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Young's Modulus and Material Comparison
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Today, we will explore how the Young's modulus helps us understand the stiffness of materials. Can anyone tell me what Young's modulus represents?
Itβs the ratio of stress to strain in a material, showing how much it stretches under force.
Exactly! Higher Young's modulus means a material is stiffer. Now, if I compare steel and rubber, which do you think has a higher Young's modulus?
Steel would, right?
Correct! Steel has a modulus around 200 GPa, while rubber is just around 0.01 GPa. Remember the acronym 'SR' for Stiffness Ratio. Let's use that to remember the comparison!
So, a higher stiffness means less strain for the same stress?
Yes! Great observation. Do any of you find it difficult to remember what affects Young's modulus?
Not really, but I sometimes mix up tensile and compressive stress.
A tip for that is to remember: 'Tensile is stretching, Compressive is squeezing.' Great job, everyone!
Stress-Strain Curves
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Now let's look at stress-strain curves. Why do engineers rely on these curves when designing structures?
To see how materials respond to different stress levels!
Exactly! The area under the curve indicates energy absorption. Can someone explain the significance of the yield point?
Itβs where the material begins to deform permanently, right?
Absolutely. We often say cross the yield point, and you're in the plastic zone now! Remember 'YP' for Yield Point!
What about the different regions like elastic and plastic?
Remember, if it's elastic, you can return to original form; in plastic, you cannot. Think of it as bouncing back versus being reshaped!
Calculating Elongation in Wires
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Let's solve an example. If a steel wire of length 2 m and cross-sectional area 0.1 cmΒ² is stretched by a force of 1000 N, how would we calculate the elongation of the wire?
We use the formula L = (F/A) * (L_0 / Y) where Y is Youngβs modulus!
Correct! For steel, Y is about 200 GPa. So what do we do next?
Calculate stress first, and then we find strain and elongation!
Great! Remember 'Safe Strain'βto always keep within elastic limits while doing this calculation.
That makes sense. I really like how the formulas connect.
Awesome! And once you get elongation, it helps visualize the physics behind this material's behavior under load.
Introduction & Overview
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Quick Overview
Standard
The exercises span a range of difficulties, targeting key concepts like Young's modulus, stress-strain relationships, and material properties under different conditions, allowing students to apply theoretical knowledge in practical scenarios.
Detailed
Exercises in Mechanical Properties of Solids
In this section, we delve into a series of exercises designed to consolidate understanding of the mechanical properties of solids covered in Chapter 8. The exercises include calculations involving Youngβs modulus, stress, strain, and the behavior of various materials under load. Each problem is structured to apply theoretical concepts to real-world situations, reinforcing learning through practical application. This methodology not only assesses comprehension but also encourages critical thinking and problem-solving skills essential in physics and engineering design.
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Exercise 8.1
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A steel wire of length 4.7 m and cross-sectional area 3.0 Γ 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 Γ 10β5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Detailed Explanation
This exercise requires you to compare the Young's modulus of two different materials, steel and copper, based on their elongation under the same load. The Young's modulus is a measure of the stiffness of a material, calculated as the ratio of stress to strain. Here, since both wires stretch the same amount under the same load, the ratio of their Young's moduli can be determined by comparing their lengths and cross-sectional areas. This is expressed mathematically.
Examples & Analogies
Think of a rubber band and a piece of string. When you pull both with the same force, the rubber band stretches much more than the string. This is similar to how steel and copper will behave under tension. Steel, being stiffer (higher Young's modulus), will stretch less compared to copper when the same load is applied.
Exercise 8.2
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Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material?
Detailed Explanation
In this exercise, you're asked to interpret a graph that illustrates the relationship between stress and strain for a material. Young's modulus can be determined from the slope of the linear portion of the curve, while the yield strength is found at the point where the curve deviates from linearity, indicating the transition from elastic to plastic deformation.
Examples & Analogies
Imagine pulling a piece of taffy. At first, it stretches easily (elastic behavior), but after a certain point, it begins to deform permanently. The slope of the initial stretch represents the Young's modulus, while the point at which it begins to deform permanently shows the yield strength.
