1.6 Determination of Stresses on Inclined Planes 13
ThenegativesignforτindicatesthattheshearstressisinthedirectionBAandnotinAB.
FromEq.(1.9)whenτxy=0,
τ=(σx−σy)(sin2θ)/2(i)
The maximum value ofτtherefore occurs when sin2θis a maximum—that is, when sin2θ=1and
θ= 45 ◦.Then,substitutingthevaluesofσxandσyinEq.(i),
τmax=(57.4− 75 )/ 2 =−8.8N/mm^2
Example 1.2
Acantileverbeamofsolid,circularcrosssectionsupportsacompressiveloadof50kNappliedtoits
freeendatapoint1.5mmbelowahorizontaldiameterintheverticalplaneofsymmetrytogetherwith
atorqueof1200Nm(Fig.1.10).Calculatethedirectandshearstressesonaplaneinclinedat60◦tothe
axisofthecantileveratapointontheloweredgeoftheverticalplaneofsymmetry.
Thedirectloadingsystemisequivalenttoanaxialloadof50kNtogetherwithabendingmoment
of50× 103 ×1.5=75000N/mminaverticalplane.Therefore,atanypointontheloweredgeofthe
verticalplaneofsymmetry,therearecompressivestressesduetotheaxialloadandbendingmoment
whichactonplanesperpendiculartotheaxisofthebeamandaregiven,respectively,byEqs.(1.2)and
(15.9):
σx(axialload)= 50 × 103 /π×( 602 / 4 )=17.7N/mm^2
σx(bendingmoment)= 75000 × 30 /π×( 604 / 64 )=3.5N/mm^2
The shear stress,τxy, at the same point due to the torque is obtained from Eq. (iv) in Example 3.1,
thatis,
τxy= 1200 × 103 × 30 /π×( 604 / 32 )=28.3N/mm^2
Fig.1.10
Cantilever beam of Example 1.2.