3 (a) In this case N(x)=(1)/(2) ρg b h(l-x)^2 where b= width of the girder. Also I=b h^3 / 12. T

step 1 Hi everyone in the given case moment of force as a function of position is given by half row GVH

3 (a) In this case N(x)=(1)/(2) ρg b h(l-x)^2 where b= width...

3 (a) In this case $N(x)=\frac{1}{2} \rho g b h(l-x)^{2}$ where $b=$ width of the girder. Also $I=b h^{3} / 12$. Then, $$ \frac{E b h^{2}}{12} \frac{d^{2} y}{d x^{2}}=\frac{\rho g b h}{2}\left(l^{2}-2 l x+x^{2}\right) $$ Integrating, $\quad \frac{d y}{d x}=\frac{6 \rho g}{E h^{2}}\left(l^{2} x-l x^{2}+\frac{x^{3}}{3}\right)$ using $\frac{d y}{d x}=0$ for $x=0 .$ Again integrating $$ y=\frac{6 \rho g}{E h^{2}}\left(\frac{l^{2} x^{2}}{2}-\frac{b x^{3}}{3}+\frac{x^{4}}{12}\right) $$ Thus $$ \begin{aligned} &\lambda=\frac{6 \rho g l^{4}}{E h^{2}}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}\right) \\ &=\frac{6 \rho g l^{4}}{E h^{2}} \frac{3}{12}=\frac{3 \rho g l^{4}}{2 E h^{2}} \end{aligned} $$ (b) As before, $E I \frac{d^{2} y}{d x^{2}}=N(x)$ where $N(x)$ is the bending moment due to section $P B$. This bending moment is clearly $$ \begin{aligned} N &=\int_{x}^{2} w d \xi(\xi-x)-w l(2 l-x) \\ &=w\left(2 l^{2}-2 x l+\frac{x^{2}}{2}\right)-w l(2 l-x)=w\left(\frac{x^{2}}{2}-x l\right) \end{aligned} $$$$ =\frac{6 \rho g l^{4}}{E h^{2}} \frac{3}{12}=\frac{3 \rho g l^{4}}{2 E h^{2}} $$ (b) As before, $E I \frac{d^{2} y}{d x^{2}}=N(x)$ where $N(x)$ is the bending moment due to section $P B$. This bending moment is clearly $$ \begin{aligned} N &=\int_{x}^{2} w d \xi(\xi-x)-w l(2 l-x) \\ &=w\left(2 l^{2}-2 x l+\frac{x^{2}}{2}\right)-w l(2 l-x)=w\left(\frac{x^{2}}{2}-x l\right) \end{aligned} $$ (Here $w=\rho g b h$ is weight of the beam per unit length) Integrating again, $E I y=w\left(\frac{x^{4}}{24}-\frac{x^{3} l}{6}\right)+\frac{w l^{3} x}{3}+c_{1}$ As $y=0$ for $x=0, c_{1}=0 .$ From this we find $$ \lambda=y(x=l)=\frac{5 w l^{4}}{24} / E I=\frac{5 \rho g l^{4}}{2 E h^{2}} $$

3 (a) In this case $N(x)=\frac{1}{2} \rho g b h(l-x)^{2}$ where $b=$ width of the girder.
Also $I=b h^{3} / 12$. Then,
$$
\frac{E b h^{2}}{12} \frac{d^{2} y}{d x^{2}}=\frac{\rho g b h}{2}\left(l^{2}-2 l x+x^{2}\right)
$$
Integrating, $\quad \frac{d y}{d x}=\frac{6 \rho g}{E h^{2}}\left(l^{2} x-l x^{2}+\frac{x^{3}}{3}\right)$
using $\frac{d y}{d x}=0$ for $x=0 .$ Again integrating
$$
y=\frac{6 \rho g}{E h^{2}}\left(\frac{l^{2} x^{2}}{2}-\frac{b x^{3}}{3}+\frac{x^{4}}{12}\right)
$$
Thus
$$
\begin{aligned}
&\lambda=\frac{6 \rho g l^{4}}{E h^{2}}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}\right) \\
&=\frac{6 \rho g l^{4}}{E h^{2}} \frac{3}{12}=\frac{3 \rho g l^{4}}{2 E h^{2}}
\end{aligned}
$$
(b) As before, $E I \frac{d^{2} y}{d x^{2}}=N(x)$ where $N(x)$ is the bending moment due to section $P B$. This bending moment is clearly
$$
\begin{aligned}
N &=\int_{x}^{2} w d \xi(\xi-x)-w l(2 l-x) \\
&=w\left(2 l^{2}-2 x l+\frac{x^{2}}{2}\right)-w l(2 l-x)=w\left(\frac{x^{2}}{2}-x l\right)
\end{aligned}
$$$$
=\frac{6 \rho g l^{4}}{E h^{2}} \frac{3}{12}=\frac{3 \rho g l^{4}}{2 E h^{2}}
$$
(b) As before, $E I \frac{d^{2} y}{d x^{2}}=N(x)$ where $N(x)$ is the bending moment due to section $P B$.
This bending moment is clearly
$$
\begin{aligned}
N &=\int_{x}^{2} w d \xi(\xi-x)-w l(2 l-x) \\
&=w\left(2 l^{2}-2 x l+\frac{x^{2}}{2}\right)-w l(2 l-x)=w\left(\frac{x^{2}}{2}-x l\right)
\end{aligned}
$$
(Here $w=\rho g b h$ is weight of the beam per unit length)
Integrating again, $E I y=w\left(\frac{x^{4}}{24}-\frac{x^{3} l}{6}\right)+\frac{w l^{3} x}{3}+c_{1}$
As $y=0$ for $x=0, c_{1}=0 .$ From this we find
$$
\lambda=y(x=l)=\frac{5 w l^{4}}{24} / E I=\frac{5 \rho g l^{4}}{2 E h^{2}}
$$

Key Concepts

Boundary Conditions

Boundary conditions are constraints applied at specific points of a beam, such as points of support or where loads are applied, and are essential for solving the integration constants in differential equations. By imposing conditions like zero deflection or zero slope at a given point, one ensures that the mathematical model accurately reflects the physical behavior of the beam.

Integration of Differential Equations

Integration is used to convert the differential equations that describe the beam's curvature into expressions for slope and deflection. These integrations yield relationships between bending moments, slopes (first derivatives of deflection), and deflections (displacements) of the beam, necessitating the determination of constants of integration through the application of boundary conditions.

Young's Modulus

Young's modulus is a material property that measures its stiffness and ability to resist elastic deformation under stress. It plays a critical role in beam theory as it appears in the governing equation of deflection, with higher values indicating a more rigid material that will deflect less under an equivalent load.

Bending Moment

Bending moment is a measure of the internal moment that causes a beam to bend when subjected to external loads. It is calculated as the integral of the applied load distribution over a given length of the beam. Understanding the distribution of the bending moment is crucial as it directly influences the curvature and overall deflection of the beam.

Euler-Bernoulli Beam Theory

This theory governs the bending behavior of beams under load. It establishes that the bending moment in a beam is related to its curvature through the equation EI d²y/dx² = M(x), where E is the Young's modulus, I is the moment of inertia, y is the deflection, and M(x) is the bending moment. This principle forms the basis for analyzing deflection and stresses in structural elements under various loading conditions.

Moment of Inertia

The moment of inertia is a geometrical property of a beam’s cross-section that quantifies its resistance to bending. For common shapes such as a rectangular section, it is calculated using I = (b h³)/12, where b is the width and h is the height of the section. A higher moment of inertia indicates a stiffer beam that is less prone to deflection under load.

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