Exam 2 – CE595 S.Ramesh

350 ft.

40 ft. 30 ft. 230 ft.

Water

1 mile

Two loading Conditions• full water head (hydrostatic pressure) no gravity. • full water head (hydrostatic pressure) with gravity.

f’c=3500psi

Finite Element Model (For both Load Cases):

Finite Element Mesh (For both Load Cases):

Explain your choice of finite element for performing the analysis

• 2D plane strain is assumed (Length of dam>>>width, height of dam). Loading Conditions are taken to be constant along the length of the dam.

• The varying hydrostatic pressure and only in one side of the dam produces non-uniform moment distribution along the height of the dam.

• Stress and as well as strain distribution is not uniform along the height of the dam.

• Non-rectangular elements have to be selected if quadrilateral elements are to be selected because of the geometry of the cross section of the dam. • Hence, Q8 Element (CPE8R)(which has quadratic strain variation within the element) is better than other 2D elements to represent this model. (Strain is constant within the element in CST, Strain is linear within the element in LST, The linear variation of the strain does not change along the length of the element in Q4 and Q6 elements).

• Reduced integration is used to reduce the effects from the over stiffness.

•Total Number of Elements : 1134

•ABAQUS/CAE Version 6.6-1 is used for FE analysis.

Load and Boundary Conditions (For Load case 1: Hydrostatic Pressure):

Load and Boundary Conditions (For Load case 2: Hydrostatic pressure with gravity):

Explain the boundary conditions used in the model.

Two loading Conditions1) full water head (hydrostatic pressure) no gravity. 2) full water head (hydrostatic pressure) with gravity. have been analyzed. The calculations of the forces are shown below.

Boundary Condition Selected:

Keys are generally constructed to prevent the movement (in 1- and 2-directions)and rotation (about 3-direction) of the dam mounted on soil profile. In this model also it is assumed to be the same and it is assumed that the keys are provided all along the base of the dam to enough depth to prevent movement and rotation of the dam. (However the actual situation is not generally like this). Fixed boundary condition is assumed all along the base of the dam.

Calculations associated with: (a) defining the material properties, and (b) defining the loading conditions.

Homogeneous, isotropic, elastic behaviorof concrete is assumed.

a) Calculations for Material Properties

Given

fc 3500psi

Ec 57fcpsi

ksi Ec 3.37217 103 ksi

For concrete, the poission ratio is in a range of 0.15~0.20

Assume a poission ratio v 0.17

Assume a unit wieght of concrete Concrete 150pcf

g 32.2ft

sec2

MassDensityOfConcreteConcrete

g

MassDensityOfConcrete 2.24652 10 7 kipsec2

in

1

in3

All the units entered in the ABAQUS are in kips and inchesHence, the results are in kips and inches.

b) Calculations for Loading Conditions

Hydrostatic Pressure

Say the unit wieght of water water 62.4pcf

hwater.datum 4200inAt the datum (base of the dam) the height of the water

HydrostaticPressure datum water hwater.datum

HydrostaticPressure datum 0.15167ksi

At dam top (at a height of 4200 in from the datum)

HydrostaticPressure damtop water 0.0

HydrostaticPressure damtop 0ksi

Gravity Load

GravityAcceleration 386.4in

sec2 Entered as -386.4 in component 2

2) Present the results from the finite element analysis of each loading condition. For each analysis, present the contour plots of all the normal, shear, and principal stresses. Indicate the maximum values (locations and magnitudes) on the plots.

Load Case 1:

S11

S22 Load Case 1:

S33 Load Case 1:

S12 Load Case 1:

Maximum in-plane principal stresses Load Case 1:

Minimum in-plane principal stresses Load Case 1:

Out-of-plane principal stresses Load Case 1:

Load Case 2:

S11

S22 Load Case 2:

S33 Load Case 2:

S12 Load Case 2:

Maximum in-plane principal stresses Load Case 2:

Minimum in-plane principal stresses Load Case 2:

Out-of-plane principal stresses Load Case 2:

350 ft.

