Thus the shear stress on the incline is expected to be positive which is

# Thus the shear stress on the incline is expected to

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Thus the shear stress on the incline is expected to be positive, which is consistent with our answer. COMMENTS 1. 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 reasoning is a quick and important check on results, but one must be cautious with the conclusions. 2. We could have balanced forces in the x and y directions, in which case we have to find the x and y components of the normal and tan- gential forces on the force wedge. After removing the common factors Δ t Δ z we would obtain Solving these two equations, we obtain the values of σ A and τ A as before. By balancing forces in the n , t directions we generated one equation per unknown but did extra computation in finding components of forces in the n and t directions. By balancing forces in the x and y directions, we did less work finding the components of forces, but we did extra work in solving simultaneous equations. This shows that the important point is to balance forces in any two directions, and the direction chosen for balancing the forces is a matter of preference. 3. Figure 8.8 is useful in reducing the algebra when forces are balanced in the n and t directions. 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, but has the advantage that the intuitive check can be conducted as one writes the equilibrium equations as follows. The normal stress σ A on the incline will be: Tensile due to σ xx ; Compressive due to σ yy ; Compressive due to τ xy ;Compressive due to τ yx . As σ xx is 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 σ xx ;Positive due to σ yy ; Positive due to τ xy ;Negative due to τ yx . We expect the net result to be positive shear stress on the incline. PROBLEM SET 8.1 Stresses by inspection 8.1 In 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. 8.2 In 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. σ A 30 ° sin τ A 30 ° cos + 50 MPa ( ) 30 ° cos 30 MPa ( ) 30 ° 28.30 MPa = sin + = σ A 30 ° cos τ A 30 ° sin 50 MPa ( ) 30 ° sin 60 MPa ( ) 30 ° cos 76.96 MPa = = Figure 8.10 Alternative force resolution.