Homework Help: Help with problem

1. Jan 6, 2005

physkid

moments problem

Hi everyone,

Just wondering if you could help with this problem a formula will do cos i don't know where to start.

A steel beam 50 feet in length is suspended with pin connections at either end and bears a load of 6000 pounds in the centre. Its elasticity is 29 000 000 pounds per square inch and its moment of inertia is 850 pounds^4.

How far is the beam deflected?

Thanks iin advance for any help.

Last edited: Jan 6, 2005
2. Jan 6, 2005

spiderbyte

I too would like the answer to this question. Please note that all loads other than the initial load of 6000lbs (ie weight of the beam itself) is to be disregarded.

I am told that there is a graph which charts information about specific modules of elasticity. Does anyone know where I could find such a graph?

3. Jan 6, 2005

Gamma

Check the units.

4. Jan 6, 2005

Gamma

Try the following:

Draw a diagram and set the angle made by the deflacted beam be $$\theta$$. Use the formula for the elasticity, $$Y = \frac{stress} {strain}$$. Stress would be the tension devided by cross sectional area. $$Strain = \frac{length change} {original length}$$ which is a function of $$\theta$$.

To find cross sectional area use the given information about moment of inertia, length of the beam and the density of steel ( which is not given) .

gamma.

Given moment of inertia is with respect to which axis?

5. Jan 7, 2005

Astronuc

Staff Emeritus
The units of moment of inertia should be in4 in the English system.

Solving the equation for the deflection of a beam (of length, L) subject to a point (local) force in the center (L/2), one finds that the deflection given as a function of distance is:

$y = \frac{Fx}{48EI}(3L^2 - 4x^2)$, where x is taken as positive from the left hand side, F is the point force at the center along, E is the elastic modulus, and I is the moment of inertia taken with respect to the beams major axis. By convention, F is negative (-) for downward forces and positive (+) for upward forces (assuming a horizontal beam).

At the hinged ends, the moments are zero.

If weight of the beam were a factor, the weight (or mass) would have to be considered as a distributed force.

Has the instructor provided the beam equation and boundary conditions?

6. Jan 7, 2005

bbg5000

What is x.. I don't get it. I understand everything else except for x. Help!?

7. Jan 7, 2005

Astronuc

Staff Emeritus
x is the distance from the left hinge, which increases from 0 to L/2.

The equation is for the deflection y from the unloaded (undeformed) axis as a function of x, i.e. y=y(x), where y is the displacement at x.

The maximum deflection is located at L/2. The deflection a the hinge is 0.

If the force is down (-), then the deflection should be down (-).

Also, some equations use P for load rather than F for force.

8. Jan 7, 2005

bbg5000

So x would be where the force is on the beam? So if the force was in the center, and the beam was 50 ft(600in) long, then x = 300in!?

F (force)= 6,000lbs
E (Modulus of elasticity) = 29,000,000
I (Moment of inertia) = 850
L (length) = 600 in

So if I understand this right, then this would be...

y = [(6,000 x 300) / (48 x 29,000,000 x 850^4)] (3 x 600^2) - (4 x 300^2)
y = 1.09533 ?!?!??!? ?? ??!?!

9. Jan 7, 2005

Astronuc

Staff Emeritus
In the English system, the units for this problem are:

Load = pounds force = lbf
Modulus of Elasticity = lbf/in2 = psi
Moment of inertia = in4
Length = in

I will allow you to determine the units of deflection.

Regarding the Modulus of Elasticity (Young's Modulus) E, this relates stress with strain in the elastic region of the stress-strain diagram. Stress has units of force per unit area (or lbf/in2, psi or ksi) just like pressure and strain is dimensionless, therefore E has units of pressure (e.g. psi or ksi).

1 ksi = 1000 psi.

10. Jan 9, 2005

Astronuc

Staff Emeritus
I have posted the solution to this problem - https://www.physicsforums.com/showthread.php?p=423455#post423455.

The above formula should reduce to $\delta = (PL^3)/(48EI)$ at L/2, point of maximum deflection.

The term I = 850 in4, but the value 850 is not raised to the power of 4.

The answer is then $\delta$ = 1.0953 in.