Case Study Number 2: Pressure of Strip Load on Retaining Wall - Verifying Jarquio s Solution

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1 pg. 1 of 8 Case Study Number : Pressure of Strip Load on Retaining Wall - Verifying Jarquio s Solution Introduction At a retaining wall site along new construction on Highway 1 in Cheatham County, Tennessee, there is a failure of a retaining wall (Wall 5). After a meeting with the wall designer, the owner and the construction company there was a dispute as to the cause of this failure. The contractor claims that the applied earth pressure force was greater than that calculated by the wall designer and that was the cause of the failure. He specifically maintained that the designer was in error when using Jarquio s formula for calculating the pressure contributed by the strip load applied above the wall. This strip load was a applied by a storage facility for the local university. The owner, then contracted with Student Calculations to estimate the pressure force applied to the wall in order to provide the owner with independent information about the site. Site Parameters The backfill behind the wall is granular material that classifies as AASHTO classification A-1-a and is free draining material. Laboratory tests, the wall designer and the construction contractor agree that there was no water pressure behind the wall and has the following characteristics. Height of Wall H = 6 m Unit weight of backfill g = 17.9 kn/m 3 Granular backfill Friction angle of soil f = 34 degrees The strip load is located 1 feet from the face of the wall and is 1 feet in width. Strip load q = 4 kn/m For all solutions, the assumption has been made that the Poisson s ratio for the soil =.5 Standard Theory Jarquios Solution The pressure distribution behind the wall caused by a strip load, according to the theory of elasticity is non-linear and appears as below: a' b q/unit area H a b The generally accepted equation for this configuration is.3qn s p = (1) H(.16 + n )

2 pg. of 8 Where: q = surcharge load n = z/h z = depth from top of wall H = total height of the wall The earth pressure behind the wall caused by the strip load can thus be expressed as: P aload = H Ú sdz () Substituting in our values we get the following equation: P load = H.3q * l(z /H) Ú dz = È H.16 + z Ê ˆ Á Î Í H Ë 6.3* 4*(z /6) Ú dz = È z Ê ˆ Á Î Í 6 Ë z Ú (3) z ( ) This was estimated by Jarquio as: ( [ ]) P load = q 9 H q -q 1 Ê b' ˆ q = tan -1 Á Ë H Ê a'+b' ˆ q 1 = tan -1 Á Ë H (4) For this wall where H = 6 m, a =. m and b = 1. m this gives a solution of: P load = 4 ( 6[ ] ) = 45.63kN 9 Numerical Methods to Solve Added Earth Pressure In addition to obtaining the actual difference in earth pressure applied by the strip footing, the point of this case study was to show how various numerical methods could be employed instead of the Jarquio estimation to obtain an estimate of the earth pressure and the evaluate the effectiveness of these methods. The following are summaries of the methods employed with commentary of their relative effectiveness. Please see Numerical Methods for Engineers: With Software and Programming Applications by Steven Chapra and Raymond Canale for details of these methods. Analytical Method MathCad The solution obtained to the integral:

3 pg. 3 of z Ú = z ( ) The Pressure distribution being integrated has the following appearance: Depth vs Pressure applied to Wall Pressure applied to wall from strip footing (kn/m^) Depth from top of Wall (m) Newton-Cotes Method: Trapezoidal Rule The multiple application of the trapezoidal rule was used to evaluate the integral given in equation 3. This was calculated with a step sizes 1. m,.5 m and.1m. The estimate of the integral is given by the following equation: I = h È n-1 f (x ) + f (x ) + f (x ) Í Â i n Î i=1 Using a step size of 1. meters the estimate obtained was: = =1.% Using a step size of.5 meters the estimate obtained was: = =.3% Using a step size of.1 meters, the estimate obtained was: = =.1% The table showing interim steps for the 1. meter step size can be seen below:

4 pg. 4 of 8 Step Size Depth 1m f(x) I Newton-Cotes Method: Simpson s 1/3 Rule The estimate of the integral according to Simpson s 1/3 rule for multiple segements can be given as: h n-1 Â n- Â f (x o ) + 4 f (x i ) + f (x j ) + f (x n ) i=1.3.5 j=,4,6 3 As with the Trapezoidal Rule, the estimate was made with step sizes of 1. meters and.5 meters. Using a step size of 1. meters the estimate obtained was: = =.1% Using a step size of.5 meters the estimate obtained was: = =.% A table showing the interim steps for the step size = 1. meter can be seen below: Step Size 1 m Depth f(xi) f(xj) I

