Quiz 26

Question 1

The shown two dimensional beam deforms such that vertical lines stay vertical lines, vertical lines keep their original length while the new horizontal position is given by $x_{1} = X_{1} + u_{1} (X_{1})$ where $u_{1}$ is function of $X_{1}$ . The centerline of the beam deforms according to a function $y = f (X_{1})$ . Then, the small strain matrix as a function of the position in the beam is given by:

Select one:
a.

ε = (\begin{matrix} \frac{d u_{1}}{d X_{1}} & \frac{1}{2} \frac{d y}{d X_{1}} & 0 \\ \frac{1}{2} \frac{d y}{d X_{1}} & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

ε = (\begin{matrix} \frac{d u_{1}}{d X_{1}} & 0 & 0 \\ \frac{d y}{d X_{1}} & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

ε = (\begin{matrix} \frac{d u_{1}}{d X_{1}} & 0 & 0 \\ 0 & \frac{d y}{d X_{1}} & 0 \\ 0 & 0 & 0 \end{matrix})

ε = (\begin{matrix} \frac{d u_{1}}{d X_{1}} + \frac{d y}{d X_{1}} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

$u = (\begin{matrix} u_{1} \\ y \\ 0 \end{matrix})$

It is important to note that the displacement functions $u_{1}$ and $y$ are functions in $X_{1}$ and their derivatives with respect to $X_{2}$ and $X_{3}$ are zero. $ε = \frac{1}{2} (\nabla u + \nabla u^{T}) = (\begin{matrix} \frac{d u_{1}}{d X_{1}} & \frac{1}{2} \frac{d y}{d X_{1}} & 0 \\ \frac{1}{2} \frac{d y}{d X_{1}} & 0 & 0 \\ 0 & 0 & 0 \end{matrix})$

The correct answer is: $ε = (\begin{matrix} \frac{d u_{1}}{d X_{1}} & \frac{1}{2} \frac{d y}{d X_{1}} & 0 \\ \frac{1}{2} \frac{d y}{d X_{1}} & 0 & 0 \\ 0 & 0 & 0 \end{matrix})$ .

Question 2

The shown axially loaded beam is made of a linear elastic material with a unit area and a varying $E$ such that $E = 200 - 20 X_{1}$ . If the distributed load is equal to zero and the beam is subjected to an end load of value $P$ , then the differential equation of equilibrium in terms of the horizontal displacement $u_{1}$ is given by:

Select one:
a.

(200 - 20 X_{1}) \frac{d^{2} u_{1}}{d X_{1}^{2}} = 0

(200 - 20 X_{1}) \frac{d^{2} u_{1}}{d X_{1}^{2}} = - P

(200 - 20 X_{1}) \frac{d^{2} u_{1}}{d X_{1}^{2}} - 20 \frac{d u_{1}}{d X_{1}} = 0

(200 - 20 X_{1}) \frac{d^{2} u_{1}}{d X_{1}^{2}} - 20 \frac{d u_{1}}{d X_{1}} = - P

The differential equation is given by: $\frac{d E A \frac{d u_{1}}{d X_{1}}}{d X_{1}} + p = 0$

The student should differentiate between the distributed horizontal load

p

which is equal to zero in this question and the applied boundary condition

P

. The boundary condition does not appear in the differential equation.

Since $E$ is variable, the product rule is used to differentiate $E A \frac{d u_{1}}{d X_{1}}$ with respect to $X_{1}$ to obtain the correct answer.

The correct answer is: $(200 - 20 X_{1}) \frac{d^{2} u_{1}}{d X_{1}^{2}} - 20 \frac{d u_{1}}{d X_{1}} = 0$ .