Dams Forces acting on Dams Gravity 1 Dead

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Dams Forces acting on Dams (Gravity) 1

Dams Forces acting on Dams (Gravity) 1

� Dead load ◦ It comprises the major resisting force ◦ It includes the

� Dead load ◦ It comprises the major resisting force ◦ It includes the weight of the concrete or masonry or both plus that of the appurtenances such weight of gates and bridge. ◦ Usually a unit length of the dam is considered for design. 2

◦ Cross section of the dam is divided into several triangles and rectangles. ◦

◦ Cross section of the dam is divided into several triangles and rectangles. ◦ Weight of each of these is the computed at their respective centre of gravity. ◦ The resultant of all these forces is then calculated. ◦ Unit weight of concrete is normally taken as 2400 kg/m 3. 3

� Reservoir and tail water loads ◦ Water pressure corresponding to full reservoir level

� Reservoir and tail water loads ◦ Water pressure corresponding to full reservoir level constitutes the major external force acting on the dam. ◦ The water pressure acts horizontally and its with zero at the water surface and equal to WH 2/2 at the base. ◦ W is the unit weight of water and H is the depth of water upstream. 4

� The resultant force WH 2/2 acts at H/3 from the base of the

� The resultant force WH 2/2 acts at H/3 from the base of the dam. � When the upstream face is partly vertical and partly inclined, the resultant water pressures comprise of ◦ Horizontal component P = WH 2/2 acting at H/3 from the base of the dam ◦ Vertical component P 1 which is the weight of the column of water ABCDA acting at CG of the area. 5

� Tail water pressure ◦ It is the pressure of the water exerted on

� Tail water pressure ◦ It is the pressure of the water exerted on the d/s face of the dam ◦ Full value is taken for non overflow sections and a reduced value is taken for overflow sections. ◦ If h is considered to be the height of the tail water, then it exerts both horizontal and vertical pressures on the d/s face of the dam. ◦ Horizontal pressure P 2=wh 2/2 acting at h/3 and vertical pressure P 3 = weight of column of water A’B’C’. 6

� Uplift pressure ◦ Considered to be the second major external force ◦ The

� Uplift pressure ◦ Considered to be the second major external force ◦ The seepage of water normally takes place through the pores, cracks and fissures of the foundation materials, through the body of the dam and through the joints between the dam body and the foundation base. ◦ This seepage exerts pressure 7

� The uplift forces act as normal pressures in pores, cracks and seems within

� The uplift forces act as normal pressures in pores, cracks and seems within the body of the dam, between the dam and its foundation and within in the foundation. � At the heel and toe of the dam, these uplift forces are considered to be equal to the hydrostatic pressures joined by a straight line in between. 8

� Drainage galleries are sometimes provided within the body of the dam to relieve

� Drainage galleries are sometimes provided within the body of the dam to relieve some amount of this uplift pressure. � So the uplift pressure at the drainage gallery is equal to the tail water pressure plus one third of the difference between the full reservoir level and tail water head. 9

� Silt pressure ◦ The weight and the pressure of the submerged silt are

� Silt pressure ◦ The weight and the pressure of the submerged silt are to be considered in addition to weight and pressure of water. ◦ The weight of the silt acts vertically on the slope and pressure horizontally, in a similar fashion to the corresponding forces due to water. ◦ It is recommended that the submerged density of silt for calculating horizontal pressure may be taken as 1360 kg/m³. ◦ Equivalently, for calculating vertical force, the same may be taken as 1925 kg/m³. 10

� Earthquake forces ◦ Earthquake or seismic activity is associated with complex oscillating patterns

� Earthquake forces ◦ Earthquake or seismic activity is associated with complex oscillating patterns of acceleration and ground motions, which generate transient dynamic loads due to inertia of the dam and the retained body of water. ◦ Horizontal and vertical accelerations are not equal, the former being of greater intensity. 11

� The effect of an earthquake is equal to imparting acceleration to the foundation

� The effect of an earthquake is equal to imparting acceleration to the foundation of the dam. � The acceleration induced is of two types ◦ Horizontal acceleration ◦ Vertical acceleration 12

� Horizontal acceleration ◦ Normally expressed in terms of acceleration due to gravity ◦

� Horizontal acceleration ◦ Normally expressed in terms of acceleration due to gravity ◦ Values of 0. 1 g to 0. 15 g are normally considered to be sufficient for high dams in earth quake zones ◦ Causes inertia force and hydrodynamic pressure. 13

