using equation 427 compute the sagged equilibrium position uexifqxt g the boundary conditions ar

Name: using equation 427 compute the sagged equilibrium position uexifqxt g the boundary conditions are uo 0 and ul0 50456
Uploaded: 2021-09-14T05:23:57-08:00
Duration: 6 min 20 s
Channel: Khoobchandra Agrawal
Description: using equation 427 compute the sagged equilibrium position uexifqxt g the boundary conditions are uo 0 and ul0 50456

To compute the sagged equilibrium position ( u_E(x) ) given the equation ( Q(x,t) = -g ) and the

Using equation 427 compute the sagged equilibrium position...

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We use the equation given above and use the result that when $y$ is small $$ \frac{1}{R} \sim \frac{d^{2} y}{d x^{2}} . \text { Thus, } \frac{d^{2} y}{d x^{2}}=\frac{N(x)}{E I} $$ (a) Here $N(x)=N_{0}$ is a constant. Then integration gives, $$ \frac{d y}{d x}=\frac{N_{0} x}{E I}+C_{1} $$ But $\quad\left(\frac{d y}{d x}\right)=0$ for $x=0$, so $C_{1}=0$. Integrating again, $$ y=\frac{N_{0} x^{2}}{2 E I} $$ where we have used $y=0$ for $x=0$ to set the constant of integration at zero. This is the equation of a parabola. The sag of the free end is $$ \lambda=y(x=l)=\frac{N_{0} l^{2}}{2 E I} $$ (b) In this case $N(x)=F(l-x)$ because the load $F$ at the extremity is balanced by a similar force at $F$ directed upward and they constitute a couple. Then $$ \frac{d^{2} y}{d x^{2}}=\frac{F(l-x)}{E I} $$ Integrating, $\frac{d y}{d x}=\frac{F\left(L x-x^{2} / 2\right)}{E I}+C_{1}$ As before $C_{1}=0 .$ Integrating again, using $y=0$ for $x=0$ $$ y=\frac{F\left(\frac{b^{2}}{2}-\frac{x^{3}}{6}\right)}{E I} \text { here } \lambda=\frac{F l^{3}}{3 E I} $$ Here for a square cross section $$ I=\int_{-a / 2}^{a / 2} z^{2} a d z=a^{4} / 12 $$

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Consider the transverse displacement of a vibrating beam of length L whose ends are in simply-supported conditions: The beam is set in motion with initial velocity g(x) and initial displacement given by f(x), 0 < x < L. The transverse displacement u(x,t) can satisfy the following governing equation, initial and boundary conditions: 0 < x < L, t > 0 where: u(0,t) = 0, u(L,t) = 0, u(x,0) = f(x), u'(x,0) = g(x), 0 < x < L where E is the bending stiffness, p is the density, and A is the cross-sectional area of the beam. Using the method of separation of variables (i.e. let u(x,t) = X(x)T(t)), derive the following ordinary differential equations: d^2X/dx^2 + a^2X = 0 dT/dt + a^2T = 0 where a is a positive real constant. Determine the general solutions of the ordinary differential equations. Determine the fundamental solutions of the partial differential equation and boundary conditions. Determine a formal series expansion for the transverse displacement u(x,t) that also satisfies the initial conditions.

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Transcript

00:03 Hi friends, we use the equation given and use the result that when y is small, we can write 1 upon r is called to d2 by up on dx2 and thus d2 y upon dx2 is called to an x upon e .i.

00:51 First part, moment of force is called to n .0 is constant.

01:05 Then integration will give you d .y upon dx is called to n, not x over e .i plus c1.

01:31 But d .y upon dx is zero for x is called to zero.

01:39 So c1 will be zero.

01:41 So above equation you can write y is equal to n -not x square upon 2 e i where we have used y -s -cultu -0 for x -s -cult -0 set the constant of integration to be 0.

02:24 This is the equation of parabola...

Question

Using Equation (4.2.7), compute the sagged equilibrium position uE(x)ifQ(x,t)= −g. The boundary conditions are u(O) = 0 and u(L)=0

Question

Using Equation (4.2.7), compute the sagged equilibrium position uE(x)ifQ(x,t)= −g. The boundary conditions are u(O) = 0 and u(L)=0

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