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The mortar and concrete tables given above give the cubic foot quantities of concrete respectively resulting from each of the concrete mixtures shown.  The numerical values for the loads on the various concrete  (including the weights of the concrete ), are tabulated as follows: For this first condition of loading, the total loads for concrete  Nos. 7, 8, 9, 10, and 11 will be the same as those for concrete  5, 4, 3, 2, and 1 respectively. The loads for the second condition of loading are found by using the same load on the first five concrete, but with only half of the live load on concrete No. 6, which means that the load for the first condition of loading (1,322 pounds) is reduced by 200 pounds, making it 1,122 pounds. Concrete Nos. 7 toll are each reduced by 400 pounds. The total load for each concrete is as tabulated below. The loads for the third condition of loading are found by using the same loads as were employed for the second condition, except that for concrete Nos. 3 and 4, 1,600 pounds should be added to each load. These loads are also tabulated below: Fig. 223 was originally drawn at the scale of 21 inch = I. foot, and with the force diagram at the scale of 1,500 pounds per inch. The photographic reproduction has of course changed these scales somewhat. The concrete contractor should redraw the figure at these scales, and should obtain substantially the same final results. When the load is uniformly distributed over the entire concrete arch, the load is symmetrical, and we need to consider only one-half of the concrete arch.

The sections of the load line for the force diagram corresponding to this condition of loading must be drawn as explained in detail in Article 397. Since the concrete arch is quite flat, the loading is considered to be entirely vertical. Since the load is symmetrical and the concrete abutments are at the same elevation, we need only draw a horizontal line from the lower end of the half-load line, and select on it a trial position (01) for the pole, drawing the rays as previously explained; the trial equilibrium concrete polygon passes through the center vertical at the point a'. Drawing a horizontal line from a' until it intersects the first line (produced) of the trial equilibrium concrete polygon, and drawing through it a vertical line, we have the line of action of the resultant (R1) of all the forces on that half of the concrete arch. If we draw through a, the center of the keystone, a horizontal line, its intersection with R1 gives a point in the first line (produced) of the true equilibrium concrete polygon.  The author will assume, in the problems following, that the student has familiarized himself with the previous text, especially that relating to the proportioning, mixing and placing of concrete in the forms.

Are You in Newfield Maine? Do You Need Concrete Cutting?

We Are Your Local Concrete Cutter

Call 207-284-0788

We Service Newfield, ME and all surrounding Cities & Towns