� Inertia force ◦ It is given by principle of mass times the acceleration

� Inertia force ◦ It is given by principle of mass times the acceleration acting through the centre of gravity of the section, ◦ Irrespective of the shape of the cross section. ◦ Acts in an opposite direction to the ground acceleration ◦ Causes an overturning moment about the horizontal section in addition to that of the hydrodynamic force 14

◦ Inertia force is given by W/g(αg)=Wα W = weight of dam g =

◦ Inertia force is given by W/g(αg)=Wα W = weight of dam g = acceleration due to gravity α = acceleration co efficient, earth quake acceleration due to gravity 15

� Hydrodynamic pressure ◦ An instantaneous horizontal pressure is exerted against the dam in

� Hydrodynamic pressure ◦ An instantaneous horizontal pressure is exerted against the dam in addition to hydrostatic forces due to the horizontal acceleration of dam and its foundation ◦ This is known as hydrodynamic pressure ◦ The direction of hydrodynamic pressure is opposite to the direction of earthquake acceleration. 16

� The hydrodynamic force is given by ◦ Pe = Cαhw h ◦ C

� The hydrodynamic force is given by ◦ Pe = Cαhw h ◦ C = coefficient that varies with the shape and depth. Value can be approximated from: 17

Pe = pressure intensity acting normal to the face of the dam at a

Pe = pressure intensity acting normal to the face of the dam at a point depth y from the reservoir water surface i. e. hydrodynamic pressure at depth y. (kg/m 2) αh = horizontal earthquake intensity assumed for the purpose of design. Varies from 0. 03 to 0. 24 at the top of the dam reduced linearly to zero. 18

w = unit weight of water (1000 kg/m 3) h = reservoir water depth

w = unit weight of water (1000 kg/m 3) h = reservoir water depth (m) i. e. reservoir level minus base level of dam at u/s face. y = vertical distance from the reservoir surface to the elevation under consideration Cm = maximum value of pressure coefficient. The force Pe acts at 4 H/3π from the base. 19

� Vertical acceleration ◦ A vertical acceleration may either act downward or upward. ◦

� Vertical acceleration ◦ A vertical acceleration may either act downward or upward. ◦ When it is acting in the upward direction, then the foundations of the dam will be lifted upward and become closer to the body of the dam and thus the effective wt. of dam will be increased and hence the stress developed will increase. ◦ Increased weight of the dam material = Wc(1+ α) ◦ Increased weight of water = W(1+ α) 20

� When the vertical acceleration is acting downward, the foundation shall try to move

� When the vertical acceleration is acting downward, the foundation shall try to move downward away from the body of the dam and thus reducing the effective wt. and stability of the dam and hence the worst case for designs. ◦ Reduced weight of the dam material = Wc(1 - α) ◦ Reduced weight of water = W(1 - α) 21

� Wave pressure ◦ Waves are generated on the surface of the reservoir by

� Wave pressure ◦ Waves are generated on the surface of the reservoir by blowing winds, which exert a pressure towards the d/s side. ◦ Wave’s pressure depends upon the wave height. ◦ Wave height is given by the equation (for F<32 km) 22

� For F>32 km hw = wave height (m) F = fetch or straight

� For F>32 km hw = wave height (m) F = fetch or straight length of reservoir (km) v = wind velocity. 23

◦ Maximum unit pressure = Pw = 2400 hw acting at 0. 125 hw

◦ Maximum unit pressure = Pw = 2400 hw acting at 0. 125 hw above still water ◦ Total wave force = Fw = 2000 hw 2 kg/m acting at 0. 375 hw above still water level. 24

Ice pressure � Considered for dams constructed in cold countries or at higher elevations

Ice pressure � Considered for dams constructed in cold countries or at higher elevations � It is exerted due to the formation of ice on the reservoir water surface � Ice expands and contracts due to change in temperature. � Ice pressure is caused due to expansion of ice. 25

� The face of the dam is subjected to force due to expansion of

� The face of the dam is subjected to force due to expansion of ice which is taken as 25 to 150 T/m 2. � It is applied to the face of the dam over its anticipated area of contact with ice. 26

Wind pressure � Wind pressure is normally ignored in the design of dams. �

Wind pressure � Wind pressure is normally ignored in the design of dams. � But if considered, it is taken as 100 to 150 kg/m 2 for the dam portion exposed to wind. 27

