Substituting the above value of c, in equation 261,

Values of TYZU and TYZD are given by equations 247 and 235, respectively.

187. Radial Shear in Arch Elements.-An equation for horizontal arch shear stress acting in a radial direction on a vertical radial plane, see figure 68, may be derived by the procedure followed in deriving equation 264. The formula obtained is

188. Horizontal Radial Stresses.-Horizontal radial stresses, acting on vertical tangential planes, see figures 67 and 68, may be computed by the formula,

Integral terms in equation 266 may be computed from values of Tyz and TX given by equations 264 and 265, respectively. Since y is measured from the downstream face, it is a function of z and must be so considered in differentiating Tyz and тxy. For example, the bracket term multiplied by y, equation 264, is the product of two functions of z and is

(267)

Following this method, Tyz is first differentiated with respect to z .

This expression is then integrated with respect to y.

3y2 Tyzu 3Y2TYZD 6Y2VC 2YT YZU

+

T2

The same procedure is followed in deriving a formula for the second integral term of equation 266, obtaining

If temperature changes are being considered in the analysis, equation 277 must be amplified to include temperature effects.

The derivation of σyD is explained in section 182, and its value is given by equation 236. When y equals T, oy becomes σyu, which is determined by equation 248.

The sum of equations 270, 274, and 236 gives equation 266.

189. Summary of Stresses.-A summary of stresses calculated on horizontal and vertical planes, with symbols and equation numbers, is tabulated below.

190. General. If a dam is to be increased in height, it may be necessary to add new concrete on the downstream face. As this introduces a possible plane of weakness, investigations of normal and shear stresses along the face are required.

TYZ,

Equations for normal and shear stresses on an inclined circumferential area, one parallel to the X axis and inclined to the Y and Z axes, are determined by using the six stresses, ox, oy, σZ, TXY, TXZ, and considering the elemental area as a plane. Forces acting on an element, see figure 69, are equated to derive stress equations. It is assumed that forces on planes BFC and AED are self balancing. It must be remembered that TXY equals TуX, TXZ equals TZX, and TYZ equals TZY.

191. Stress Equations.-Considering the forces in figure 69b, the following stress equations are found:

Horizontal shear stress on inclined plane,

τι sec θ Δχ Δε = TXY AX AZ + Txz tan e Ax Az

T1 = TXY COS e + Txz sine

(278)

Shear stress on inclined plane, acting at right angles to horizontal

The preceding formulas for stresses on an inclined circumferential area may be checked by substituting direction cosines, given in section 192, in equations 282 and 280.

STRESSES ON OBLIQUE PLANE

192. General.-Equations for normal and shear stresses on an oblique plane are derived so that stresses in any direction can be determined at any point in the dam.

Consider the tetrahedron OABC cut from the dam as shown in figure 70. The three mutually perpendicular planes are taken as coordinate. planes, and the unit stresses, not the total forces, are shown on the faces. If O is the point where stresses on the XY, XZ, and YZ planes are known, stresses on the plane ABC approach the stresses on a plane through O parallel to plane ABC if the tetrahedron OABC is made infinitesimal. Since the element is made very small, body forces can be neglected, and stresses over the sides can be assumed as uniformly distributed.