Exercise 8.3
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The stress-strain graphs for materials A and B are shown in Figure 8.10. (a) Which of the materials has the greater Young's modulus? (b) Which of the two is the stronger material?
Detailed Explanation
This exercise focuses on analyzing two different materials based on their stress-strain curves. The material with the steeper initial slope has the greater Young's modulus, indicating itβs stiffer. To determine which material is stronger, you must check which material has a higher ultimate tensile strengthβthe maximum stress it can withstand before failing.
Examples & Analogies
Consider a race between two cars. One that can accelerate quickly represents a material with high Young's modulus (less stretch for the same stress), while the car that can carry a heavier load without collapsing represents the stronger material.
Exercise 8.4
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Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young's modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus.
Detailed Explanation
In this task, you analyze two statements regarding material properties. The first statement is false; rubber, which is highly elastic but not stiff, has a much lower Young's modulus than steel. The second statement is also false; the stretching of a coil is primarily determined by its Young's modulus rather than its shear modulus, which pertains to how materials deform under shear stress.
Examples & Analogies
Think about stretching a thick rope versus a stretchy rubber band. The rubber band can stretch a lot, but it is still not as 'stiff' as the rope, which doesn't stretch as easily. This illustrates how different materials respond to forces.
Exercise 8.5
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Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Figure 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Detailed Explanation
You need to calculate the elongation of two different wires under the same load. Using the relationship of stress, strain, and Young's modulus, you can solve for the elongations using the equations related to material properties. Remember, the elongation will differ due to the materials having different Youngβs moduli even though they are subjected to the same forces.
Examples & Analogies
Imagine stretching two different types of slinkies. One made of metal and another made of plastic. When both are stretched with equal force, the metal one wonβt stretch as much as the plastic one because of its greater stiffness (Young's modulus).
Exercise 8.6
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The edge of an aluminum cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminum is 25 GPa. What is the vertical deflection of this face?
Detailed Explanation
This problem involves applying an external force that induces shear stress on the aluminum cube. Given the shear modulus, you can calculate the amount of deflection experienced by the face of the cube due to the weight applied. The shear strain and stress relation will guide the calculation.
Examples & Analogies
Think of a thick book resting on a soft table. If a heavy object is placed on top of the book, itβll bend slightly under the weight, similar to how the aluminum cube twists slightly when the weight is applied.
Exercise 8.7
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Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.
Detailed Explanation
In this exercise, you're calculating how much each column compresses under a known load. The load is shared equally among the four columns, and you can find the compressive stress applied to each. Then, relating this stress to strain through Young's modulus gives the compressional strain.
Examples & Analogies
Consider a multi-story building. The pillars at the base support the weight of the whole structure above. Just like the steel columns, these pillars compress slightly under the weight, ensuring stability and safety of the building.
Exercise 8.8
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A piece of copper having a rectangular cross-section of 15.2 mm Γ 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Detailed Explanation
This task involves calculating strain produced by a known force applied to a copper piece with a specific cross-section. Using the definition of strain (change in length/original length), you can find the elongation cause by the force and then compute the strain.
Examples & Analogies
Imagine pulling a thick rubber band. The more you pull, the more it stretches. The ratio of how much it stretches to its original length is analogous to strain in the copper piece.
Exercise 8.9
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A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N mβ2, what is the maximum load the cable can support?
Detailed Explanation
To find the maximum load supported by the cable without exceeding the allowable stress, use the formula for stress and solve for load. Here you will be calculating stress as force divided by area, allowing you to rearrange and find the maximum force that can be applied before reaching the stress limit.
Examples & Analogies
Think of a tightrope walker. The rope can only hold so much weight before it snaps. Similarly, the steel cable can only support a certain load before reaching its maximum allowable stress.
Exercise 8.10
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A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension.
Detailed Explanation
Here, youβll use Youngβs modulus to create a relationship between the materials and their respective diameters, ensuring equal tension. Since copper and iron have different properties, their diameters will differ if the tension equals in each wire.
Examples & Analogies
It's like using different sized ropes to hold a load. If one rope is weaker (like copper), it has to be thicker to bear the same load as a stronger rope (like iron).
