Transcript Slide 1
Shear stress versus shear strain For homogeneous & isotropic materials pure shear causes angular distortion of a material as shown above. Pure shear is most often studied by torsional tests of thin bars. Behavior is completely analogous to normal stresses. Shear stress-strain diagram G is shear modulus or modulus of rigidity G E G 2(1 ) The shear stress-strain diagram for a titanium alloy bar tested in torsion is shown in the Figure. Determine i) Shear modulus ii) Proportional Limit iii) Ultimate stress iv) Yield stress v) Fracture stress A bearing pad used to support machines is shown. The following formulae can be used to determine shear stress, shear strain and displacement. aver aver G V ab V abG V d h tan h tan abG For small angles d h hV abG Problem (4): The plastic block shown of 50 mm depth is bonded to a rigid support and to a vertical plate to which a 240-kN load P is applied. Knowing that for the plastic used G=1000 MPa, determine the deflection of the plate. 80 mm 120 mm Rigid support P Spring 2006 Mid Term Question Problem (4): The plastic block shown of 50 mm depth is bonded to a rigid support and to a vertical plate to which a 240-kN load P is applied. Knowing that for the plastic used G=1000 MPa, determine the deflection of the plate. 80 mm aver 120 mm V 240 ,000 40 MPa ab 0.05 * 0.12 aver 40 0.04 rads G 1000 Rigid support P Spring 2006 Mid Term Question d h 80* 0.04 3.2mm Allowable Stress Factor of safety fail F.S. allow fail F .S. allow allow allow F.S.=1 fail F.S. fail F.S. F.S.=3 Designing for allowable stress Bearing (compression) A P allow Tension A V allow Shear Problem (6): Determine the intensity w of the maximum distributed load that can be supported by the hanger assembly so that an allowable shear stress of allow = 15 ksi is not exceeded in the 0.40-in.-diameter bolts at A and B, and an allowable tensile stress of σallow = 25 ksi is not exceeded in the 0.5-in.-diameter rod AB. Spring 2006 Mid Term Question Solve at home – will cover in help session on Sept 7th