Thus the shear stress on the incline is expected to be positive, which is consistent with our answer.COMMENTS1.In the intuitive check, three of the components gave one answer, whereas the fourth gave an opposite answer. What happens if two intui-tive deductions are positive, two intuitive deductions are negative, and the stress components are nearly equal in magnitude? The ques-tion emphasizes that intuitive reasoningis a quick and important check on results, but one must be cautious with the conclusions.2.We could have balanced forces in the xand ydirections, in which case we have to find the xand ycomponents of the normal and tan-gential forces on the force wedge. After removing the common factors Δt Δzwe would obtainSolving these two equations, we obtain the values of σAandτAas before. By balancing forces in the n, tdirections we generated oneequation per unknown but did extra computation in finding components of forces in the nand tdirections. By balancing forces in the xand ydirections, we did less work finding the components of forces, but we did extra work in solving simultaneous equations. Thisshows that the important point is to balance forces in any two directions, and the direction chosen for balancing the forces is a matter ofpreference.3.Figure 8.8 is useful in reducing the algebra when forces are balanced in the nand tdirections. But you may prefer to resolve compo-nents of individual forces, as shown in Figure 8.10, and then write the equilibrium equations. The method is a little more tedious, buthas the advantage that the intuitive check can be conducted as one writes the equilibrium equations as follows.The normal stress σAon the incline will be:•Tensile due to σxx; Compressive due to σyy; Compressive due to τxy;Compressive due to τyx.As σxxis the smallest stress component, it is not surprising that the total result is a compressive normal stress on the inclined plane.The shear stress on the incline will be:•Positive due to σxx;Positive due to σyy; Positive due to τxy;Negative due to τyx.We expect the net result to be positive shear stress on the incline.PROBLEM SET 8.1Stresses by inspection8.1In Figure P8.1, determine by inspection (a) if the normal stress on the incline AA is in tension, compression, or cannot be determined; (b)if the shear stress on the incline AA is positive, negative, or cannot be determined. 8.2In Figure P8.2 determine by inspection (a) if the normal stress on the incline AA is in tension, compression, or cannot be determined; (b)if the shear stress on the incline AA is positive, negative, or cannot be determined. σA30°sinτA30°cos+50 MPa()–30°cos30 MPa()30°28.30 MPa–=sin+=σA30°cosτA–30°sin50 MPa()–30°sin60 MPa()30°cos–76.96 MPa–==Figure 8.10Alternative force resolution.