Report

Shear Stress and Strain Shear Stress, Shear Strain, Shear Stress and Strain Diagram 1 Shear Stress ave V V: shear load A A: area of the plane ( the force is acted on the center of designated area) Lecture 1 2 Shear Stress: Load (V) 3 Shear Stress: pin and rivet ave VA d 2 4 VA= 0.5 FA 4 Shear Stress: pin and rivet What is the value of V and A? 5 Shear Stress: pin and rivet What is the value of V and A? Sketch its FBD to show the value of V V= 1 bolt: V= 2 parallel rivet V = 3 parallel rivet V = Vx= Vy = Total VR = 6 Shear Stress: torque Shear load due to torsion V T V Equilibrium equation: T 0 T VD 0 V T D 7 Shear Stress: torque Calculate the V and A. Sketch the FBD. T= 10 Nm ccw Bolt d = 10mm Position of the bolt R = 100 mm 8 Allowable Shear Stress • The allowable stress is the maximum shear stress allowed due to the properties of the material. all ave Considered as SAFE all ave Considered as FAIL F .S all ave Safety Factor Safe F.S >= 1 Fail F.S < 1 9 Shear Strain Shear strain is detected the changes in angle is detected. 3 Normal strains (ex, ey, ez ) and 3 shear strains (gxy, gyz, gxz) are detected. 10 Approximate length of the edges (1 e x ) x (1 e y ) y (1 e z ) z Approximate angles between three sides 2 g xy 2 g yz 2 g xz 11 The Shear Stress–Strain Diagram • For pure shear, equilibrium requires equal shear stresses on each face of the element. • When material is homogeneous and isotropic, shear stress will distort the element uniformly. The Shear Stress–Strain Diagram • For most engineering materials the elastic behaviour is linear, so Hooke’s Law for shear applies. Gg G = shear modulus of elasticity or the modulus of rigidity • 3 material constants, E, and G are actually related by the equation G E 2 1 v Example A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram is shown. Find the shear modulus G, the proportional limit, and the ultimate shear stress. Also, find the maximum distance d that the top of a block of this material could be displaced horizontally if the material behaves elastically when acted upon by a shear force V. What is the magnitude of V necessary to cause this displacement? Example Discuss the approach? G: slope of the elastic curve Yield shear stress: from curve Ultimate shear stress: from curve tan (0.008 rad) = d/50 V A V A Solution: The coordinates of point A are (0.008 rad, 360 MPa). Thus, shear modulus is G 360 0 . 008 MPa 45 10 3 (Ans) By inspection, the graph ceases to be linear at point A. Thus, the proportional limit is pl 360 MPa (Ans) This value represents the maximum shear stress, point B. Thus the ultimate stress is u 504 MPa (Ans) Since the angle is small, the top of the will be displaced horizontally by tan 0 . 008 rad 0 . 008 d d 0 . 4 mm 50 mm The shear force V needed to cause the displacement is avg V A ; 360 MPa V 75 100 V 2700 kN (Ans) Solve it! The shear stress-strain diagram for a steel is shown in the figure. If a bolt is having a diameter of 20 mm is made of this material and used the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield. Take u = 0.3. Discuss the approach??? 1) FBD of the bolt V P 2 2) Ssy = 420 MPa, calculate the P y 420 P/2 d / 4 2 P 263 . 9 N 3) G is slope of the curve G 420 / 0 . 00545 77 . 064 GPa 4) Calculate E based on value G and u G E 2 1 v Review Question A load W = 1000 kg is supported by a 3‐cable system as shown in figure 1. The cable run along 2 pulleys F and D which is attached to the solid wall by a round beam EF and CD both of which are 30mm diameter and 300mm long. Ignore the mass of cable and beam. All cables are the same size. The sizes of the pulleys at D & F can be ignored. Gravity = 9.81m/s2 Answer the following questions: Calculate the tension of cable AB, BD, and BF Calculate the diameter of the cable is the allowable stress is 140 MPa, assume all the cables have the same diameter. Find the internal stress at the mid point of round beam FE 20 Equilibrium TBF TBD F X 0 T BD cos( 29 . 5 ) T BF cos( 45 . 5 ) F Y 0 T BD sin( 29 . 5 ) T BF sin( 45 . 5 ) 9810 0 T BF 8840 N W T BD 7118 N Stresses: the highest will be cable BA as it has the highest load all A T A T all 9810 70 . 1mm 2 140 d 9 . 44 mm 21 Internal forces at middle of BE (150mm) TBF F Y 0 V T BF T BF sin( 45 . 5 ) 0 P V V 2535 N M M F 0 M V * 150 TBF 380250 Nmm F x 0 T BF cos( 45 . 5 ) P 0 P 6196 N Lecture 1 22 Example The joint is fastened together using two blots. Determine the required Diameter of the bolts if the failure shear stress of the bolt is 350 Mpa. Use factor of safety of F.S = 2.5. 23 V for each bolt V 80 , 000 / 4 20 , 000 kN Shear stress resulted from V 20 , 000 d / 4 2 Allowable shear stress all F .S Therefore: 350 140 MPa 2 .5 all d 13 . 5 mm d 14 mm 24 Example The A-36 steel wire AB has a cross-sectional area of 10 mm2 and unstretched when q =45o. Determine the applied load P needed to cause q = 44.9o. 25 Initial length LAB L AB 400 400 ) 2 2 565 . 68 mm Final elongation L ' AB 400 400 2 ( 400 )( 400 ) cos( 90 . 2 ) 2 2 566 . 67 mm Strain e 566 . 67 565 . 68 1 . 75 (10 ) 3 mm / mm 565 . 68 Normal Stress on cable E e 200 (10 ) (1 . 75 )(10 ) 3 3 350 MPa 26 The tension of the cable T A T A 350 (10 ) 3500 N FBD when the cable is stretched O M O 0 P cos( 0 . 2 )( 400 ) T sin( 44 . 9 )( 400 ) 0 P 2 . 46 kN P 27