Considering plane BCO in figure 63, OXD AY, sec BD = σXD AY, COS BD + TXYD AY, tan Bo cos Bo + TXYD AY, sin BD + OYD AY, tan Bo sin Bo ་ OXD XD COS2 Bo+ ovo sin2 Bo + 2 TxYD sin Bo cos Bo D (239) Considering figure 64, the shear stress parallel to CB, on plane BCO, is CBD = TzXD COS Bo - TZYD Sin Bo Considering figure 65, the shear stress parallel to AD is the resultant shear stress on ADO because the shear stress perpendicular to it is zero. Considering figure 66, two of the principal stresses are given by the Where (o'zDo'xD) > 0, use + for radical sign. Where (o'zDσ'xD) < 0, use for radical sign. ZD tan 2 αp XD . Στίχο στο σχο If tan 2 a, is +, 0 < a < 45° If tan 2 a, is-, -45°<α。 <0 (242) Positive angles are measured clockwise from the Z' axis on the left part of the dam and counterclockwise on the right part of the dam, looking upstream. The two values of opp given by equation 241 and the normal pressure Po are the three principal stresses at the face of the dam. The maximum shear stress acts on a plane that bisects the angle between the largest and the smallest of the principal stresses. It is equal to half the difference between the principal stresses. A summary of stresses calculated at the downstream face of the dam, with symbols and equation numbers, is tabulated below. If the extrados and intrados curves are concentric circular arches, the angle ' becomes D and equation 238 gives the cantilever stress parallel to the downstream face. D 183. Stresses at Upstream Face.-Stresses at the upstream face are derived in the same way as for the downstream face. The resulting equations are STRESSES ON HORIZONTAL AND VERTICAL PLANES 184. Procedure. After stresses have been evaluated for the faces of a dam, equations may be derived for stresses on horizontal planes and vertical radial and tangential planes at any interior point in the dam. Since horizontal arch stresses vertical cantilever stresses oz, and tangential cantilever shear stresses 7zx, are assumed to have a linear variation from the upstream face to the downstream face, equations can be derived for radial cantilever shear stresses τzy, radial arch shear stresses TMxy, and radial stresses normal to the vertical tangential planes σy. Since TXY equals Tуx, Txz equals тzx, and Tyz equals Tzy; it is evident that the six stresses, ox, σy, σz, txy, тxz, and τyz, determine all stresses acting on three mutually perpendicular planes. These planes are a vertical radial plane YZ, a vertical tangential plane XZ, and a horizontal plane XY. TYZ 185. Normal Cantilever and Arch Stresses.-Vertical cantilever stresses on a horizontal plane, see figure 67, may be calculated by the equation, W M M Ic [Y-(T-19)] = 020 + Y Ic (255) Horizontal arch stresses on a vertical radial plane, see figure 68, may stresses, acting in tangential directions on horizontal planes, may be Formulas for horizontal cantilever shear stresses, acting in radial directions on horizontal planes, see figure 67, may be derived in the following manner: assume a parabolic distribution of shear and let The total shear on a parallel-side horizontal plane, one foot wide, between the origin and any point y is, In considering horizontal planes of cantilevers with radial sides, it is assumed that the total shear on a plane, one foot wide at the arch center line, is equal to the total shear on a horizontal plane with parallel sides one foot apart. Substituting the above value of b, in equation 259, (260) (261) (262) |