40 ft. 30 ft. 230 ft.

Water

1 mile

A

B C

3]. Calculation of Applied forces and Comparison with Reactions got from FEM Analysis

Reactions Load case 1

NodeDistance from B (1_direction) (in) RF1 (kip) RF2 (kip)

1 0 -1.01E+06 -1.59E+06 1217 1900 -1.27E+06 -2.51E+062 100 -7.98E+05 -7.86E+05 1221 2000 -1.25E+06 -1.34E+063 200 -5.40E+05 -4.84E+05 1224 2100 -1.04E+06 -8.41E+054 300 -4.88E+05 -3.39E+05 1227 2200 -9.24E+05 -5.55E+055 400 -4.35E+05 -2.16E+05 1230 2300 -8.32E+05 -3.34E+056 500 -3.96E+05 -1.18E+05 1233 2400 -7.62E+05 -1.48E+057 600 -3.65E+05 -3.04E+04 1236 2500 -7.06E+05 1.91E+048 700 -3.41E+05 4.96E+04 1239 2600 -6.61E+05 1.74E+059 800 -3.20E+05 1.25E+05 1242 2700 -6.24E+05 3.20E+0510 900 -3.04E+05 1.96E+05 1245 2800 -5.96E+05 4.59E+0511 1000 -2.92E+05 2.63E+05 1248 2900 -5.73E+05 5.89E+0512 1100 -2.82E+05 3.26E+05 1251 3000 -5.57E+05 7.09E+0513 1200 -2.76E+05 3.83E+05 1254 3100 -5.47E+05 8.16E+0514 1300 -2.72E+05 4.33E+05 1257 3200 -5.41E+05 9.04E+0515 1400 -2.70E+05 4.71E+05 1260 3300 -5.38E+05 9.66E+0516 1500 -2.68E+05 4.95E+05 1263 3400 -5.33E+05 9.91E+0517 1600 -2.65E+05 4.97E+05 1266 3500 -5.18E+05 9.60E+0518 1700 -2.42E+05 4.65E+05 1269 3600 -4.76E+05 8.22E+0519 1800 -7.51E+04 1.59E+05 Total -20180601 2306349

Reactions are reported in the “Reaction” Files also.

RF1-Load case 1

RF 2-Load case 1

RF Magnitude-Load case 1

Reactions Load case 2

Node

Distance from B

(1_direction) (in) RF1 (kip) RF2 (kip)

1 0 -7.08E+05 -7.45E+05 1217 1900 -8.16E+05 4.18E+042 100 -5.49E+05 4.45E+05 1221 2000 -8.74E+05 1.12E+063 200 -3.90E+05 7.29E+05 1224 2100 -7.79E+05 1.57E+064 300 -3.79E+05 8.60E+05 1227 2200 -7.44E+05 1.82E+065 400 -3.63E+05 9.55E+05 1230 2300 -7.18E+05 1.97E+066 500 -3.54E+05 1.02E+06 1233 2400 -7.04E+05 2.08E+067 600 -3.49E+05 1.06E+06 1236 2500 -6.96E+05 2.15E+068 700 -3.46E+05 1.09E+06 1239 2600 -6.92E+05 2.20E+069 800 -3.46E+05 1.11E+06 1242 2700 -6.93E+05 2.22E+0610 900 -3.47E+05 1.12E+06 1245 2800 -6.96E+05 2.23E+0611 1000 -3.50E+05 1.11E+06 1248 2900 -7.03E+05 2.22E+0612 1100 -3.53E+05 1.10E+06 1251 3000 -7.11E+05 2.19E+0613 1200 -3.58E+05 1.08E+06 1254 3100 -7.21E+05 2.13E+0614 1300 -3.63E+05 1.05E+06 1257 3200 -7.29E+05 2.04E+0615 1400 -3.67E+05 9.97E+05 1260 3300 -7.35E+05 1.92E+0616 1500 -3.68E+05 9.27E+05 1263 3400 -7.32E+05 1.76E+0617 1600 -3.63E+05 8.33E+05 1266 3500 -7.08E+05 1.53E+0618 1700 -3.31E+05 7.03E+05 1269 3600 -6.47E+05 1.19E+0619 1800 -1.02E+05 2.16E+05 Total -20180602 48044323

