Uniformly Distributed Load

Uniformly distribnted load is not tested typically at testing facilities because of some technical difficulties. Commonly, tests are conducted using one of those 3-pt or 4-pt loads, described above, and the nniform load is calculated using standard eqnations. 

For a nniformly distributed load on a straight beam (elastically stressed) with its left and right ends simply supported [1], the maximum fiber stress, or flexural strength, can be expressed by the following formnla  

Coefficient 144 just reflects a transition from square inches to sqnare feet. 

Code Section 1606.1 of the BOCA National Building Code/1999 reqnires the minimum uniformly distributed live load to be 100 Ib/fC for main floors, exterior balconies, and other structural systems. In order to evalnate the related properties of composite decking components, the following are calculations of stress imposed on said components at their flexural failure under superimposed uniformly distributed load. 

It was shown above that the flexural stress at center-point load is described by the formula (7.20) 

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If the stress in the composite beam in the previous question is not to exceed 7 MN/m estimate the maximum uniformly distributed load which it could carry over its whole length. Calculate also the central deflection after 1 week under this load. The bending moment at the centre of the beam is lVL/24. 

Ashton observed that skew (parallelogram-shaped) isotropic plates under uniform distributed load Po as shown in the orthogonal X-Y coordinates in Figure 5-9 are governed by the equilibrium differential equation... 

Consider a bar of length L and uniform cross-sectional area A to which an axial, uniformly distributed load with a magnitude, P, is applied at each end (Figure 2-25). Then within the bar there is said to be uniaxial stress o, defined as the load, or force per unit area... 

Now consider a block to which a uniformly distributed load of magnitude P is applied parallel to opposed faces with area A (Figure 2-27). These loads produce a shear stress within the material T. 

Floor loading capacity (point load, uniformly distributed load, line load)... 

Loading (uniformly distributed loads and point loads)... 

SIMPLY SUPPORTED BEAM UNIFORMLY DISTRIBUTED LOAD... 

Compare the thickness required for a 2 m diameter flat plate, designed to resist a uniform distributed load of 10 kN/m2, if the plate edge is ... 

Where q is the uniformly distributed load, a is the side of the square plate, E is its elastic modulus, and h is the plate thickness. Substituting the measured values in this expression, we obtain E=71 N/mm2 referring to table (1), we observe that this value is about twice the value obtained by the direct measurements. This difference could be due to geometrical effects which are present in panel testing, but are not too active in tension tests. 

From the above equation, a break load (an ultimate load) for a standard Trex board equals to 509 lb. This would translate to an ultimate uniformly distributed load of 1667 Ib/ftl... 

Similar calculations for GeoDeck deck show that at flexural strength of the board (hollow boards of 5.5" width and 1.25" thickness and moment of inertia of 0.733 in. ) of 2782 psi, a break load at 16" span (center point load) would be 816 lb. This would translate to an ultimate uniformly distributed load of 2670 Ib/fC, which is more than 10 times higher than the ICC required load including the necessary safety factor. 

Deflection under uniformly distributed load is determined by the following formula ... 

For Trex boards E = 175,000 psi as reported by Trex, with / = 0.895 in. ), deflection under uniformly distributed load of 100 Ib/in. at 16-in. span would be 0.021 in., which is within the building code requirements. However, at support span of 24 in., deflection at the same conditions would be 0.105 in. that significantly exceeds the allowable limitation (24 7360 = 0.0667 in.). 

PI = specimen overall depth, in in. h = specimen s hollow part depth, in in. and W = uniformly distributed load, Ib/ft. ... 

At the moment of break, load P equals to the ultimate load, or load at failure, and S equals to the flexural strength value. In order to reach the same maximum stress in the outer surface of the board, the uniformly distributed load (W) should be as follows ... 

One can see that quarter-point load (in psi) is indeed equivalent to uniformly distributed load (in psf) with respect to stress in the outer surface. That is why quarter-point load is often employed as a proxy for uniformly distributed load. 

These requirements to deck boards used as stair treads are much more severe compared to requirements to regular deck board, among them an ability to hold a uniformly distributed load of 100 Ib/fC multiplied by a safety factor of 2.5, hence, 250 Ib/fC. 

As it was indicated above, to apply a uniformly distributed load testing is a technically difficult task therefore, testing is usually done using three- or four-point load. As it was shown above, Eq. (7.42)-(7.44) describe equivalency between these loads in terms of their equal flexural stress at the outer surface ... 

Therefore, if load at failure for uniformly distributed load is 250 Ib/ft for a deck board of 5-1/2 in. wide and 16 and 20 in. long, the same stress will be reached for values, shown in Table 7.27. 

Load Load in lb, equivalent to 250 Ib/fE at the uniformly distributed load at support span ... 

As it was described above, a deck should satisfy a building code requirement of the uniformly distributed load of 250 Ib/fF. (100 Ib/fF of the requirements itself, with a factor of safety of 2.5). In some snowy areas there is a more stringent requirement for a deck of the uniformly distributed load of 400 Ib/fC. Let us consider a joist support span rating for this latter case with an example of GeoDeck. 

An alternative two sets of data (Tables 7.18, 7.19 and 7.39, 7.40) give us flexural strength and modulus for GeoDeck at ambient temperature, at 10 and at -20°F. Linear extrapolation of these numbers (flex strength and modulus separately, of course) to 32°F gives us flexural strength values of 2430 and 2820 psi, and flexural modulus of 445,000 and 360,000 psi, respectively (from two separate sets of data). From Eq. (7.41) we get that the uniformly distributed load at failure is equal to... 

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