Project at hand:
The project is to analysis the deformations and stresses of a low rise building structure with different door placements by increasing the amount of finite elements. The problem is to solve which door placement is the best... meaning to have the lowest stresses in the wall around the door. The software used to complete the project is AnSys. There are 3 model geometry's: a building with door locations in the middle (fig 1a), left (fig 1b) and right (fig 1c) of a wall. The building length, width and height are 40, 30 and 15 ft respectively. The 1 ft walls of the building are masonry with a Young's Modulus of 1.08e8 psf. The roof plate is wood (E = 0.66e8 psf). The layer thickness of the roof is 6 inches. And, the beam (lintel) above the door is concrete (E = 5.19e8 psf). The beam width and height is 1 ft and 1.25 ft respectively. See fig 2b for the location of the beam. As for constraints, the bottom of the walls (boundary conditions) are fixed... meaning the bottom cannot move in the x, y and z direction (DX=DY=DZ=0. See fig 3).

The wind loading is designed using code procedure of a low-rise building. The location of the building is Miami, Florida with Exposure C, which is open terrain with scattered obstructions having heights generally less than 30 feet. The basic wind speed in Miami is 150 mph. In viewing fig 2, the wind pressure in red is +44.56 psf. This pressure is the direction of the wind. The wind pressure in blue is –32.46 psf. Also, the light blue wind pressure is –22.28 psf. A positive pressure is toward the internal surface. While, the negative pressure is away from the internal surface.

There are 3 element types in the building: Solid92, Beam4 and Shell93. The masonry walls has the type Solid92, which is a Tetrahedral Solid having 10 nodes in 3D space. The Solid92 has three DOF (Degrees of Freedom): UX, UY, UZ. The beam (lintel) is the Beam4 element, which is a structural elastic 3D beam with 2 nodes and 6 DOF: UX, UY, UZ, ROTX, ROTY, ROTZ. The roof plate is Shell93 element, which is a structural shell with 8 nodes in 3D space and 6 DOF: UX, UY, UZ, ROTX, ROTY, ROTZ. The Solid92 is used because of the opening in the building. If, for example, Solid45 is the element type for the walls, an error will occur. As for the other element types, they are used because of their 3D properties.

To verify the model is working properly, 5 different finite element sizes were used to analysis the centered door model: 5.0, 4.0, 3.0, 2.0 and 1.5. Respectively, the deflections to the different element sizes are 0.00104, 0.001193, 0.00135, 0.001383 and 0.001398 ft. As the element size becomes smaller... the distance between deflections of the element sizes lessens. This means as the element size decreases the deflection converges and becomes more accurate to the theoretical solution.

To apply the theoretical to the model, only a wall is used to verify the comparison with the theoretical (fig 10 and 10b). The theoretical equivalent to a wall with a uniform load and fixed boundaries is a cantilever beam with a uniform load. 3 element sizes (1.5, 1.0 and 0.75) were used to compare maximum deflection to the theoretical. In order (1.5, 1.0, 0.75), the maximum deflections are 0.02867 ft, 0.02882 ft and 0.02888 ft respectively. The theoretical solution for the maximum deflection of a cantilever, which is located at the free end, is equal to [w*(length)⁴]/(8*E*I) = [(44.56 psf)*(15 ft)*(40 ft)⁴]/[8*(1.08e8 psf)*(1.25 ft)⁴] = 0.031331 ft where w = unit load, E = Young's Modulus and I = Moment of Inertia = (base)*(height)⁴/12 = (15 ft)*(1 ft)⁴/12 = (1.25 ft)⁴. As for verification, there is a convergence toward the theoretical answer in the deflections of the conresponding element sizes. As the element size lessens, the deflection becomes more accurate to the theoretical solution.

The only way to see which door placement is the best; the stresses around the door have to be looked at to make a determination. When looking at the normal stresses of the 3 doors (fig 11a, 12a and 13a), the left door has the lowest stresses. As for the shear stresses (fig 11b, 12b and 13b), the middle door has the lowest stresses. Last, the left door has the lowest principal stresses (fig 11c, 12c and 13c). Are shear stresses more important than normal and principal stresses? Yes, so since the middle door has the lowest shear stresses. It's the best placement out of the three. Also, the most logical placement is the middle because in this project... the wind load is only applied to one side, which makes the right door placement the worst. But if the wind was applied to the opposite side, the worst placement would be the left.

List of Figures:
· Fig 1a – Centered Door Model
· Fig 1b – Left Door Model
· Fig 1c – Right Door Model
· Fig 2 – Wind Loading
· Fig 2b – Beam (Lintel)
· Fig 3 – Boundary Conditions
· Fig 5 – Element Size 5.0
· Fig 6 – Element Size 4.0
· Fig 7 – Element Size 3.0
· Fig 8 – Element Size 2.0
· Fig 9 – Element Size 1.5
· Fig 10 – Cantilever Wall Detail (Model vs Theoretical)
· Fig 10b – Cantilever Wall Deflection (Model vs Theoretical)
· Fig 11a-11c – Normal (S_Y), Shear (S_XY), Principal (S₁) Stresses (Mid Door)
· Fig 12a-12c – Normal (S_Y), Shear (S_XY), Principal (S₁) Stresses (Left Door)
· Fig 13a-13c – Normal (S_Y), Shear (S_XY), Principal (S₁) Stresses (Right Door)