332 Mechanics of Materials §13.6 Consider once again the equilibrium of a triangular block of material of unit depth (Fig.13.8);this time AC is a principal plane on which a principal stress p acts,and the shear stress is zero (from the property of principal planes). ÷0 Fig.13.8. Resolving forces horizontally, (ox×BC×1)+(txy×AB×1)=(op×AC×1)cos8 Ox+txy tan0=op tan0=p-ox (13.14) 下y Thus we have an equation for the inclination of the principal planes in terms of the principal stress.If,therefore,the principal stresses are determined and substituted in the above equation,each will give the corresponding angle of the plane on which it acts and there can then be no confusion. The above formula has been derived with two tensile direct stresses and a shear stress system,as shown in the figure;should any of these be reversed in action,then the appropriate minus sign must be inserted in the equation. 13.6.Graphical solution-Mohr's stress circle Consider the complex stress system of Fig.13.5 (p.330).As stated previously this represents a complete stress system for any condition of applied load in two dimensions. In order to find graphically the direct stress and shear stress te on any plane inclined at 0 to the plane on which a acts,proceed as follows: (1)Label the block ABCD. (2)Set up axes for direct stress (as abscissa)and shear stress (as ordinate)(Fig.13.9). (3)Plot the stresses acting on two adjacent faces,e.g.AB and BC,using the following sign conventions: direct stresses:tensile,positive;compressive,negative; shear stresses:tending to turn block clockwise,positive;tending to turn block counterclockwise,negative. This gives two points on the graph which may then be labelled AB and BC respectively to denote stresses on these planes.332 Mechanics of Materials 513.6 Consider once again the equilibrium of a triangular block of material of unit depth (Fig. 13.8); this time AC is a principal plane on which a principal stress up acts, and the shear stress is zero (from the property of principal planes). UY t Fig. 13.8. Resolving forces horizontally, (1 3.14) Thus we have an equation for the inclination of the principal planes in terms of the principal stress. If, therefore, the principal stresses are determined and substituted in the above equation, each will give the corresponding angle of the plane on which it acts and there can then be no confusion. The above formula has been derived with two tensile direct stresses and a shear stress system, as shown in the figure; should any of these be reversed in action, then the appropriate minus sign must be inserted in the equation. 13.6. Graphical solution - Mohr’s stress circle Consider the complex stress system of Fig. 13.5 (p. 330). As stated previously this represents a complete stress system for any condition of applied load in two dimensions. In order to find graphically the direct stress and shear stress tg on any plane inclined at 8 to the plane on which CT, acts, proceed as follows: (1) Label the block ABCD. (2) Set up axes for direct stress (as abscissa) and shear stress (as ordinate) (Fig. 13.9). (3) Plot the stresses acting on two adjacent faces, e.g. AB and BC, using the following sign conventions: direct stresses: tensile, positive; compressive, negative; shear stresses: tending to turn block clockwise, positive; tending to turn block counterclockwise, negative. This gives two points on the graph which may then be labelled AB and E respectively to denote stresses on these planes