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Strength of materials STRENGTH OF MATERIALS deals with the relations between external applied loads and their internal effects on bodies. STRENGTH OF MATERIALS Moreover, the bodies are no longer assumed to be ideally rigid; the deformations however small, are the major interest. The properties of the materials of which a structure or machine is made affect both its choice and the dimensions that will satisfy the requirements of strength and rigidity. Mechanics of Deformable Bodies It includes the study of the strength capabilities and characteristics of selected materials. Strength of materials Strength of materials The subject matter includes discussions of the fundamental concepts of stresses and strains experienced and/or developed by different materials in their loaded state and subjected to different conditions of constraint. Understanding of how bodies or materials respond to applied load is the main area of emphasis. Study of the relationship between externally applied loads and their internal effects on rigid bodies. RIGID BODY – bodies which neither change in shape and size after the application of forces. FREE BODY DIAGRAM – Sketch of the isolated body showing all the forces on it. THREE MAJOR DIVISIONS OF MECHANICS 1. Mechanics of Rigid Bodies – Engineering Mechanics 2. Mechanics of Deformable Bodies – Strength of Materials 3. Mechanics of Fluids - Hydraulics Strength of materials Strength of materials The strength of a material is its ability to withstand an applied stress without failure Two categories -> Yield Strength and Ultimate Strength Yield Strength - the stress level at which a material begins to deform plastically. Yield strength refers to the point on the engineering stress-strain curve beyond which the material begins deformation that cannot be reversed upon removal of the loading Ultimate Strength - It is the maxima of the stress-strain curve. It is the point at which necking will start. Ultimate strength refers to the point on the engineering stress-strain curve corresponding to the maximum stress. A material's strength microstructure. is dependent on its Fracture Strength - The stress calculated immediately before the fracture. 1 Analysis of internal forces Analysis of internal forces By setting up the equilibrium conditions, the inner forces of a member subjected to an external load situation can be determined. So far neither the material nor the type of cross section applied for the member are being taken into account. But both material and type of cross section obviously have an impact on the behavior of the member subjected to load. y Mxy Pxy Pxx Mxz z Analysis of internal forces Pxx (Axial Force) – The component measures the pulling (or pushing) action over the section. A pull represents a tensile force which tends to elongate the member whereas a push is a compressive force which tends to shorten it. It is often denoted by P. Analysis of internal forces Mxx (Torque) – This component measures the resistance to twisting the member and is commonly given the symbol T. Mxy, Mxz (Bending Movements) – These components measure the resistance to bending the member about the Y or Z axes and are often denoted by My or Mz. x Pxz To design the member therefore a closer look on how the internal forces act along its cross section needs to be taken. Analysis of internal forces Mxx Pxy,Pxz (Shear Force) – These are components of the total resistance to sliding the portion to one side of the selection pass the other. The resultant shear force is usually designated by V and its components by Vy and Vz to identify their directions. Fundamental concept of stress When a force is transmitted through a body, the body tends to change its shape or deform. The body is said to be