# 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.