Fundamental Concepts of Stress PDF

Fundamen tal Concept of stress  LEARNING OUTCOMES: By the end of this lecture, students should be able to:  Define stress and classify load/force  Solve problems involving normal stress  Solve problems involving shearing stress  Solve problems involving bearing stress ...

Fundamen tal Concept of stress  LEARNING OUTCOMES: By the end of this lecture, students should be able to:  Define stress and classify load/force  Solve problems involving normal stress  Solve problems involving shearing stress  Solve problems involving bearing stress  Solve problems involving tangential & longitudinal stress in thin-walled vessel What is Stress? Stress is the a measure of what the material feels from externally applied forces. It is simply a ratio of the external forces to the cross sectional area of the material. Watch this! (8) An Introduction to Stress and Strain - YouTube  stress 𝑭 𝝈= 𝑨 Where: = average normal stress F = Normal force A = cross section area  NORMAL TYPES STRESS is the stress developed when force applied is acting normally to the surface. OF SHEARIN SIMPL G STRESS the stress developed due to a shear force that tends to slide one section pass the other E section of a body. BEARING STRES STRESS the contact pressure S between two surfaces of bodies.  NORMAL TYPES STRESS is the stress developed when force applied is acting normally OF to the surface. Compressiv tensile SIMPLe STRESS the stress developed due to the STRESS E compressive force that compress or shorten the tends to the stress developed due to the tensile force that tends to elongate the body STRESmember S  Problem 1  The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown.  solution  The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown. a b c a b c solution Section a-a 𝑷 𝑨𝑩 (T) Section b-b 𝑃 𝐵𝐶 =12 𝑘𝑁 +9 𝑘𝑁 + 9 𝑘𝑁 𝑷 𝑩𝑪 (T) solution Section c-c 𝑷 𝑪𝑫 𝑃 𝐶𝐷 =12 𝑘𝑁 + 9 𝑘𝑁 +9 𝑘𝑁 − 4 𝑘𝑁 − 4 𝑘𝑁 (T) Solve for the maximum average normal stress 𝑃 𝐵𝐶 30 ( 10 3 ) 𝑁 𝜎= ¿ Ans 𝐴 (0.035 𝑚)(0.010 𝑚)  Problem 2  Member AC shown is subjected to a vertical force of 3 kN. Determine the position x of this force so that the average compressive stress at the smooth support C is equal to the average tensile stress in the tie rod AB. The rod has a cross-sectional area of 400 mm2 and the contact area at C is 650 mm2.   solution Member AC shown is subjected to a vertical force of 3 kN. Determine the position x of this force so that the average compressive stress at the smooth support C is equal to the average tensile stress in the tie rod AB. The rod has a cross-sectional area of 400 mm2 and the contact area at C is 650 mm2. Consider FBD 1, Eq. 1 + ↺+ FBD 1   solution Member AC shown is subjected to a vertical force of 3 kN. Determine the position x of this force so that the average compressive stress at the smooth support C is equal to the average tensile stress in the tie rod AB. The rod has a cross-sectional area of 400 mm 2 and the contact area at C is 650 mm2. For average normal stress, By Substitution with Eq. 1 Eq. 1, 𝐹 𝐶 =1.625 𝐹 Solve 𝐴𝐵 for x from Eq.2, Ans  Shear 𝑽 𝝉= stress 𝑨 Where: = average shear stress V = shear force A = sheared area  Shear STRESS Shear stress is tangent or parallel to the plane TYPES on which it acts. Shear stress arises whenever the applied loads cause one section of a body to slide past its adjacent section. Shearing stress is OF PUNCHI also known as tangential stress. Single Double NG SIMPLshear shear shear E STRES S  Problem 3  Determine the average shear stress in the 20-mm-diameter pin at A and the 30-mm-diameter pin at B that support the beam in the figure shown.   Determine the average shear stress in the 20-mm- diameter pin at A and the 30-mm-diameter pin at B that solution support the beam in the figure shown. Determine the reaction at A and the ↺+ tension FB = 12.5 kN + = 7.50 kN + = 20 kN 𝐹 𝐴= √ 𝐴𝑥+ 𝐴𝑦 =√(7.50𝑘𝑁) +(20𝑘𝑁) =21.36𝑘𝑁 2 2 2 2   Determine the average shear stress in the 20-mm- diameter pin at A and the 30-mm-diameter pin at B that solution support the beam in the figure shown. Determine values of shear Double Shear @ A = Ans Single Shear @ B = =12.50 kN = Ans  Problem 4  The inclined member shown is subjected to a compressive force of 600 lb. Determine the average compressive stress along the smooth areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by DB.   