strained. STRESS – is defined as the strength of material per unit area. It is the force on a member divided by the area which carries the force. In symbol; = Where: σ = stress P = force A = area 2 Fundamental concept of stress Units σ, Stress P, Force A, Area Note: 1 1 1 1 English psi (lbs/in2), ksi (kips/in2) pounds, kips sq. in. (m2) Metric Pa (N/m2) N, kN (mm2), m2 MPa = 1x106 Pa = 1x106 N/m2 kip = 1000 pounds (lbs) ksi = 1000 psi N/mm2 = 1 MN/m2 = 1 MPa Kinds of stresses Axial Stress – the type of stress wherein the force applied is perpendicular or normal to the area. It can be tensile or compressive stress. Shearing Stress – the type of stress wherein the force applied is parallel to the area Bearing Stress – is the constant pressure between separate bodies. It differs from the compressive stress as it is an internal stress caused by the compressive force. Torsional Stress – stress produced due to torque Bending Stress – stress developed due to bending of the member STRENGTH OF MATERIALS Mechanics of Deformable Bodies AXIAL Stress AXIAL Stress problem 1 Axial stress may be tensile, or compressive, and result from forces acting perpendicular to the plane of the cross-section Determine the axial stress on members BD, CE & CD of the truss shown. Assume cross-sectional area of each member is 900mm². Tension Compression conclusion: the normal stress acting along a section of a member only depends on the external load applied (e.g. a normal force F) and the geometry of its cross section A (true for statically determinant systems). 3 AXIAL Stress problem 1 AXIAL Stress problem 1 FBD Av To get the force AV: ME = 0 AV(6) – 200(3) – 100(1.5) + 50(3) = 0 AV(6) = 200(3) + 100(1.5) – 50(3) Ev Av AV = 100kN AXIAL Stress problem 1 AXIAL Stress problem 1 To get the stress of member BD, use the formula: PBD A 100kN = 900mm2 σBD = B BD CD CE 3m C A 200kN Av=100kN σBD Av To get force BD, MC = 0 AV(3) – BD(3) = 0 100(3) – BD(3) = 0 BD = 100kNcompression σBD = 111.11MPa compression AXIAL Stress problem 1 AXIAL Stress problem 1 To get the stress of member CE, use the formula: PCE A 50kN = 900mm2 σCE = B BD=100kN D Av To get force CE, MD = 0 BD(4.5) – 200(1.5) – CE(3) = 0 100(4.5) – 300 = CE(3) CE BD= =50kN 1 tension A 3m Av=100kN C CE 100kN σCE σCE = 55.55MPa tension 200kN 4 AXIAL Stress problem 1 B To get the stress of member CD, use the formula: BD=100kN CE=50kN 3m σCD C 200kN Av Av=100kN θ = tan−1 3m θ 1.5m θ = 63.43° 3 1.5 PCD A 111.81kN = 900mm2 = 124.23MPa tension σCD = CD A AXIAL Stress problem 1 To get force CD, MA = 0 200(3) – BD(3) – CDV(3) = 0 200(3) – 100(3) – CD(sin63.43)(3) = 0 CD = 111.81kNtension AXIAL Stress problem 2 σCD AXIAL Stress problem 2 An aluminum tube is rigidly fastened between a bronze rod and a steel rod as shown. Axial loads are applied at the positions indicated. Determine the stress in each material. For Bronze, FX = 0 PB = 20kN PB 20,000N σB = = AB 700mm2 σB = 28.57MPa compression AXIAL Stress problem 2 For Aluminum, FX = 0 20kN − 15kN = PA PA = 5kN 5,000N PA = σA = AA 1,000mm2 σA = 5MPa compression AXIAL Stress problem 2 For Steel, FX = 0 20kN − 15kN − 15kN + PS = 0 PS = 10kN PS 10,000N σS = = AS 800mm2 σS = 12.5MPa tension 5 AXIAL Stress problem 3 A 12in. square steel bearing plate lies between an 8in. diameter wooden post and a concrete footing. Determine the maximum value of the axial load P if the stress in wood is limited to 1800psi and in concrete is limited to 650psi. AXIAL Stress problem 3 For concrete, PC σC = AC For wood, PW σW = AW 1800 lbs = in2 PW 8in 4 650 2 PC = 93,600lbs PW = 90,477.87lbs AXIAL Stress problem 4 lbs PC = in2 12in 12in Pmax = 90,477.87lbs AXIAL Stress problem 4 A homogenous 150kg bar AB carries a 2kN force as shown. The bar is supported by a pin at B and a 10mm diameter cable CD. Determine the stress in