Exercise 8.11
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A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.
Detailed Explanation
This problem involves centripetal force and tension as the mass moves in a circle. At the lowest point, the tension in the wire will be affected by the weight of the mass and the centripetal force. Using the known values and stress-strain relationships will yield the elongation.
Examples & Analogies
Picture a swing at the playground. At the lowest point of the swing, the chains tighten due to the weight of the person. Similar principles of tension and elongation apply to the steel wire here as it supports the mass.
Exercise 8.12
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Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 Γ 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.
Detailed Explanation
For this exercise, you will calculate the bulk modulus based on the change in volume due to a change in pressure. After finding the bulk modulus of water, compare it to that of air to highlight the differences in compressibility between the two states of matter.
Examples & Analogies
Imagine trying to compress a balloon filled with air versus a solid ball made of rubber. You can easily squeeze the balloon, showing that air is far more compressible (lower bulk modulus) than solid rubber, which barely changes volume under pressure.
Exercise 8.13
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What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 Γ 103 kg mβ3?
Detailed Explanation
This exercise involves applying hydrostatic pressure principles to find how the density of a fluid changes under pressure. It utilizes the formula relating pressure, density, and gravity to arrive at the density at a specified depth.
Examples & Analogies
Think of stacking books on a table; as you add books, the bottom one gets heavier due to the weight on top. Similarly, the deeper water layers are 'heavier' due to the pressure from all the layers above.
Exercise 8.14
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Compute the fractional change in volume of a glass slab when subjected to a hydraulic pressure of 10 atm.
Detailed Explanation
You will calculate how much the volume of the glass slab changes under a certain pressure, using the definition of bulk modulus and the relationship between pressure and volume change.
Examples & Analogies
Consider the difference in size between a full and an empty soda can. When you apply pressure to the full can, the volume will change slightly compared to the empty oneβthis illustrates how hydraulic pressure can affect volume.
Exercise 8.15
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Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 Γ 106 Pa.
Detailed Explanation
In this exercise, you will determine the change in volume of a copper cube when subjected to considerable hydraulic pressure. This follows the definition of bulk modulus and the formula relating volume change to the applied pressure.
Examples & Analogies
Imagine a sponge under water pressure. It squeezes down and becomes smaller. Similarly, when you apply pressure to copper, it will have a small change in volume while under pressure.
Exercise 8.16
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How much should the pressure on a litre of water be changed to compress it by 0.10%?
Detailed Explanation
Calculating this requires understanding the relationship between pressure change and volume change. This will involve applying the concept of compressibility and bulk modulus to find the pressure needed for a specific volume change.
Examples & Analogies
Consider a sponge again; if you want to compress it significantly, you need to apply more pressure. The same is true for water, but since water is less compressible, you'll need more pressure for a small volume change!
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Stress: The restoring force per unit area, crucial for understanding material behavior.
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Strain: The relative change in shape or size, directly proportional to stress within limits.
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Young's Modulus: Indicates material stiffness, directly impacting engineering designs.
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Yield Point: Key indicator of when a material transitions from elastic to plastic deformation.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Example of calculating elongation in a steel wire subjected to a tensile force.
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Comparison of Young's modulus between steel and rubber, illustrating different stiffness levels.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
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Stress and strain, apply the force, Young's modulus is the guiding course!
π Fascinating Stories
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Imagine pushing a slinky; it stretches, but a rubber band snaps back. The slinky's elasticity shows Young's modulusβit's all about keeping that shape!
π§ Other Memory Gems
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Remember YS for Yield Strength, itβs where elastic turns to permanent!
π― Super Acronyms
Use YSRβYield Strength Rules for remembering the yield point!
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
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Term: Young's Modulus
Definition:
A measure of the stiffness of a material, defined as the ratio of stress to strain.
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Term: Stress
Definition:
The restoring force per unit area, measured in Pascals (Pa).
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Term: Strain
Definition:
The deformation of a material measured as a fraction of its original length.
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Term: Yield Point
Definition:
The point on a stress-strain curve where a material begins to deform plastically.
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Term: Elastic Limit
Definition:
The maximum extent to which a material can be deformed and still return to its original shape.