I'm stuck on the following question. I feel like I have computed the correct results but because my units have cancelled I'm a bit lost.
Anyway, here's the question:
If a beam of length L is fixed at the ends and loaded in the centre of the beam by a point load of F Newtons the deflection at a distance x from one end is given by:
\( y = \frac{F}{48EI} (3L^{2}x - 4x^{3})\)
Where E = Young's Modulus and I = Second Moment of Area of the beam. Find the position and the value of the maximum deflection of the beam if:
L = 10m
EI = \( 30_{10^{6}}Nm^{2} \)
F = 100kN
\( y = \left(\frac{100kN}{(48)(30_{10^{6}}Nm^{2})}\right) ((3)(10m)^{2}x - 4x^{3})\)
\( y = \left(\frac{1}{14400}\right) (300x - 4x^{3}) \)
\( y = \frac{1}{48} (x) - \frac{1}{3600} (x^{3}) \)
\( \frac{dy}{dx} = \frac{1}{48} - \frac{1}{1200} (x^{2}) = 0\ for\ a\ maximum\ or\ minimum\ value \)
\( x = \sqrt{\frac{1200}{48}} = \pm5\)
\( \frac{d^{2}y}{dx^{2}} = -600x \)
\( \frac{d^{2}y}{dx^{2}} = - (600)(5) = -3000\ for\ a\ maximum\ value \)
Substituting x= (+5) into the initial equation for a maximum deflection:
\( y = \left(\frac{1}{14400}\right) ((300)(5) - (4)(5^{3})) = \frac{5}{72} \)
If I envisage this geometrically, I'd say the position of the beam is (+5) for a maximum deflection, could some one please confirm this.
Thanks in advance
Anyway, here's the question:
If a beam of length L is fixed at the ends and loaded in the centre of the beam by a point load of F Newtons the deflection at a distance x from one end is given by:
\( y = \frac{F}{48EI} (3L^{2}x - 4x^{3})\)
Where E = Young's Modulus and I = Second Moment of Area of the beam. Find the position and the value of the maximum deflection of the beam if:
L = 10m
EI = \( 30_{10^{6}}Nm^{2} \)
F = 100kN
\( y = \left(\frac{100kN}{(48)(30_{10^{6}}Nm^{2})}\right) ((3)(10m)^{2}x - 4x^{3})\)
\( y = \left(\frac{1}{14400}\right) (300x - 4x^{3}) \)
\( y = \frac{1}{48} (x) - \frac{1}{3600} (x^{3}) \)
\( \frac{dy}{dx} = \frac{1}{48} - \frac{1}{1200} (x^{2}) = 0\ for\ a\ maximum\ or\ minimum\ value \)
\( x = \sqrt{\frac{1200}{48}} = \pm5\)
\( \frac{d^{2}y}{dx^{2}} = -600x \)
\( \frac{d^{2}y}{dx^{2}} = - (600)(5) = -3000\ for\ a\ maximum\ value \)
Substituting x= (+5) into the initial equation for a maximum deflection:
\( y = \left(\frac{1}{14400}\right) ((300)(5) - (4)(5^{3})) = \frac{5}{72} \)
If I envisage this geometrically, I'd say the position of the beam is (+5) for a maximum deflection, could some one please confirm this.
Thanks in advance