RF1-Load case 2

RF2-Load case 2

RF Magnitude-Load case 2

Calculation of Applied Forces:

HydrostaticPressure datum 0.15167ksi HydrostaticPressure damtop 0ksi

Total Force due to hydrostatic Pressure:

Say the forward face of the dam (the sloped face near to the water) make an angle withhorizontal

Ldam 5280ftLength of the Dam

BC 40ft AC 350ft

AB BC2 AC2 AB 352.2783ft

SinACAB

TanACBC

CosBCAB

FHydrostaticPressure12

HydrostaticPressure datum HydrostaticPressure damtop AB Ldam

FHydrostaticPressure 2.03115 107 kip

acting in an angle with vertical

Total Force due to Gravity:

Concrete 150pcf

CrossSectionalAreaofDam12

30ft 300ft( ) 350 ftCrossSectionalAreaofDam 5.775 104 ft2

Ldam 5280ftLength of the Dam

FGravity Concrete CrossSectionalAreaofDam Ldam

FGravity 4.5738 107 kip acting downward

Comparison of Applied Forces with the Reactions got from the FEM Analysis for LoadingCase1

Total Horizontal Froce

FH1 FHydrostaticPressure SinFH1 2.01802 107

kip

From the FEM Analysis

RF1 2.01806 107 kipThe Total Reaction in the 1-direction

For equlibrium the Applied Forces (FH1)+The Reactions(RF1)=0 O.K

Total Vertical Froce

FH2 FHydrostaticPressure CosFH2 2.3063 106

kip

From the FEM Analysis

RF2 2.30635 106 kipThe Total Reaction in the 2-direction

For equlibrium the Applied Forces + The Reactions=0

Forces and reactions are almost equal in magnitude and opposite in direction O.K

Comparison of Applied Forces with the Reactions got from the FEM Analysis for LoadingCase 2

Total Horizontal Froce

FH1 FHydrostaticPressure SinFH1 2.01802 107

kip

From the FEM Analysis

RF1 2.01806 107 kipThe Total Reaction in the 1-direction

For equlibrium the Applied Forces (FH1)+The Reactions(RF1)=0 O.K

Total Vertical Froce

FH2 FHydrostaticPressure Cos FGravity

FH2 4.80443 107 kip

From the FEM Analysis

RF2 4.80443 107 kipThe Total Reaction in the 2-direction

For equlibrium the Applied Forces + The Reactions=0

Forces and reactions are almost equal in magnitude and opposite in direction O.K

Are there any other simple calculations you can suggest or perform to check the analysis results?

• For a point in a horizontal section along the length of the dam (can be base of the dam), stress due to gravity of the part above that section (P/A + Pey/I) and the stress due to moment coming from the hydrostatic pressure (My/I) can be calculated and checked with S22 obtained from the FE Analysis.

• One of the principal stresses would be zero at the free boundary.

• One of the principal stresses would be equal to (-) Hydrostatic pressure at the boundary adjacent to the Water. Hydrostatic pressure can be calculated along the face of the dam and checked with the principal stresses.

• Can pick one element and check for equilibrium. This is normally approximately satisfied.

• Displacements at fixed points are zero. For example at Node 1: U1=725.362E-33 in, U2=1.61748E-30 in

Checking the Finite Element model for convergence

Load Case 1: Mesh is refined with 4752 Elements (almost 4 times)instead of 1134 elements (which were earlier) and the displacements are compared.

