For the shown state σx = 18 MPa, σy = 12 MPa, τxy = +4 MPa, select the free-body diagram showing maximum in-plane shear and its associated normal stresses.
The mean normal stress is $(18+12)/2=15$ MPa. Mohr-circle radius is $\sqrt{[(18-12)/2]^2+4^2}=5$ MPa, so maximum in-plane shear magnitude is 5 MPa. For the relevant clockwise rotation, $\tan(2\theta_s)=-(18-12)/(2\times4)=-3/4$, giving θs = −18.4°. Applying the traction signs at this orientation gives B: outward 15 MPa normal tractions and the indicated 5 MPa shear sense.
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