 # PRINCIPAL STRESSES AND SHEAR STRESS

• The vertical stress intensity is not the maximum direct stress intensity within the dam section.
• It is therefore necessary to determine the intensity of the maximum normal stress.
• The maximum normal stress occurs on the major principal plane.
• The D/S face of the dam which is inclined at an angle α to the vertical is a principal plane, as no shear stress acts on it.
• If any tail water is there it contributes a pressure p’ entirely normal to the plane and is one of the principal stresses.
• The second principal plane will be inclined at right angles to the face of the dam and hence at an angle α to the horizontal.
• The stress acting on such plane is the major principal stress Pmax.
• All the stresses can be represented by Mohr’s circle. See Fig. 13.6.
• The vertical stress Pn at toe of the dam in terms of principal stresses may be given as follows.

$P_{n}= P+(P_{max}-P)(1+cos2\Theta )$

$P_{n}= P+(P_{max}-P)cos^{2}\Theta$

$P_{max}= \frac{P_{n}}{cos^{2}\Theta }-\frac{P(1-cos^{2}\Theta )}{cos^{2}\Theta }$

$= P_{n}sec^{2}\Theta -p.tan^{2}\Theta$

For maximum value of compressive stress, the tail water depth is assumed as zero. Hence p=0

$P_{max}= p_{n}sec^{2}\Theta$

Shear stress pt from Mohr’s circle is given by

$P_{t}= \left ( \frac{P_{max}-P}{2} \right )sin2\alpha$

$= (P_{max}-p)sin\Theta cos\Theta$

Putting

$p= 0-and-p_{max}= p_{n}sec^{2}\Theta -in-this-equation-we-get$

$p_{t}= p_{n}sin\alpha cos\Theta sec^{2}\Theta$

$= p_{n}tan\Theta$

Equation (1)

• gives the maximum intensity of direct stress which will occur near the D/S toe and which will act on a plane inclined at an angle θ to the horizontal.
• Equation (2) given intensity of shear stress on a horizontal plane near Toe.
• The maximum intensity of stress at heel can also be similarly computed. For U/S face let

$P_{max}$

Major principal stress on a plane inclined at βº with the horizontal. (Maximum intensity of direct stress). ## Similar Posts

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