A line load can be thought of as a point load that is applied repeatedly, in a uniform manner, along the y-axis as illustrated in Figure 3.15. The line load is applied infinitely along they-axis. The units of a line load are given as force per
∆σz
X
Z x z
(x,z) Y
q (kN/m)
FIGURE 3.15 Stresses caused by a line load.
unit length, such as kN/m. Due to the nature of line load, the resulting stresses in thex –z plane are independent of y (i.e., we will get the same stresses in any x –z plane as we travel along the y-axis). This type of loading–geometry is termed plane strain. The vertical stress increase at any point (x,z) is given as
σz= 2qz3
π(x2+ z2)2 (3.10)
whereq is the line load (force/unit length) and x and z are the coordinates at which the stress increase is calculated.
Example 3.5 A vertical line load of 10 kN/m is applied at the surface of a semi-infinite soil mass. (a) Calculate the increase in vertical stress directly under the applied load forz= 0 to 0.3 m. (b) Repeat your solution using the finite element method and assuming that the soil is linear elastic with E= 1 × 107 kPa and ν = 0.3.
SOLUTION: (a) To calculate the increase in vertical stress directly under the applied load forz= 0 to 0.3 m, we substitute x = 0 into (3.10):
σz= 2q πz
Using this equation, we can calculate the increase in vertical stress as a function of z. The equation is plotted in Figure 3.16. According to this equation, σz is linearly proportional to the intensity of the line load and inversely proportional toz.
This means thatσz is very large near the point of load application but decreases rapidly with depth, as shown in Figure 3.16.
(b) Finite element solution (filename: Chapter3 Example5.cae) A plane strain condition is assumed in which the semi-infinite soil mass is represented by a 1 m× 2 m (x –z) plane as shown in Figure 3.17. The purpose of the analysis is to calculate
0 100 200 300 400 500 600 Vertical Stress (kPa)
0.00
0.10
0.20
0.30
Depth (m)
FEM, x= 0 m Boussinesq, x= 0 m
X
Z
Y
q (kN/m)
FIGURE 3.16 Stresses caused by a line load: FEM compared with the analytical solution.
q= 10 kN/m
1 m
2 m
FIGURE 3.17 Plane strain finite element mesh of the line load problem.
the increase in vertical stress within the soil mass due to use of the line load and to compare with the analytical solution presented in part (a).
The two-dimensional plane strain finite element mesh used has 20 elements in the x-direction and 40 elements in the z-direction, as shown in Figure 3.17. The finite element mesh is made finer in the zone around the line load, where stress concentration is expected. The element chosen is a four–node bilinear plane strain quadrilateral element. The increase in vertical stress is plotted as a function of depth, as shown in Figure 3.16 forx= 0 m. The figure shows excellent agreement between the stresses calculated using the analytical solution (3.10) and the finite element solution.
Example 3.6 Two parallel line loads, 10 kN/m each, are applied at the surface of a semi-infinite soil mass as shown in Figure 3.18a. This type of loading can be thought as a load caused by the rails of a train while the train is standing still. (a) Calculate the increase in vertical stress directly above the crown of the underground tunnel located in the vicinity of the railroad, as shown in Figure 3.18a.
(b) Repeat your solution using the finite element method and assuming that the soil is linear elastic withE= 1 × 107 kPa andν = 0.3.
SOLUTION: (a) To calculate the total increase in vertical stress at the crown of the tunnel due to the line loadsq1 andq2, we can calculate the stress increase caused by each line load separately and then combine the two. Called superimposition, this is permitted only when the loaded medium is linear elastic, which is the case in the present problem.
With the assistance of Figure 3.18b we can calculate (σz)1, which is caused byq1. For that we substitutex= 0.51 m and z = 0.272 m into (3.10):
(σz)1= (2)(10)(0.272)3
π(0.512+ 0.2722)2 = 1.148 kPa
Also, with the help of Figure 3.18c we can calculate (σz)2, which is caused by q2. Substitute x= 1.49 m and z = 0.272 m into (3.10):
(σz)2= (2)(10)(0.272)3
π(1.492+ 0.2722)2 = 0.024 kPa Therefore,
σz= (σz)1+ (σz)2 = 1.148 + 0.024 = 1.172 kPa
Note that the stress increase caused byq1at the crown of the tunnel is much greater than that caused byq2. This is becauseq1 is closer thanq2 to the tunnel.
(b) Finite element solution (filename: Chapter3 Example6.cae) A plane strain condition is assumed in which the semi-infinite soil mass is represented by a
q1= 10 kN/m q2= 10 kN/m
0.98 m 0.51 m
0.272 m ∆σz
(a) q1= 10 kN/m
x1= 0.51 m
x2= 1.49 m z1= 0.272 m
z2= 0.272 m x
z (∆σz)1
(∆σz)2 (b)
q2= 10 kN/m x
z (c)
FIGURE 3.18 Two-parallel-line-loads problem.
2.8 m× 3.4 m (x–z) plane as shown in Figure 3.19. The purpose of the analy-sis is to calculate the increase of vertical stress at the crown of the tunnel due to the application of the two line loads and to compare with the analytical solution presented in part (a).
The two-dimensional plane strain finite element mesh used for this analysis has 22 elements in the x-direction and 50 elements in the z-direction, as shown in Figure 3.19. The element chosen is a four-node bilinear plane strain quadrilateral element. The underground tunnel is not modeled in this simplified analysis. For more elaborate analysis, the underground tunnel can be modeled as a cavity. The presence of an underground cavity will affect the stress distribution, especially around the tunnel.
Position of Tunnel (tunnel is not modeled)
2.8 m
3.4 m q1= 10 kN/m q2= 10 kN/m
0.98 m 0.51 m
0.272 m ∆σ
FIGURE 3.19 Finite element mesh of the two-parallel-line-loads problem.
The calculated increase in vertical stress at the crown of the tunnel is 1.188 kPa.
This is in excellent agreement with the stress increase of 1.172 kPa calculated using the analytical solution (3.10).
3.3.3 Stresses Under the Center of a Uniformly Loaded Circular Area