Design Parameters 28

Design Parameters 28

� The elementary profile of a gravity dam subjected only to water pressure on

� The elementary profile of a gravity dam subjected only to water pressure on the upstream face is a right angled triangle having zero width at the top and a maximum base width B, where the water pressure is maximum. � So it can be said that the elementary profile resembles the hydrostatic pressure distribution. � When the reservoir is empty, the only force acting is the self weight of the dam W at a distance B/3 from the heel. 29

Vertical stress distribution � Forces acting on the dam are � Weight of the

Vertical stress distribution � Forces acting on the dam are � Weight of the dam kg/m 3) is the unit weight of the dam material (2400 is the unit weight of water (1000 kg/m 3) 30

� Water pressure acting at H/3 from the base. � Uplift pressure c is

� Water pressure acting at H/3 from the base. � Uplift pressure c is the coefficient of uplift pressure intensity and is taken as 1. 31

� Normal stress without considering uplift pressure 32

� Normal stress without considering uplift pressure 32

� Normal stress distribution with uplift pressure considered 33

� Normal stress distribution with uplift pressure considered 33

For empty reservoir � Under general conditions, the maximum compressive stress occurs at the

For empty reservoir � Under general conditions, the maximum compressive stress occurs at the toe and can be calculated using e=B/6 and will be equal to 2 W/B. and the corresponding normal stress at heel will be zero. � The resultant will act at inner third (B/3 from toe). 34

For full reservoir � For no tension to develop, the resultant should act at

For full reservoir � For no tension to develop, the resultant should act at inner third point i. e. B/3 from the toe (L 2). � Taking zero. moments about L 2 and equating to 35

� Multiply both sides by 6/w. H When uplift pressure intensity is not considered,

� Multiply both sides by 6/w. H When uplift pressure intensity is not considered, 36

� For stability against sliding, the horizontal forces causing sliding (ΣH=P) are balanced by

� For stability against sliding, the horizontal forces causing sliding (ΣH=P) are balanced by the frictional forces opposing it (μΣV or μΣ(Wu). Hence P= μ(W-u) 37

when uplift is not considered. 38

when uplift is not considered. 38

Base width B is the maximum of the two given by above equations. Normal

Base width B is the maximum of the two given by above equations. Normal stress (at toe) is given by Under limiting condition e=B/6 then 39

At heel 40

At heel 40

Principal stress near the toe is the maximum normal stress in the dam and

Principal stress near the toe is the maximum normal stress in the dam and is given by 41

42

42

Shear stress at a horizontal plane near the toe 43

Shear stress at a horizontal plane near the toe 43

The principal and shear stress at the heel is zero since normal stress at

The principal and shear stress at the heel is zero since normal stress at the heel is zero. 44

Low and high gravity dams �A low dam is of limiting height such that

Low and high gravity dams �A low dam is of limiting height such that the resultant of all forces passes through the middle third and the maximum compressive stress at the toe does not exceed the permissible limit i. e. � Limiting side) height H is (ignoring uplift on safe 45

� The limiting height for the usual stresses of dam material w = 1000

� The limiting height for the usual stresses of dam material w = 1000 kg/m 3 ρ = 2400 kg/m 3 f = 30 kg/cm 2 or 300 ton/m 2 gives H=300/(1(2. 4+1))=88 m. 46

�A dam that exceeds the limiting height of a low dam is termed as

�A dam that exceeds the limiting height of a low dam is termed as a high dam. � In this case the resultant of water pressure and self weight of the dam passes through the middle third at certain sections of the dam body indicating the development of tension. 47

� The dam profile has to be adjusted by adding extra slope or batter

� The dam profile has to be adjusted by adding extra slope or batter on the upstream and downstream sides to limit the compressive stresses within the allowable limits. 48

Stability analysis � Gravity method ◦ The preliminary analysis of all gravity dams can

Stability analysis � Gravity method ◦ The preliminary analysis of all gravity dams can be made easily by isolating a typical x-section of the dam of a unit width. ◦ This section is assumed to behave independently of the adjoining section. In other words, the dam is considered to be made of a number of cantilevers of unit width each, which act independently of each other. 49

� The dam transfers the load to the foundation through cantilever action. � The

� The dam transfers the load to the foundation through cantilever action. � The loads are resisted entirely by the weight of individual cantilevers of unit length. � Other assumptions made are ◦ The dam is considered to be composed of a number of cantilevers each of 1 m thick and each of which are independent of the other. 50

� Hence for wide- U shaped valleys where transverse joints are not generally grouted

