Capítulo II Marco teórico1
2.4 Sistemas Informáticos
2.4.1 Ciclo de vida del desarrollo de sistemas
In the design of a bridge foundation, the structural engineer determines sets of external loads consisting of resultant vertical load and bending moment. These sets of loading may be plotted on the capacity curve to determine whether these loads are inside the envelope of the capacity curve. If these loads are located inside the envelope, the particular footing is adequate for adoption in design. Otherwise, a larger footing can be tried until all the sets of loading are inside the envelope of the footing capacity curve.
The size of a footing can be directly solved when the total vertical load and biaxial bending moments are given. This method uses the standard flexure formula, which is applicable only when the whole area of the footing is in compression. However, we know from the capacity curve that the most efficient use of the footing to resist bending is when part of the footing is in tension. Under this condition it is possible to develop the maximum uplift moment to resist the applied external loads on the footing.
To design a rectangular footing with biaxial bending, we need to deter-mine the overburden forces on top of the footing. Applying the same pro-cedures as in our previous derivations, we can derive the equations for calculating the overburden pressure and the forces and bending moments resulting from this overburden on top of the footing.
Draw Figure 1.6 to show a circular and rectangular column on top of the footing foundation. Label the dimensional parameters of these columns.
Using basic calculus, derive the equations for the forces and their corre-sponding bending moments.
The variables considered in the calculation of the forces due to the overburden material on top of the footing are the column dimensions and the uniform load w.
The equation of the circular column is given by
(1.84)
Figure 1.6 Overburden on rectangular footing.
Source: Jarquio, Ramon V. Analytical Method in Reinforced Concrete. Universal Publishers, Boca Raton, FL. p. 164. Reprinted with Permission.
x2+z2=R12
h/2 h/2 − x2
Z
x2 1 X
V2
V1
2 2
O
θ
V4 R1
w b
X d’
1 d
Y Footing capacity axis
b’
O
The force due to the uniform load w is given by
(1.85)
(1.86)
The moment of this force is given by the expression
(1.87)
(1.88)
The overburden forces for the rectangular footing are given by the fol-lowing expressions:
V1 = − w(cot θ + tan θ) (1.89)
V1 = (1/2)(cot θ + tan θ){h2/4 + x2 (x2 − h)} (1.90)
V1x1 = − w(cot θ + tan θ) (1.91)
(1.92)
V2 = w(h tan θ + 2 zo) (1.93)
V2 = wx2(h tan θ + 2zo) (1.94)
V2x2 = w (h tan θ + 2 zo) (1.95)
(1.96) V4=2w
∫ (
R12−x2 1 2)
/ dxV4=( / )1 2w Rπ 12
V x4 4 w R12 x2 3 dx
1 2
=2
∫ { (
−) }
/V x4 4=( / )2 3wR13
(x h− / )dx
∫
2{x2−( / ) }h 2x dx
∫
V x1 1=( / ) (cot1 12w θ+tan ) ( / )θ
{
h3 4 +x22(4x2−3h)}
∫
dx∫
xdx V x2 2=( / ) ( tan1 2w h θ+2z xo) 22The moment due to overburden material is
Mo = V1x1 + V2x2 − V4x4 (1.97) When the column is rectangular, use Equations 1.89 to 1.97 by replacing d with d′, b with b′, h with h′, x2 with x2′, and zo with zo′ in which b′ and d′
are the rectangular column dimensions. For a square column, b′ = d′.
The moment due to overburden material is then calculated as
Mo = (V1x1 + V2x2) − (V1x1’ + V2x2’) (1.98)
Example 1.3
Determine the total amount of reinforcing steel bars for the example rectan-gular footing above when the column dimension is 36 in. (0.91 m) in diam-eter, fc′ = 5 ksi (34.5 MPa), fy′ = 60 ksi (414 MPa), and w = 1.26 kips per square foot (60.9 kPa).
Solution:
Step 1: Determine the maximum moment uplift from the footing capacity curve for this footing. Figure 1.3 is the footing capacity curve for the allowable soil-bearing pressure of 345 kPa. From this curve we read MR = 971 kN-m. This is the maximum resultant uplift moment resistance of the footing foundation when the ca-pacity axis is along the diagonal of the rectangular footing. From the Excel worksheet, Mz = 180 kN-m and M = 954 kN-m. The inclination of the resultant moment from the horizontal axis is equal to 0.6947 - arctan(180/954) = 0.6947 − 0.1865 = 0.5082 radians (29.12°).
