4- A thin rectangular plate of unit thickness is loaded along the edge y=+d by a linearly varying distributed load of intensity w=px with corresponding equilibrating shears along the vertical edges at x=0 and I. As a solution to the stress analysis problem, an Airy stress function varnothing is proposed as emptyset =(P)/(120d^3)[5(x^3-l^2x)(y-2d)-3xy(y^2-d^2)^2] Show that sb satisfies the internal compatibility conditions and obtain the distribution of stresses within the plate Determine also the extent to which the static are satisfied. Hint: sigma _(xx)=(partial ^2phi )/(partial y^2) etc. and biharmonic eq. (partial ^4emptyset )/(partial x^4)+2(partial ^4emptyset )/(partial x^2)partial y^(2)+(partial ^4emptyset )/(partial y^4)=0 (25 points)

To show that the Airy stress function $\varnothing$ satisfies the internal compatibility conditions, we need to verify that it satisfies the biharmonic equation:<br /><br />$\frac{\partial^{4}\varnothing}{\partial x^{4}} + 2\frac{\partial^{4}\varnothing}{\partial x^{2}\partial y^{2}} + \frac{\partial^{4}\varnothing}{\partial y^{4}} = 0$<br /><br />Let's start by finding the fourth partial derivatives of $\varnothing$ with respect to $x$ and $y$.<br /><br />First, we find the first partial derivatives of $\varnothing$ with respect to $x$ and $y$:<br /><br />$\frac{\partial\varnothing}{\partial x} = \frac{P}{120d^{3}}[15(x^{2}-l^{2})(y-2d)-3y(y^{2}-d^{2})^{2}]$<br /><br />$\frac{\partial\varnothing}{\partial y} = \frac{P}{120d^{3}}[5(x^{3}-l^{2}x)(1)-6xy(y^{2}-d^{2})]$<br /><br />Next, we find the second partial derivatives of $\varnothing$ with respect to $x$ and $y$:<br /><br />$\frac{\partial^{2}\varnothing}{\partial x^{2}} = \frac{P}{120d^{3}}[30(x-l^{2})-3y(y^{2}-d^{2})^{2}]$<br /><br />$\frac{\partial^{2}\varnothing}{\partial x\partial y} = \frac{P}{120d^{3}}[15(x^{2}-l^{2})-6xy(y^{2}-d^{2})]$<br /><br />$\frac{\partial^{2}\varnothing}{\partial y^{2}} = \frac{P}{120d^{3}}[5(x^{3}-l^{2}x)-6y(y^{2}-d^{2})]$<br /><br />Now, we find the third partial derivatives of $\varnothing$ with respect to $x$ and $y$:<br /><br />$\frac{\partial^{3}\varnothing}{\partial x^{3}} = \frac{P}{120d^{3}}[30]$<br /><br />$\frac{\partial^{3}\varnothing}{\partial x^{2}\partial y} = \frac{P}{120d^{3}}[15-12y(y^{2}-d^{2})]$<br /><br />$\frac{\partial^{3}\varnothing}{\partial y^{2}\partial x} = \frac{P}{120d^{3}}[15-12y(y^{2}-d^{2})]$<br /><br />$\frac{\partial^{3}\varnothing}{\partial y^{3}} = \frac{P}{120d^{3}}[-12(y^{2}-d^{2})]$<br /><br />Finally, we find the fourth partial derivatives of $\varnothing$ with respect to $x$ and $y$:<br /><br />$\frac{\partial^{4}\varnothing}{\partial x^{4}} = 0$<br /><br />$\frac{\partial^{4}\varnothing}{\partial x^{2}\partial y^{2}} = \frac{P}{120d^{3}}[-12(y^{2}-d^{2})]$<br /><br />$\frac{\partial^{4}\varnothing}{\partial y^{4}} = \frac{P}{120d^{3}}[144(y^{2}-d^{2})]$<br /><br />Now, we can substitute these derivatives into the biharmonic equation:<br /><br />$0 + 2(-\frac{12P}{120d^{3}})(y^{2}-d^{2}) + \frac{144P}{120d^{3}}(y^{2}-d^{2}) = 0$<br /><br />Simplifying, we get:<br /><br />$0 = 0$<br /><br />This shows that the Airy stress function $\varnothing$ satisfies the biharmonic equation.<br /><br />To obtain the distribution of stresses within the plate, we can use the stress function to find the normal and shear stresses. The normal stress $\sigma_{xx}$ is given by:<br /><br />$\sigma_{xx} = \frac{\partial^{2}\varnothing}{\partial y^{2}}$<br /><br />Substituting the expression for $\varnothing$, we get:<br /><br />$\sigma_{xx} = \frac{P}{120d^{3}}[5(x^{3}-l^{2}x)-6y(y^{2}-d^{2})]$<br /><br />Similarly, the shear stress $\tau_{xy}$ is given by:<br /><br />$\tau_{xy} = \frac{\partial^{2}\varnothing}{\partial x\partial y}$<br /><br />Substituting the expression for $\varnothing$, we get:<br /><br />$\tau_{xy} = \frac{P}{120d^{3}}[15(x^{2}-l^{2})-12y(y^{2}-d^{2