The inclined member shown is subjected to a Solution compressive force of 600 lb. Determine the average compressive stress along the smooth areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by DB. Consider FBD 1 and solve for the compressive forces + = 360 lb + = 480 lb Consider FBD 2 and solve for shear + = 360 lb FBD 2 FBD 1   The inclined member shown is subjected to a compressive force of 600 lb. Determine the average compressive stress along the smooth Solution areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by DB. The average compressive stresses = = Ans Ans The average shear stress Ans FBD 2 FBD 1  𝑷𝒃 Bearing If two bodies are pressed against each 𝝈 𝒃= Stress other, compressive forces are developed on the area of contact. The pressure caused by 𝑨𝒃 these surface loads is called bearing stress. Where: = bearing stress Pb = bearing force Ab = bearing area  Problem 5  The lap joint shown is fastened by four rivets of 3/4-in. diameter. Find the maximum load P that can be applied if the working stresses are 14 ksi for shear in the rivet and 18 ksi for bearing in the plate. Assume that the applied load is distributed evenly among the four rivets, and neglect friction between the plates.   The lap joint shown is fastened by four rivets of 3/4-in. diameter. Find Solution the maximum load P that can be applied if the working stresses are 14 ksi for shear in the rivet and 18 ksi for bearing in the plate. Assume that the applied load is distributed evenly among the four rivets, and neglect friction between the plates. From shearing stress of rivet From bearing stress of plate The safe load P as governed by shearing in rivet is P = 24 700 lb Ans Thin- walled cylinders  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. Two stresses developed: 1. Tangential stress, σt (Circumferential Stress) 2. Longitudinal stress, σL  The forces acting are the total pressures TANGENTIAL STRESS, caused by the internal pressure p and the σt (Circumferential Stress) total tension in the walls T. Consider the tank shown being subjected to an internal pressure p. The length of the tank F=pA=pDL is L and the wall thickness is t. Isolating the T=σtAwall=σttL right half of the tank: ΣFH=0 F=2T pDL=2(σttL) σt=pD/2t If there exist an external pressure p o and an internal pressure pi, the formula may be expressed as: Where: =tangential stress P=internal pressure D=internal diameter T=thickness LONGITUDINAL STRESS, σL The total force acting at the rear of the tank F must equal to the total longitudinal stress on the wall PT = σLAwall. Since t is Consider the free body diagram in the transverse so small compared to D, the area of the wall is close to πDt section of the tank: ΣFH=0 PT=F = If there exist an external pressure po and an internal pressure pi, the formula may be expressed as: It can be observed that the tangential stress is twice that of the longitudinal stress. σt=2σL  Problem 6  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/m 2. (a) Calculate the tangential and longitudinal stresses in the steel.  (b) To what value may the internal pressure be increased if the stress in the steel is limited to 120 MN/m2?  (c) If the internal pressure were increased until the vessel burst, sketch the type of fracture that would occur.  Solution b) Since tangential stress is the crit σt=pD/2t a) Tangential stress (longitudinal section): 120=p(400)/2(20) p=12MPa answer F=2T pDL=2(σt tL) σt=pD/2t=4.5(400)/2(20) σt=45MPa answer Longitudinal Stress (transverse section): c) Since tangential stress governs, fracture is expected F=P as shown. ¼ πD2p=σl(πDt) fracture σl=pD/4t=4.5(400)/4(20) σl=22.5MPa answer  Solution Total internal pressure: P=p(1/4 πD2) Resisting wall: F=P σA=p(1/4 πD2) σ(πDt)=p(1/4 πD2) σ=pD/4t 8000=p(4×12)4(516) p=208.33psi answer  Problem 7 The wall thickness of a 4-ft-diameter spherical tank is 5/16 inch. Calculate the allowable internal pressure if the stress is limited to 8000 psi.  Problem 8 A water tank, 22 ft in diameter, is made from steel plates that are 1/2 in. thick. Find the maximum height to which the tank may be filled if the circumferential stress is limited to 6000 psi. The specific weight of water is 62.4 lb/ft3. Solution Solution Given a free surface of water, the actual pressure distribution on the vessel is not uniform. It varies linearly from zero at the free surface to γh at the bottom (see figure below). Using this actual pressure distribution, the total hydrostatic pressure is reduced by 50%. This reduction of force will take our design into critical situation; giving us a maximum height of 200% more than the h above. EXERCIS ES solution 2.0 solution REFERENCES/ wEBSITES