the cable. MB = 0 4 150 ∗ 9.81 3 PCD 3 −2 6 − =0 5 1000 PCD = 6.84kN tension AXIAL Stress problem 4 A= SCD SCD SCD π 10 2 = 25πmm2 4 PCD 6.84kN 103 N/kN = = A 25πmm2 = 87.09N/mm2 tension = 87.09MPa tension AXIAL Stress problem 5 Determine the largest weight W which can be supported by the two wires shown. The stresses in wires AB and AC are not to exceed 100MPa and 150MPa respectively. The cross-sectional areas of the two wires are 400mm2 for wire AB and 200mm2 for wire AC. B C 30o A 45o W 6 AXIAL Stress problem 5 AXIAL Stress problem 5 FBD P = AS Y = 0 ACY = WY AC sin 75 = W sin 60 AC = 0.897W H = 0 ABH = ACH AB cos 30 = AC cos 45 AB cos 30 = 0.897W cos 45 AB = 0.732W AXIAL Stress problem 6 The Bell Crank shown is in equilibrium. Determine the required diameter of the connecting rod AB if its axial stress is limited to 100MPa. For AC: AC = (200)(150) = 30000N 0.897W = 30,000 W = 33,444.82N For AB: W = 33,444.82N AB = 400(100) = 40,000 0.732W = 40,000 W = 54644.81N AXIAL Stress problem 6 ΣMD = 0 30kN sin 60 240mm − P 200mm = 0 P = 31.18 kN P A AB N 31.18 x 103 N 100 = mm2 πd2 4 d = 19.92mm Say: d = 20mm σAB = STRENGTH OF MATERIALS Mechanics of Deformable Bodies 7 shearing Stress Shearing Stress is produced whenever the applied load cause one section of a body to tend to slide past its adjacent section. τ= Where τ – shear stress V – shear force A – area in shear V A shearing Stress Shear Stresses are produced by equal and opposite parallel forces not in line. The forces tend to make one part of the material slide over the other part. Shear Stress is tangential to the area over which it acts. Shear Stress is a measure of the internal resistance of a material to an externally applied shear load. shearing Stress The rivet resists shear across its cross-sectional area. Single Shear shearing Stress Bolt resists shear across two cross-sectional areas. Double Shear ΣFy = 0 P = 2V P V= 2 shearing Stress A circular slug is about to be punched out of a plate. Punching Shear ΣFy = 0 P=V Ashear = C * t Ashear = πdt shearing Stress problem 1 A hole is to be punched out of a plate having an ultimate shearing stress of 300MPa. If the compressive stress in the punch is limited to 400MPa. Determine the maximum thickness of the plate from which a hole of 100mm in diameter can be punched. If the plate is 10mm thick, compute the smallest diameter hole that can be punched. 8 shearing Stress problem 1 shearing Stress problem 1 For punching force, P P σc = Ac N P 400 = π(100mm)2 mm2 4 = . First Situation τp = 300MPa σc = 400MPa d = 100mm For shear force, V ΣFy = 0 P=V . = shearing Stress problem 1 P Ac P σc = πd2 4 Solving for thickness, t V Av V τp = πdt N 3.14 x 106 N 300 = 2 mm π 100 mm t t = 33.33 mm τp = First Situation τp = 300MPa σc = 400MPa d = 100mm shearing Stress problem 1 ΣFy = 0 P=V σc = πd2 4 N πd2 P = 400 mm2 4 N πd2 P = 400 mm2 4 P = σc Second Situation τp = 300MPa σc = 400MPa t = 10mm = shearing Stress problem 2 The end chord of a timber truss is framed into the bottom chord as shown in the figure. Neglecting friction, compute dimension b if the allowable shearing stress is 900kPa. P=50KN b 30o c 100πd2 = V V τp = Av 100πd2 τp = πdt N 100πd2 300 = mm2 πd 10 mm d = 30 mm Second Situation τp = 300MPa σc = 400MPa t = 10mm shearing Stress problem 2 P=50KN b 30o c τ= V Av V = PH V = 50cos30 V = 43.3kN N 43.3x103 N 0.90 = 2 mm 150 mm (b) b = 320.74mm 9 shearing Stress problem 3 shearing Stress problem 3 The Bell Crank shown is in equilibrium. Determine the shearing stress in the pin at D if its diameter is 20mm. ΣMD = 0 30kN sin 60 240mm − P 200mm = 0 P = 31.18 kN ΣFH = 0 DH = P + 30kN cos 60 DH = 31.18 kN + 30kN cos 60 DH = 46.18 kN DH ΣFV = 0 DV DV = 30kN sin 60 Dv = 25.98kN = . shearing Stress problem 3 τ= V A 52,980N π 2 ∗ 20mm 4 τ = 84.33MPa τ= 2 59 10