Deflection of a Statically Determinate Beam by Integration

Now we will use the method of integration to `solve' for curved shape (deflections) of a statically determinate beam.

Background

We will assume that the deflections are small, and thus the slopes of the curved shape are very shallow, and thus, the Euler-Bernoulli Beam Equation applies, namely,

- w / EI = d⁴ v / dx^4,,

where,

w is the distributed loading (positive downward),

E and I are (in this case) constants (the Modulus of Elasticity and the Moment of Inertia),

v is the amount of deviation of the axis of the beam from the straight (un-bent) shape (positive upward), and

x is the distance along the beam (positive starting from the left).

The negative sign in the equation derives from w positive downward and v positive upward.

The beam is `flexing' within the `linear elastic range' of the material (no plastic hinges, etc.). And, for us to `integrate', the load is some algebraic expression in terms of `x'.

Procedure

We will integrate the load equation to get shear, shear to get moment, moment to get slope, and slope to get curved shape. Since the beam is statically determinate, our `bounding' shears and moments are known.

Example

As an example let us use a beam with span of 10 ft, simply supported (free to rotate) at ends, and loaded with a triangular load starting with 0 plf at one end and ending at w_o = 500 plf at the other. We will use a beam with EI = 810,000,000 lb-in². The self weight of the beam will be neglected.

Solution

Step 1 - Algebraic Expression for the Load.

The load may be expressed as

w(x) = 50 x,

( ... w = - EI d⁴ v / dx⁴ ...)

where the 50 has units of plf/ft. For convenience the units will not be carried through all the calculation steps.

Step 2 - Integrate Load to get Shear.

V(x) = ∫ (-) w(x) dx + C₁ ...

= - ∫ 50 x dx + C₁

= - 25 x ² + C₁

We can find out from `statics' what the shear values are at the ends ... and thus we can solve for C₁ in the above equation.

(Maybe this should be Step 1 ... anyway ...)

The shear values at the ends are ...

V (x=0 ft) = 833 lb, and

V (x=10 ft) = - 1667 lb.

The moments at the ends are ...

M (x=0 ft) = 0 lb-ft, and

M (x=10 ft) = 0 lb-ft.

So,

Letting V (x=0) = 833 lb ...

833 lb = - 25 (0) ² + C₁

C₁ = 833 lb.

(The constant of integration here becomes the beginning shear.)

And we can check ...

Does V(x=10) = - 1667 lb?

Let's see ...

... - 25 (10)² + 833 =? -1667 lb.

... - 1667 lb = - 1667 lb ... Good!

So,

V = - 25 x² + 833 ...

( ... V = EI d³ v / dx³ ... )

Step 3 - Integrate Shear to Get Bending Moment

M(x) = ∫ V(x) dx + C₂

=∫ ( - 25 x ² + 833) dx + C₂

= - 25 x³ / 3+ 833 x + C₂

= - 8.33 x³ + 833 x + C₂

But we know what M is at x = 0 ... so let's put that to use and solve for C₂ ...

M (x=0) = 0 = - 8.33 x³ + 833 x + C₂ ...

C₂ = 0 ... that's nice.

And, yeah, the constant of integration is the starting moment, which is zero.

Let's check ...

M (x=10) =? 0 ...

... does - 8.33(10)³ + 833 (10) = 0? ... Yes! Great!

With confidence we continue!

M = - 8.33 x³ + 833 x ... = ... EI d² v / dx² ...

Step 4 - Integrate Moment (M/EI) to get Slope (θ = dv/dx) ...

EI θ (x) = ∫ M(x) dx + C₃

= ∫ (- 8.33 x³ + 833 x) dx + C₃

= - 8.33 x⁴ / 4 + 833 x² / 2 + C₃

= - 2.0833 x⁴ + 416.7 x² + C₃

We don't know what the initial slope is ... the simple beam allows the ends to rotate (albeit with real beams the slopes are shallow) ... so we can't solve for C₃. We will leave it as an unknown for now.

Step 5 - Integrate Slope to get Curved Shape.