Mesh with 1134 Elements Mesh with 4752 elements

Max U1= 0.3579 in Max U1= 0.3579 in

Max U2= 0.08462 in Max U2= 0.08464 in

Max U magnitude = 0.3662 in Max U magnitude = 0.3662 in

Load Case 2: Mesh is refined with 4752 Elements (almost 4 times)instead of 1134 elements (which were earlier) and the displacements are compared.

Mesh with 1134 Elements Mesh with 4752 elements

Max U1= 0.1984 in Max U1= 0.1985 in

Max U2= -0.1224 in Max U2= -0.1224 in

Max U magnitude = 0.2301 in Max U magnitude = 0.2301 in

The displacement pattern and the values are seemed to be very similar. This implies the convergence of the model.Stresses and reactions are calculated from the displacements by FEM. They are generally approximately similar.

4) From your analysis results, identify the maximum compressive and tensile stresses in the concrete dam. Determine whether the structure will crack (in tension) or crush (in compression) under the given loading conditions?

Let us use Rankine Theory for concrete (Brittle Material).

Compressive strength of the concrete = 3.5 ksi

Modulus of Rupture of the concrete =7.5*(3500)^0.5*1/1000 ksi =0.44 ksi

For Load case 1:

From the FE Analysis the maximum maximum principal stress =0.713 ksi (Tension) >0.44 ksi cracking (0.889 ksi (Tension) from refined mesh>0.44 ksi)

max. (in magnitute ) minimum principal stress =0.1925 ksi (compression) <3.5 ksi No Crushing (0.1944 ksi (Com.) from refined mesh <3.5 ksi)

For Load case 2:

From the FE Analysis the maximum maximum principal stress =0.3457 ksi (Tension) <0.44 ksi No cracking (0.4737 ksi (Tension) from refined mesh>0.44 ksi cracking )

max. (in magnitute ) minimum principal stress =0.2927 ksi (compression) <3.5 ksi No Crushing (0.2927 ksi (Com.) from refined mesh <3.5 ksi)

[Note: Tensile strength of the concrete = 6(f’c)^0.5. ACI 318-2005 suggest to consider the modulus of rupture to check for cracking.US Army Corps of Engineers Manual also recommends to consider modulus of rupture for checking for cracking.]

5) From the analysis results, discuss which of two loading conditions (i) or (ii) dominates the problem. For mesh with 1134 elements (coarse mesh)

Loading Condition

σp1 (max)(ksi)

σp3(max in mag.) (ksi)

U1(max)(in)

U2(max)(in)

U(mag)(max)(in)

1 0.713 0.1925 0.3579 0.08462 0.3662

2 0.3457 0.2927 0.1984 -0.01224

0.231

The tensile stresses and deflections are higher for loading condition 1. Hence, we could say that the loading condition 1 dominates the Problem.

Due to the gravity, compressive stress increases and tensile stress decreases. Concrete can take up to 3.5 ksiof compressive stress (with out no load factors and reduction factors). Hence the loading case 2 is more safe than loading case 1.

US Army Corps of Engineers Manual for gravity dam design list the following loading cases to be considered.

(1) Uplift.(2) Forces due to temperature effects.(3) Earth and silt pressures.(4) Ice pressure.(5) Earthquake forces.(6) Wind pressure(7) Wave pressure. (hydro dynamic forces).(8) Reaction of foundation.

Citation: US Army Corps of Engineers Manual for gravity dam design (1995) http://www.usace.army.mil/publications/eng-manuals/em1110-2-2200/toc.htm

6) Identify some of the other loading conditions or design criteria that must be checked for the concrete dam structure. Provide a reference (citation only) that you used to determine these. Will the finite element model you developed be useful for these additional cases?

If there is life load during construction or after construction, vehicle loads if therewill be a road along the top of the dam they also should be considered.

The Finite element model developed here will be useful for these additional cases. Dynamic analysis For earthquake and hydrodynamic forces, and thermal analysis for forces due to temperature effects should be done.