5 pg. 5 of 8 Newton-Cotes Method: Simpson s 3/8 Rule The estimate of the integral according to Simpson s 3/8 for a single application of the rule can be given as follows: 3h ( 8 f (x ) + 3 f (x ) + 3 f (x ) + f (x ) 1 3 ) When applied a single time it yields a n estimate of the integral as = =.94% When applied twice, to achieve a step size of 1., it yields an estimate of = =.7% When applied three times, to achieve a step size of.5, it yields an estimate of = =.4% A table showing the interim calculations for two applications of the rule (thus using a step size of 1. meters) is shown below: Step Size 1m Depth f(xi) f(xj) I I Sum I

6 pg. 6 of 8 Romberg Integration The estimate of the integral given by the Romberg Integration method can be given as follows: I 4 k-1 I j +1,k-1 - I j,k-1 4 k-1-1 The results were calculated for a 4 th order, 6 th order and 8 th order solution. Using a 4 th order solution (k=) the estimate obtained was: = =.% Using a 6 th order solution (k=) the estimate obtained was: = =.4% Using an 8 th order solution (k=) the estimate obtained was: = =.3% A table showing the interim calculations is shown below: Trap Rule Est Romberg Estimations k=1 O^4 O^6 O^8 No. SegementsI k= k=3 k=

7 pg. 7 of 8 Gauss Quadrature using Gauss Legendre Formula To use the Gauss Legendre formula we must perform a variable change so that the limits that are integrated are between 1 and 1, rather than to 6 for this wall. The new value of z is given by the following formula: z = (b + a) + (b - a)z d = (6 + ) + (6 - )z d = 3+ 3z d The new value of dz is given by: z d = b - a dz d = 6 dz d = 3dz d These values are then substituted into the original equation (3) to yield: P load = (3+ 3z d ) 1 Ú-1 ( ) (3+ 3z d ) dz d The integral is then estimated by the following equation: c f (x ) + c 1 f (x 1 ) c n-1 f (x n-1 ) Where the values of c and x are from tables given Table.1 of Numerical Methods for Engineers. Solutions were obtained for point, 4 point and 6 point estimates and yielded the following results: Using a point estimate the solution obtained was: With an error of et = =1.3% Using a 4 point estimate the solution obtained was: = =.1% Using an 8 th order solution (k=) the estimate obtained was: = =.% A table showing interim calculations for the 4 point estimate is given below: ci zi f(z) ci*f(zi) c x c x c.65145x c x S

8 pg. 8 of 8 Discussion and Conclusions Firstly, all of the solution methods illustrated in this case study converge on an answer that is near kn for the force applied by the pressure distribution caused in the backfill behind the retaining wall by the strip load. This is much smaller than the estimate provided by Jarquio s solution, which gave 45.63kN. As this is much larger than the one obtained by the conventional solution, and the wall was designed to withstand this force, the failure of the wall may require further investigation as it s cause has not been identified by this limited case study. Barring further research that disproves the solution obtained by equation 1, which has been supported by to date empirical evidence, it must be concluded that the designer was not negligent in using Jarquio s solution. All of the methods illustrated in this case study can be used to supply an estimate for the force acting on the wall as a result of the strip load applied above the wall. Even the methods with the highest error, 1.3% for the point estimate of the Gauss Legendre formula, give an acceptable tolerance for the retaining wall design. Certainly all are far closer to the actual integral than Jarquio s estimation. As all of the methods illustrated differ significantly only in their calculation difficulty, any of these methods could be used to predict the force added by the strip load. For all the methods, other than obtaining a direct solution through MathCad, Simpson s 1/3 rule seems to give the best accuracy for the lowest computational difficulty, followed closely by small step size applications of the trapezoidal rule. Of course, were the trapezoidal rule of Simpson s 1/3 rule applied as one segment, and not to multiple segments, their accuracy would be far poorer and unacceptable for this application. Finally, it must be stressed that all of these illustrated methods supplied results that are in excess of the significant figures for the problem. It is unlikely that we know all the actual soil parameters to the accuracy obtained with these solutions.