� Hence for wide- U shaped valleys where transverse joints are not generally grouted this assumption is nearly satisfied. � No loads are transferred to the abutments by beam action � The foundation and the beam behave as a single unit, the joint being perfect. 51

� No movements of the foundations coursed due to transference of loads. are �

� No movements of the foundations coursed due to transference of loads. are � The material in the foundation and body of dam are homogeneous. � Small opening made in the body of dam do not affect the general distribution of stresses and they only produce local affects 52

� Analytical method of analysis � Consider unit length of the dam � Find

� Analytical method of analysis � Consider unit length of the dam � Find out ΣV and ΣH � Calculate the lever arm of all the forces from the toe. � Calculate the sum of overturning moments ΣMo and righting moments (ΣMr) at toe. � Calculate the algebric sum of all those moments (ΣMr-ΣMo) 53

� Calculate toe the location of the resultant from � Calculate the eccentricity e

� Calculate toe the location of the resultant from � Calculate the eccentricity e of the resultant R from the centre of the base width B (<B/6 for no tension. 54

� Calculate heel the normal stress at the toe and � Calculate the maximum

� Calculate heel the normal stress at the toe and � Calculate the maximum normal stress and shear stress at the toe and heel. 55

neglect tail water 56

neglect tail water 56

� Pe is used while considering hydrodynamic pressure exerted by tail water during earthquake.

� Pe is used while considering hydrodynamic pressure exerted by tail water during earthquake. � Calculate the factor of safety against sliding (should be greater than unity) 57

� Calculate the shear friction factor (ranges from 3 -5) 58

� Calculate the shear friction factor (ranges from 3 -5) 58

Safety criteria for gravity dams � Safety factor against Overturning ◦ The dam has

Safety criteria for gravity dams � Safety factor against Overturning ◦ The dam has to be safe against overturning at any plane within the dam at the base or at any plane below the dam ◦ The overturning of the dam may take place if the resultant of all the forces acting on the dam passes outside the base. ◦ But practically speaking, before overturning takes place, other failures such as crushing of toe material, cracking of upstream material and increase in uplift and sliding may occur. 59

� The ratio of righting (stabilizing) moments about toe (anti-clockwise) to the over turning

� The ratio of righting (stabilizing) moments about toe (anti-clockwise) to the over turning moments about toe (clock-wise) is called factor of safety against over turning. � Its value generally varies between 1. 5 to 2. 5. 60

� Safety factor against sliding ◦ The horizontal loads including horizontal components of the

� Safety factor against sliding ◦ The horizontal loads including horizontal components of the loads acting on a dam are resisted by frictional or shearing forces along horizontal or nearly horizontal planes ◦ The total magnitude of the forces tending to induce sliding shall be less than the total available resistance along the sliding path. ◦ The resistance depends upon the cohesion and the angle of internal friction of the soil material. 61

� For cases where cohesion is insignificant, friction is the only force that resists

� For cases where cohesion is insignificant, friction is the only force that resists sliding. � The factor of safety should not be less than 2 and is given by W is the total weight of the dam, u is the uplift force P is the horizontal force. tanɸ is the coefficient of internal friction (0. 7 to 0. 8) 62

� If we consider cohesion too then Where C is the cohesion of the

� If we consider cohesion too then Where C is the cohesion of the material and A is the area in sq. meters. For full reservoir conditions and maximum flood discharge, it should not be less than 3 and should not be less than 1. 5 under maximum flood discharge with extreme uplift condition 63

� Safety against crushing ◦ It is ensured if the compressive stresses produced are

� Safety against crushing ◦ It is ensured if the compressive stresses produced are within the allowable limits ◦ Equations for normal stresses for toe and heel have been already described. ◦ When pressures at both toe and heel are compressive, the max. compressive stress occurs at the toe when e = B/6. ◦ Excessive stresses at toe and heel can be brought into limits by providing fillets at slopes 1: 1 on u/s and 2: 1 on d/s at heights 64

h = fillet height (m) H = dam height (m) The maximum allowable compressive

h = fillet height (m) H = dam height (m) The maximum allowable compressive stress is taken as 30 kg/cm 2. 65

� Safety against tension ◦ A gravity dam is usually designed such that no

� Safety against tension ◦ A gravity dam is usually designed such that no tension is developed in the dam body and for that the resultant must lie within the middle third. ◦ If the eccentricity e is greater than B/6, tension is developed at the heel. 66