Materials - Tensile Force Notes PDF

# Materials ## Tensile Force Suppose a solid shaped in the form of a rod, wire, or strip is pulled length-wise. Then, the solid is under a state of tension, and the force acting is called tensile force. * stretch = tensile force * compress = compressive force ## Experiment To study the stretchi...

# Materials ## Tensile Force Suppose a solid shaped in the form of a rod, wire, or strip is pulled length-wise. Then, the solid is under a state of tension, and the force acting is called tensile force. * stretch = tensile force * compress = compressive force ## Experiment To study the stretching of a copper wire/ rubbery string/ thick rubber cord. To find the Young's Modulus of a Copper Wire ## Apparatus - Safety Glasses - Metal Wire Under Test - Travelling Microscope with vernier scale - Ruler - Marker - Pulley - Paper Flap - G Clamp - Load - Mass Catcher ## Method The original length (l_o) of the wire was measured using a metre ruler. The initial vernier reading (l₁) was recorder using a vernier scale. The load was increased in steps of 0.2 kg. Therefore, the tensile force increased in steps of 2N. The new vernier reading (l₂) was then noted. The procedure was repeated until the wire broke. ## Table of Results | Tensile Force (Load) / N | Initial Vernier Reading, l₁/m | Final Vernier Reading, l₂/m | Extension (l₂-l₁)/m | |---|---|---|---| ## Graph - CD: Plastic Deformation - OP: Permanent Extension - OA: Proportional limit - A: Proportional Limit - B: Elastic Limit - OB: Elastic Deformation - C: Lower yield stress point - D: Ultimate stress point - E: Breaking or rupture point ## For straight line OA The extension is directly proportional to the load up to A, therefore A is called the Proportional Limit. Hooke's Law is obeyed between O and A. Hooke's Law states that the extension is directly proportional to the applied load provided that the proportional limit is not exceeded. * Extension ∝ Force * △L ∝ F * F = k ∆L * F = Force (Load) / N * △L = Change in length (extension or compression) / m * k = Stiffness Constant or Force Constant / Nm^-1 The Stiffness Constant is defined as force per unit change in length. ## Elastic Limit The force at B is the Elastic Limit. Along OB, the wire undergoes Elastic Deformation. Between AB, Hooke's Law is not obeyed (not a straight line). ## Yield Point As the load increases, deformation remains elastic until C. When the load is removed, the wire recovers along CP (permanent extension OP). The force at C is called Yield Point. The yield point is the point at which a large extension occurs with no increase in the load. ## Beyond C The wire suffers Permanent Deformation; the wire retains some of its extension after the load is removed. The wire is work-hardened. ## Increasing the Load The extension, and at D, the wire thins forming a neck, ultimately breaking at E. ## Breaking Stress The force at D is called the **Breaking Stress** or **Ultimate Tensile strength** as it measures the strength of the material. It has the narrowest cross-section. It is the point at which the material breaks. ## Ductile & Brittle - Ductile substance (ex. copper wire) lengthen and undergo elastic and plastic deformation until they break. - Brittle substances (ex. glass) break just after the elastic limit. They only undergo elastic deformation. (no plastic deformation takes place) ## Tensile & Compressive Strain - A tensile stress is produced when a deforming force produces an increase in length. - A compressive stress is produced when deformation produces a decrease in length. - Deformations are called tensile strain and compressive strain. ## Stress Stress is defined as the force per unit cross-sectional area of the material. - σ = F / A - σ = Stress (Pa / Nm²) - F = Force (N) - A = Area of cross-sectional (m²) ## Strain Strain is defined as the change in length per unit original length. - ε = Strain - △L = Change in length (m) - l_o = Original length - ε = △L / l_o ## Young's Modulus Young's Modulus of a material is defined as the ratio of stress to strain. It's the stress per unit strain as well as the stiffness of a material. Young's Modulus is a constant provided that the proportional limit is not exceeded. - e _Y = σ / ε - E = Young's Modulus (Pa /Nm^-2) - σ = Stress (Pa / Nm²) - ε = Strain ## Calculations - e_Y = F/A - e_Y = ΔL/l_o - e_Y = F l_o / A ΔL - F = e_Y A ΔL / l_o - F = e _Y A l_o / ΔL - y = m...x - ε = △L / l_o - σ = F / A ## Area Area = πd² / 4 Diameter is found by a micrometer screw gauge. ## A value for Young's Modulus A value for Young's Modulus for the material of the wire was found by substituting for the original length, slope and are in Equation 2. ## Stress - Strain Graph **Gradient**: Δy/ Δx = σ/ε = e_Y **Hooke's Law**: F = k ΔL and F = e_Y A ΔL/l_o ## Young's Modulus & Stiffness Constant k = e_Y A/ l_o ## Two Springs or Wires in Series > Consider two springs or wires, having force constants k₁ and k₂,connected in series and having a load F at one end. **Hooke's Law**: F = k ΔL - F = Total Force (N) - k = Total Force Constant of the system (Nm^-1) - ΔL = Total Extension (m) - F = k ΔL = ΔL/K **For spring 1**: F = k₁ Δl₁ - (k₁ = force constant for spring 1) - Δl₁ = F/k₁ *** **For spring 2**: F = k₂ Δl₂ - (k₂ = force constant for spring 2) - Δl₂ = F/k₂ *** ΔL = Δl₁ + Δl₂ Therefore: F/K = F/k₁ + F/k₂ **Divide by F**: 1/K = 1/k₁ + 1/k₂ ## Two Springs or Wires in Parallel > Consider two springs or wires, having the same force constants k, connected in parallel, and having a load F at one end. **Hooke's Law**: F = k ΔL - F = Total Force (N) - K = Total Force Constant of the system (Nm^-1) - ΔL = Total Extension (m) Since the load is shared: F/2 = k ΔL Therefore, for the whole system: F/2 + F/2 = (k + k) ΔL **K = (k + k₂)** ## Load - Extension Graph - The force-extension graph is a straight line graph passing through the origin. Therefore, Hooke's Law is obeyed. ## Energy stored in a Stretched Wire > Consider an extremely small extension such that the force acting during the extension is constant. Then, the shaded strip approximates a rectangle. Therefore, work done in producing extension: - W = F. ΔS - W = L x B - W = Area of rectangle **Total work done in stretching wire ΔL by force F**: - Area between graph and extension axis - Area of triangle - = 1/2 bh = 1/2. Δl. F - = Elastic Energy = Elastic P.E. - = Strain Energy = Gain in P.E. (atoms to increase displacement from average position) *** **But, 1/2 FΔL**: = Average force x extension, and the actual force is F - 1/2 F stretches wire while 1/2 is borne (held) by support - Energy stored in a wire - E = 1/2 F△] *** **By substituting other formulas**: - Energy in wire E = 1/2 E_YA Δl²/l_o - Energy in wire E = 1/2 . k. △1² ## Energy per unit volume - Energy stored in a wire - E = 1/2 F△] - ε = ΔL/l_o - Δ] = ε l_o - σ = F/A - F = σ A - E = 1/2 (V) (σε) - V = A l_o **Energy per unit volume**: - E = 1/2 (σ ε) / V **Area under graph** = Area of triangle - = 1/2 bh - = 1/2 εσ **Area under graph** = Energy per unit volume ## Experiment 2 To determine the Young's Modulus of a Copper (Steel) Wire (2m or longer) ## Apparatus - Control wire or reference wire P - Test wire Q (~ 2m) - Vernier arrangement + spirit-level to measure extension Δl of Q - Small weight to keep P taut (stretched) - Variable load ## Method The original length of wire Q was noted. Weights were attached to the two wires so that they were free of kinks. The vernier was adjusted so that the spirit-level was horizontal. The first vernier reading was recorded. A load of 5N was added to Q, and the vernier was readjusted so that the spirit-level was again horizontal. The second vernier reading was recorded. The extension of Q under a load of 5N was given by subtracting the 1st vernier reading from the 2nd. The experiment was repeated, adding 5N every time and noting the extension. This procedure was repeated for unloading. The elastic limit was not exceeded since the extensions for loadings and unloading were the same, and the material was obeying Hooke's Law. ## Table | Force F/N | Loading Extension/m | Unloading Extension/m | Average Extension /m | |-----|-----|-----|-----| ## Graph - Graph of Force/N against Extension/m ## Calculations - e_Y = F l_o / A ΔL - F = e_Y A ΔL / l_o - F = e _Y A l_o / ΔL - y = m...x - m = E_Y A l_o - e_Y = m l_o/A ## Precautions - Wires were thin, so a moderate load produces a large tensile stress - A vernier scale and a spirit-level were used for an accurate reading. - Wires were long so that a measurable extension was produced. - The elastic limit was not exceeded. - Wires were free of kinks. ## Metal - Stress σ / 10⁸ Pa - Elastic Deformation - Ex. **Copper** - Ductile Material (undergoes elastic and permanent deformation) - e_Y = 13 x 10¹⁰ Pa Metal Sping is easy to compress but difficult to break. - **Strenght** - high - **Stiffness** low ## Glass - Stress σ / 10⁸ Pa - Elastic Deformation - Glass breaks here at B - Elastic Limit - e_Y = 7 x 10¹⁰ Pa Ex. **Glass** - Brittle Material (undergoes elastic deformation only. There is no plastic deformation) Glass is difficult to bend but easy to break. - **Strenght** - low - **Stiffness** - high ## Deductions Since metal undergoes plastic deformation, it's tougher than glass. - **Copper** - e_Y = 13 × 10¹⁰ Pa - **Glass** - e_Y = 7 x 10¹⁰ Pa Copper is twice as stiff as glass (resists deformation by twice as much). The steeper the gradient (Slope = e_Y), the greater the value of Young's Modulus. ## Maximum Breaking Stress - **Copper** = 4 × 10⁸ Pa - **Glass** = 0.7 x 10⁸ Pa Maximum Breaking Stress for copper is greater, therefore stronger. ## Strength and Stiffness - Stress σ / 10⁸ Pa - Wire A has a steeper gradient, therefore a greater Young's Modulus than wire B. - Wire A is stiffer than wire B. - Wire B has a bigger maximum breaking stress than Wire A. - Wire B is stronger than wire A. **Strength** is the ability to withstand an applied load without getting plastically deformed or ruptured. **Stiffness** is the degree to which an object resists its deformation to an applied load. * **Strenght** - hard to be broken * **Stiffness** - hard to be deformed ## Elastomer **Properties of rubber**: - Great range of elasticity - e_Y of rubber is 10⁴ smaller than most solids - e_Yincreases with an increase in temperature. - Stress σ /Pa - Rubber comes back to original length when the force is removed - Typical extension 5 to 10 times natural length of sample - Strain e If the elastic limit is exceeded, the curve will not meet the origin but on the x-axis to the right, creating a permanent strain. **Deductions:** 1. Rubber doesn't obey Hooke's Law, therefore not a straight-line graph. 2. It stretches easily at first but stiffer at large extensions (e_Y increases) 3. Area enclosed by OABC and strain axis represents the energy supplied to cause stretching. 4. Area enclosed by CDEO and strain axis represents the energy given up during contraction. 5. Shaded area = Area 1 - Area 2 = The total heat gained by the rubber in one expansion/contraction cycle. It's called the Hysteresis Loop. 6. Once rubber is streched, it becomes warmer. When stress is removed, temperature falls but remains warmer than initially. 7. Car tyres have a small hysteresis loop, otherwise they would disintegrate. 8. The strain for a given stress is greater when unloading than when loading. Therefore, the unloading strain lags the loading strain. This is called Elastic Hysteresis. ## Experiment 3 To study the Elastic Characteristics of Rubber ## Apparatus - Rubber cord - Clamp - Ruler - Stand ## Method The original length of the rubber cord was measured and noted. It was then hung from the clamp, and the corresponding position of the end of the cord was read against the ruler and noted (A). A mass was hung to the rubber cord, and the new position was recorded (B). So, the extension was calculated as B - А. This procedure was repeated for different values of load, both for loading and unloading of the rubber cord. ## Table of Results | Load/N | Loading Extension/m | Unloading Extension/m | |-----|-----|-----| ## Graph - Force/N - loading - unloading - Extension/m - Elastic limit is not exceeded ## Calculations - Shaded Area = Energy retained by rubber cord during a loading/ unloading cycle. - Shaded Area = (Area of 1 box) x (No. of boxes) ## Example Ex. A mass of 3.5kg is gradually applied to the lower end of a vertical wire and produces an extension of 0.80mm. Calculate: **(i) Energy Stored** Energy stored in wire = 1/2 FΔl = 1/2 (3.5 x 10)(0.8 x 10^-3) = 0.014] **(ii) Loss in P.E. of mass during loading.** Loss in P. E = mgh = 3.5 × 10 × 0.8 x 10^-3 =0.028] ## The energy stored The energy stored is only half the loss in gravitational potential energy because the wire needs a gradually increasing load to extend it. The remaining gravitational P.E. is given to the loading system. If the load is suddenly applied, the initial extension would be 1.6mm, twice the equilibrium extension.