EI v (x) = ∫ EI θ (x) dx + C₄

= ∫ ( - 2.0833 x⁴ + 416.7 x² + C₃ ) dx + C₄

= - 2.0833 x⁵ / 5+ 416.7 x³ / 3 + C₃ x + C₄

= - 0.4167 x⁵ + 138.89 x³ / 3 + C₃ x + C₄

We can solve quickly and easily for C₄ by knowing that the displacement at the left end is zero.

EI v (x=0) = 0 = = - 0.4167 (0)⁵ + 138.89 (0)³ + C₃ (0) + C₄

So C₄ = 0.

Our shape of the bent beam is, therefore,

EI v(x) = - 0.4167 x⁵ + 138.89 x³ + C₃ x

We still have C₃ unknown. However, we can employ our other boundary condition on deflection, namely at the other end of the beam the displacement is zero.

Step 6 - Impose additional boundary condition(s) to solve for remaining unknown constant(s) of integration.

At the far end the displacement is zero ... so,

EI v(x=10) = 0 ...,

0 = - 0.4167(10)⁵ + 138.89 (10) ³ + C₃ (10)

Gives,

C₃ = ( -138,890 + 41,670 ) / 10 = - 9,722.

So,

EI v (x) = - 0.4167 x⁵ + 138.89 x³ + - 9722 x.

All of these terms have units of lb-ft³. (We started with lb/ft and integrated four times with respect to length.)

We could also write,

v (x) = ( - 0.4167 x⁵ + 138.9 x³ +- 9722 x ) / EI

Step 7 - Plug in values for EI and now we have our deflection equation.

We started with w and it was in units of lb/ft.

We then integrated four times with respect to distance giving us lb/ft times ft⁴ ... or lb-ft³.

Thus, if we put in units of x in ft, and multiply by 12 in.³ / ft³ ... we'll get deflection in inches ... (lb-in.³ / lb-in.² = in.).

12³ = 1728 ...

So ... v(x) in in. = (- 0.4167 x⁵ + 138.9 x³ - 9722 x ) 1728 / 810,000,000 in. ... x in ft.

Let's calc some values ...

x (ft), v (in.)

0, 0.00

1, -0.20

2, -0.39

3, -0.54

4, -0.65

5, -0.69

6, -0.67

7, -0.59

8, -0.43

9, -0.23

10, 0.00

It looks like the maximum deflection is somewhere between (x = ) 5 and 6 ... we can find it by plugging in more numbers ... or we could take the equation of the slope curve, set it equal to zero, find out where the slope goes to zero, and plug that value of x into our v(x) equation.

We've done a lot of calculus already, so I'm just going to find the max v by plugging in various values of x and looking at v ...

... v _max = - 0.70 in. at x = 5.2 ft.

Note that the beam bends a bit heavy to the right ... which is what we would expect, since it's loaded more to the right.

Step 8 - Check

We always want to find a way to check, either approximately, or exactly, our solution.

As an approximate check, let's take the same amount of total load ... ½ (500 plf)(10 ft) = W = 2500 lb ... and spread it uniformly over the beam ... 2500 lb / 10 ft = 250 plf. We can find a published solution rather easily for that case ...

... Δ = (5/384) w L⁴ / EI ... (Case 1, Simple Beam - Uniformly Distributed Load, Appendix A.2 Beam Diagrams and Formulas, Timber Construction Manual, 5^th Edition, American Institute of Timber Construction.)

So, ...

Δ = (5/384) (250 /12 lb / in.) (120 in.)⁴ / 81,000,000 lb-in.² = ...

... 0.69 in.

Whoa ... nice! (And it would be at x = 5.0 ft.)

This may not be apparent to the beginner ... but as long as the supports are the same, and the beam is the same (same EI) ... then how (where) the load is distributed does not have a strong effect on (changing) the deflection. (All that integration kind of `smooths' things out.) I won't say this for slope, and especially for Moment, Shear, and Reactions ... and there are several reasons (for not saying it). But I will not get into that here. (I mean, look, that should be apparent: on the left the maximum shear is 833 lb, where on the right it is 1667 lb ... way different!)

References

Timber Construction Manual, 5^th Edition, American Institute of Timber Construction, Wiley, 2005.