The torques shown are exerted on pulleys A and B. Knowing that the shafts are solid and made of

step 1 Once again, welcome to a new problem. This time we are dealing with forces and stresses. For the

The torques shown are exerted on pulleys A and B. Knowing...

Key Concepts

Torsion in Shafts

Torsion refers to the twisting of an object due to an applied torque. In mechanical engineering, when a torque is applied around the longitudinal axis of a shaft, it induces shear stresses and strains within the material, leading to an angular deformation along the length of the shaft. The analysis of torsion is crucial for ensuring that shafts and rods can withstand twisting loads without failing.

Shear Modulus

The shear modulus, also known as the modulus of rigidity, is a material property that measures a material's ability to resist shear deformations. It plays a vital role in torsion calculations by directly controlling the amount of twist a material will experience under a given torque. Higher shear modulus values indicate materials that are more resistant to twisting.

Polar Moment of Inertia

The polar moment of inertia is a geometric property that quantifies the distribution of a cross?section's area relative to the axis of rotation. In torsional analysis, it is a key factor that influences the angle of twist: a larger polar moment of inertia results in a smaller angle of twist for a given applied torque, making it a crucial parameter in the design of circular shafts.

Angle of Twist

The angle of twist is a measure of the rotational displacement that occurs along a shaft’s length due to applied torque. It is calculated based on the torque, the length of the shaft, the polar moment of inertia, and the material's shear modulus. Understanding and controlling the angle of twist is important for ensuring proper mechanical functionality and structural integrity in systems subject to torsional loading.

Composite Shaft Torsion

Composite shaft torsion involves the analysis of shafts made from materials with different mechanical properties, which are bonded together to form a single structural element. When a composite shaft is subjected to torsion, each material will experience its own deformation characteristics based on its shear modulus. The overall angle of twist of a composite shaft is determined by the combined behavior of its individual components, making it essential to consider compatibility and continuity conditions between the different materials.

Transcript

00:01 Once again, welcome to a new problem.

00:04 This time we are dealing with forces and stresses.

00:11 For the most part, you can have tension forces and also you can have compressive forces.

00:21 When you're dealing with tension forces, what happens is you could have a rod like this.

00:27 And then it's extended.

00:29 And in between, it extends and get bigger, and so it's undergoing tension.

00:36 Okay, it's undergoing tension, and there are certain materials that are stronger when it comes to tensions.

00:43 For example, metals like, say, iron or aluminum.

00:50 So metals like iron or aluminum.

00:55 When it comes to compressive forces, we tend to deal with things like concrete.

01:04 Concrete is very strong when it comes to compressive forces, but now we have another type called reinforced concrete which pretty much has metals inside of it.

01:20 It has metals inside of it and therefore it can be used to span.

01:25 It can be used to span columns.

01:28 Columns such that even when spanning columns, two columns, for example, it can undergo bending forces and therefore experience some tension and at the same time it can undergo compression due to the forces from the column.

01:50 So in this context, let's think about an assembly system.

01:58 So here we're dealing with an assembly system.

02:01 And in the system, we do have a wall.

02:07 Excuse me, we have a wall.

02:10 And on the wall, on the wall, on the wall, on the wall, we have a metal.

02:32 In different sections.

02:33 So these are balls drilled onto the wall, screwed onto the wall, and then there's a steel rod that runs between two joints.

02:48 So the wall joint and then there's another joint at b.

02:51 So this portion is c and this is a steel rod, c, b.

02:58 And then we have another aluminum rod on this side so this is this is steel and this is aluminum and it ends at a so b a is aluminum and c b is a steel rod and the diameter the diameter of each one of these is the same as 12 millimeters so the diameter becomes the same as 12 millimeters and so the radius obviously is going to be six so half of that the radius is is going to be six so both of these steel roads have a diameter of 12 millimeters and and then at point a at point a we have a load, a loading system at point a and there's a force of 18 kiloons, 18 kilonutons.

04:19 The distance between this section and the midsection of point b happens to be 3 meters, and then the distance from the midsection of b up until the end of the road, is 2 meters, so it's 2 meters.

04:43 And then there's going to be axial loading at point a.

04:52 So this is axial, meaning that it's moving parallel to the aluminum together with the steel roads.

05:01 At point b, there's coupling that takes place.

05:05 So at this point we have coupling taking place at point b.

05:13 And so the goal of this particular problem is to determine, we want to determine the displacement of coupling b and coupling b, and the end, and the end, and end a.

05:44 So we're saying coupling b and end a.

05:47 Those are the two things we're looking at.

05:52 If you look at the figure, it's still unstretched.

06:01 The figure is still unstretched and we want to find the displacement.

06:09 Of course the connections, these connections we neglect the joints or the connections in terms of distances, neglect connections or joints, or joints in computations, connections or joints in computations, in terms of sizes.

06:40 So we don't really care about the sizes of these connections.

06:47 And so the these are rigid.

06:54 These are rigid.

06:59 And so the stress that they can take, the stress that they can take, is going to be different numbers.

07:21 So the stress that they can take is going to be different numbers.

07:32 Seeing right here so we have two numbers we have two numbers we're dealing with this particular problem and those two numbers we're going to use them to resolve to resolve the problem so the volume energy density we have the volume energy density e for steel is 200 gigapascals and then the volume energy density for aluminum is 70.

08:11 So these are magic numbers that we're going to use to solve the problem...

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