Read Structural Loading text version

Structural Loading

Objectives:

1) Set up the concentrated load vector for a structure 2) Calculate equivalent loads when loads not at DOF 3) Calculate the equivalent loads for distributed loading Remember the stiffness equation:

Kr = R

The only type of loads allowed are concentrated loads acting directly on the DOF. For loads are applied NOT at the DOF, we can use a work equivalence method for finding the equivalent load value. Work of Real Loads = Work from Equivalent Loads As an example, for the structure shown only two DOF are considered, b and . Mapplied b c

b

a

Example of Load not at DOF If a concentrated moment load is applied, we need to find the equivalent load for the DOF considered. Using the displaced shape for b,

Direct Stiffness using Full a matrix - Page 1 of 6

we see that the real work is: W real = M c * c Notice that for the displaced shape, c has a non-zero value. This is because it is not one of the DOF and must be left free to rotate. The equivalent work is the desired load for DOF b. This is calculated by; W equiv. = M b * 1 This is the only unknown force since the other DOF, , is not allowed to move and hence its load generates no work. Now c can be calculated from the slope deflection formula based on a unit rotation of b. Since it is known that the moment Mcb is equal to zero, we can solve for c. The required equation is; 0 = M cb = 2EI Lcb (2 b + c + 3 Lcb )

But = 0 and b = 1, therefore solving for c we get c = -1/2. This gives: Mc R1 = 2 We can do the same for the DOF. By applying a unit we can see that no rotation occurs at c. The final load vector is therefore: Mc R= 2 0

Direct Stiffness using Full a matrix - Page 2 of 6

Distributed Loads

First convert all distributed loads into concentrated loads. Second use equivalent work to convert to an equivalent loading. We need an equivalent set of concentrated loads that cause the same deformation as the uniform load.

W (force/length)

Uniformly Loaded Simple Beam

Both ends rotate an amount . The fixed end moments are the exact forces required to cause no rotation at the ends. We can show this using the following structure.

W (force/length)

Uniformly Loaded Fixed End Beam

Using consistent deformations, we can remove the moments and then solve for the moments that push the displacement back to zero (the redundant support moments).

Direct Stiffness using Full a matrix - Page 3 of 6

wL2 12

wL2 12

Equivalent Fixed End Moments

As can be seen, the final displacements are identical between these loads and the distributed loads.

Distributed Load Example

Find the equivalent concentrated load vector:

For a triangularly distributed load the fixed end forces are:

Direct Stiffness using Full a matrix - Page 4 of 6

W (force/length)

wL2 - 30 wL2 20

7wL 20 3wL 20

Fixed End Forces for Triangular Load

From these reaction forces we convert the distributed load on the structure to the concentrated loads as shown:

Now use the displaced shapes to find the work equivalent loads. From the displaced shape with a unit rotation at DOF 1 we have the work equation:

From the displaced shape for DOF 2, we have the work equation:

Direct Stiffness using Full a matrix - Page 5 of 6

From the displaced shape for DOF 3, we have the work equation: 3( 2.5 )96( .6 ) 12 = 1.8 K 20

R3 = -

3wL [.6 (1)] = 20

Notice how the vertical displacement has a work component as a result of the horizontal displacement. The slanted member linking the vertical and horizontal DOF causes this. This makes the final concentrated load vector:

- 96 R = 64 1.8

Direct Stiffness using Full a matrix - Page 6 of 6

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Structural Loading

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