advertisement

SIMPLE STRESS JANNA FEB B. SIMENE, CE SHEAR STRESS -force applied is parallel to the resisting area τ = Shear Force (V) Area EXAMPLES What force is required to punch a 20-mmdiameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2 . SOLUTION: τ= V A 350MPa = V (π)(20)(25) V = (350)( (π)(20)(25) = 549,778 N The bell crank shown is in equilibrium. (a) Determine the required diameter of the connecting rod AB if its axial stress is limited to 100MN/m2. (b) Determine the shearing stress in the pin at D if its diameter is 20mm. a) Determine the required diameter of the connecting rod AB if its axial stress is limited to 100MN/m2. SOLUTION: ∑MD=0 (P)(200)=(30sin60)(240) P=31.18KN σ=P A 100MPa = (31.18KN)(1000) (π)(D²/4) D = 19.92mm 30cos60 30sin60 (b) Determine the shearing stress in the pin at D if its diameter is 20mm. SOLUTION: ∑Fx=0 P + 30cos60= RD RDH=46.18KN ∑Fy=0 RDV = 30sin60 RDV=25.98KN RD² = RDV² + RDH² = 52.986KN RDH = 46.18KN RDV = 25.98KN (b) Determine the shearing stress in the pin at D if its diameter is 20mm. SOLUTION: τ= = V A (52.986KN)(1000) (π)(20)²(2) 4 = 84.33MPa 30cos60 RD 30sin60 BEARING STRESS -developed when there’s contact pressure between two bodies EXAMPLES In the figure shown, assume that a 20mm diameter rivet joins the plates which are each 100mm wide. (a) If the allowable stresses are 140MN/m² for bearing in the plate material and 80MN/m² for shearing of the rivet, determine the minimum thickness of each plate. (b) Under the conditions specified in part (a) , what is the largest average tensile stress in the plate. 100mm In the figure shown, assume that a 20mm diameter rivet joins the plates which are each 100mm wide. (a) If the allowable stresses are 140MN/m² for bearing in the plate material and 80MN/m² for shearing of the rivet, determine the minimum thickness of each plate. P A 140MN/m² = P (20)(t) τPLATE = = τRIVET 80MN/m² = t P A P (π)(20)² 4 P = 25,133N P = 2800t D=20mm SHEAR ON PLATE P P D=20mm 25,133N = 2800t t= 8.98mm SHEAR ON RIVET (b) Under the conditions specified in part (a) , what is the largest average tensile stress in the plate. D=20mm τPLATE = P A = 25,133N (100x8.98)-(20x8.98) = 35MPa t= 8.98mm W=100mm SHEAR ON PLATE The lap joint shown in the figure is fastened by three 20mm diameter rivets. Assuming P= 50KN , determine (a) The shearing stress in each rivet (b) The bearing stress in each plate (c) The maximum average tensile stress in each plate. Assume that the axial load P is distributed equally among the three rivets. (a) The shearing stress in each rivet τRIVET = P A = (50KN)(1000) (π)(20)² (3) 4 = 53.05MPa 50KN 50KN (b) The bearing stress in each plate τPLATE = P A = (50KN)(1000) (20) (3)(25) = 33.33MPa 50KN 50KN (c)The maximum average tensile stress in each plate. τPLATE = P A = (50KN)(1000) (130-20) (25) = 18.18 MPa D=20mm t= 25mm W=130mm SAMPLE PROBLEMS THIN-WALLED CYLINDERS OR THIN-WALLED PRESSURE VESSELS -A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections. TANGENTIAL STRESS/ CIRCUMFERENTIAL STRESS LONGITUDINAL STRESS SPHERICAL SHELL TANGENTIAL/CIRCUMFERENTIAL STRESS - Consider the tank shown being subjected to an internal pressure p. The length of the tank is L and the wall thickness is t. σt = inside pressure(p) x radius (r) = (p) (D) thickness (t) 2(t) = (pi - po) (r) (t) where: T are total tension in the walls F are the total forces from the internal pressure, p Take Note! T = (σt)(Awall) = (σt)(tL) F = pA = pDL EXAMPLES Calculate the minimum wall thickness for a cylindrical vessel that is to carry a gas at a pressure 1400psi. The diameter of the vessel is 2ft, and the stress is limited to 12ksi. SOLUTION: σt = (p) (r) t (12)(1000) = (1400)(1 x 12) t t= 1.4in 1400psi AXIAL OR LONGITUDINAL STRESS The total force acting at the rear of the tank F must equal to the total longitudinal stress on the wall PT. Due to the forces on the ends of the cylinder, walls will stretch along the axial direction and this will induce longitudinal stress on the cylindrical walls. σL = pressure on the wall (p) x radius (r) = (p) (D) 2 x thickness (t) 4t Take Note! Take Note! σt =2(σL) PL= (σL)(Awall) =(σL)(Dt) F = pA = pπD² 4 =pπr² EXAMPLES A cylindrical steel pressure vessel 400 mm in diameter with a wall thickness of 20 mm, is subjected to an internal pressure of 4.5 MN/m2 . Calculate the longitudinal stresses in the steel. SOLUTION: σL = (p) (D) 4t = (4.5MPa)(0.4) (4)(0.02) = 22.5MPa SPHERICAL SHELL σs = internal pressure(p) x radius (r) 2 x thickness (t) Take Note! PT= (σt)(Awall) =(σt)(Dt) F = pA = pπD² 4 =pπr² EXAMPLES The wall thickness of a 4ft diameter spherical tank is 5/16inch. Calculate the allowable internal pressure if the stress is limited to 8000psi. SOLUTION: σs = (p) (r) 2(t) 8000psi = (p) (2 x 12) (2)(5/16in) = 208.33psi SAMPLE PROBLEMS END OF CHAPTER 1