Step 2: Calculate the moment due to the overburden material from Equations 1.84 to 1.88 as applicable for a circular column. The moment due to overburden material for the circular column when w = 60.9 is from Equation 1.88, given by
V4x4 = (2/3)(60.9)(0.46)3 = 3.88 kN-m
Determine the moment due to the overburden material on top of rect-angular footing as follows:
θ = arctan (3.05/3.66) = 39.81° or 0.6947 radian h = 3.66{cos(39.81°)} + 3.05{sin(39.81°)} = 4.764 m zo = (1/2){3.05[cos(39.81°)] − 3.66[sin(39.81°)]} = 0
x2 = (1/2){3.66[cos(39.81°)] − 3.05[sin(39.81°)]} = 0.43 m V2x2 = (1/2)(60.9){4.764 tan(39.81°)}(0.43)2 = 22.35 kN-m V1x1 = (1/12)(60.9){cot(39.81° + tan(39.81°)}{(1/4)(4.764)3
+ (0.43)2[4(0.43) − 3(4.764)]} = 255.0 kN-m Mo = 255.0 + 22.35 − 3.88 = 274 kN-m
Step 3: Determine the net moment for the design of reinforcing bars.
Net M = 971cos (10.68°) − 274 = 680 kN-m (6,019 in-kips) along the capacity axis. The net resultant moment is 680/cos (10.68°) = 692 kN-m.
(6,125 in-kips).
Step 4: Solve for the total reinforcing bars and resolve the corresponding amount along the 1-1 and 2-2 axes (Jarquio, 2004, pp. 9–13).
K = 49(51 + 60)(5)/(87 + 60)2 = 1.258 Minimum depth to rebars
= {6,125 cos (29.12°)/(1.258)(120)}1/2 = 5.95 in. (151 mm) Assume we are using 24 in. (610 mm) of total concrete thickness and 3.50 in. (90 mm) of cover to reinforcing bars. Depth to rebars = 24 − 3.50 = 20.50 in. (521 mm). c = 87(20.50)/(87 + 60) = 12.13 in. (308 mm).
As1 = 5,351/{60[20.50 − 0.417(12.13)]}
= 5.78 in.2 (38 cm2) reinforcing steel bars along axis 2−2 As2 = 2,981/{60[20.50 − 0.417(12.13)]}
= 3.22 in.2 (21 cm2) reinforcing steel bars along axis 1−1.
The total equivalent area of reinforcement = {(5.78)2 + (3.22)2}1/2 = 6.62 in.2 (43 cm2), which is equal to 6,125/{60[20.50 − 0.417(12.13)] = 6.62 in.2 (43 cm2). This amount of steel reinforcement is the maximum to put into the rectangular footing. Any more than this is a waste for the given allowable soil-bearing pressure and size of the rectangular footing in the above example.
The punching shear force can be calculated at key point R in which the value of V = 1922 kN less the overburden of 60.9{(3.05)(3.66) − 3.1416(0.46)2} to obtain 1922 − 639 = 1283 kN.
Example 1.4
Determine the amount of reinforcing steel in the circular footing example above if the column is 48 in. (1.22 m) in diameter with the same amount of overburden and stresses as the rectangular footing example.
Solution:
Step 1: Determine the maximum uplift moment from the capacity curve as 4464 kN-m. Since a circular section is symmetrical, any diameter can be a capacity axis. Hence, MR = M.
Step 2: Calculate the overburden moment as follows:
V4x4 = (2/3)(60.9)(0.61)3 = 9.20 kN-m. for the column V4x4 = (2/3)(60.9)(3.05)3 = 1,152 kN-m Net overburden = 1,152 − 9.20 = 1,143 kN-m
Step 3: Determine the net moment to design the reinforcing steel bars.
Net design moment = 4,464 − 1,143 = 3,321 kN-m. (29,395 in-Kips) Step 4: Solve for the amount of reinforcing steel bars. Use a steel mesh formation with equal amounts in either direction.
Minimum depth to rebars
= {29,395/(1.258(192)}1/2 = 11.04 in. (280 mm)
Assume we are using 30 in. depth of footing and 4 in. of cover to the center of the rebars. The depth to rebars is 30 − 4 = 26 in.
c = 87(26)/(87 + 60) = 15.388 in. (391 mm) As = 29,351/{60[26 − 0.417(15.388)]} = 24.98 in.2 (161 cm2) This is the maximum amount of steel reinforcing bars in one direction and the same amount in the perpendicular direction. At any direction θ, the amount of reinforcement provided is equal to
As cos θ + As sin θ = As (cos θ + sin θ) (1.99) Equation 1.99 is always greater than unity, and hence adequate rein-forcement is provided throughout the area of the circular footing foun-dation. Table 1.1 shows the numerical proof of Equation 1.99
The punching shear force can be calculated as 5031 − {0.50(60.9) (3.1416)[(3.05)2 − (0.61)2]} = 4177 kN. Figure 1.7 shows the sche-matics of the layout of the reinforcing steel bars for the example footings.