 | Change Point |
| R | - | -1.655 | - |