Beam Deflections by Integration: Statically Indeterminate Beam with One End Displaced and Rotated

Now we will look at the case of a beam with fixed but non-zero displacement and rotation at one end. Consider, for example, the beam of the previous example (here), but now with the far end displaced downward 0.0062 in. and also rotated clockwise 0.00011 radian. (Suppose, for example, the beam is part of a larger frame that was `hit' by a forklift or something during construction.) We want to determine the curved (bent) shape of the beam, the maximum deflection, and also the maximum Moment(s) and Shear(s) for the beam.

Recall from our earlier discussions that in general our engineering materials are stiff, and that is good, in that deflections resulting from normal loading conditions will produce relatively small deflections. However, in the case of statically indeterminate beams, relatively small displacements and or rotations can (because of the stiffness of the materials used) generate huge internal stresses.

Procedure

We will proceed as before, and when we come to imposing the end constraints to solve for constants of integration, we will put in the non-zero imposed conditions. In so doing we will need to be careful with our units.

From before (here) ...

w(x) = 50 x

V(x) = - 25 x ² + C₁

M(x) = - 8.33 x³ + C₁ x + C₂

EI θ (x) = - 2.0833 x⁴ + 0.50 C₁ x² + C₂ x + C₃

EI v (x) = - 0.4167 x⁵ + 0.1667 C₁ x³ + 0.5 C₂ x² + C₃ x + C₄

In this example the left end is taken to have no displacement or rotation, thus C₃ = 0 and C₄ = 0.

So,

EI θ (x) = - 2.0833 x⁴ + 0.50 C₁ x² + C₂ x

EI v (x) = - 0.4167 x⁵ + 0.1667 C₁ x³ + 0.5 C₂ x²

We will now solve for C₁ and C₂ using the imposed non-zero conditions.

Solution

Leaving, for now, our equation in units of ft, we first substitute the imposed rotation at the end of the beam.

The clockwise rotation of 0.00011 radians is a negative slope(ft / ft) .

Therefore,

EI θ (x=10) = - 2.0833 (10)⁴ + 0.50 C₁ (10)² + C₂ (10) = EI (-0.00011 rad)

Substituting EI = 810,000,000 lb-in.² = 5,625,000 lb-ft²

(This is different than our previous example where we went down to in. units.)

5,625,000 lb-ft² (-0.00011 rad) = - 618.8 lb-ft² = - 20,833 + 50 C₁ + 10 C₂ lb-ft²

- 618.8 = - 20,833 + 50 C₁ + 10 C₂

20,214 = 50 C₁ + 10 C₂

Now let's substitute the imposed displacement (getting in. into ft) ...

EI v (x=10) =

- 0.4167 (10)⁵ + 0.1667 C₁ (10)³ + 0.5 C₂ (10)² = - 0.0062 in. (1 ft/12 in.) EI = - 0.000517 ft EI

- 0.000517 ft (5,625,000 lb-ft² ) = - 2906 lb-ft³

- 2906 = - 41,667 + 166.7 C₁ + 50 C₂

38,761 = 166.7 C₁ + 50 C₂

Solving these two equations simultaneously for

C₁ and C₂ gives

C₁ = 748 lb

and

C₂ = - 1719 lb-ft

Therefore our deflected shape, slope, moment, and shear equations are

w(x) = 50 x

V(x) = - 25 x ² +748 lb

M(x) = - 8.33 x ³ + 748 x + -1719 lb-ft

EI θ (x) = - 2.0833 x⁴ + 0.50 (748) x² + C₂ x = - 2.0833 x⁴ + 374 x² - 1719 x lb-ft²

EI v (x) = - 0.4167 x⁵ + 0.1667 (748) x³ + 0.5 (-1719) x² = - 0.4167 x⁵ + 124.7 x³ - 860 x²

Let's check equilibrium.

V(0) = 748 lb ... so the left hand Reaction is 748 lb upward.

V(10) = - 25(10)² + 748 = - 1752 lb ... and thus the right hand Reaction is 1752 lb upward.

Σ F_y = 748 + 1752 - 2500 = 0? ... yes, good.

Moment equilibrium ...

M(0) = - 1719 lb-ft

M(10) = - 2573 lb-ft

Σ M _{left end} (CCW positive) = 1719 - 2/3 (10) (2500) + 10 (1752) - 2573 = 0? ... yes, good.

Let's look at Critical Values ... (calculus solution or simple crunching out the numbers not shown) ...

V _max = 1752 lb at the right support.

M _{max pos} = about 1010 lb-ft at 5.5 ft from the left

M _{max neg} = about 2570 lb-ft at the right support.

Deflection

Now we have an equation describing the (bent) shape of the beam ... v(x) = ( - 0.4167 x⁵ + 124.7 x³ - 860 x² )/EI.

It is not a whole lot different than the equation for the beam shape with right end fixed but not displaced or rotated. And this is because the magnitudes of the rotation and displacement are very small. Increase the imposed displacement and rotation values by factors of ten, and the resulting reactions and internal loads for the beam will be greatly affected.

This equation gives us v which is the departure of the beam (axis) from an originally straight (and presumably horizontal line). The question is how do we want to measure deflection ??? ... since the starting condition of the beam in this problem is not horizontal.

If we want to measure deflection as being the difference between the curved shape and a line between the ends, then we should subtract from the above equation an equation for the straight line between the ends ... (and watch our units).

Comparing the maximum shear and moment values with those of the shape without displacement and rotation of the right end likewise shows differences. These differences are not huge, but likewise not insignificant. Recall that the imposed displacement and rotation values are very small. Also, in a sense the clockwise rotation tends to lessen the maximum postive moment and net deflection.

References

Beam Deflections by Method of Integration: Statically Indeterminate Beams, Jeff Filler, Associated Content.