Average shear stress in going between a spliced rod?

[kal854]

Problem statement:
Two 3/4-in.-diameter nylon rods are spliced together by gluing a 2-in. section of plastic pipe over the rod ends, as shown in the figure. If a tensile force of P = 500 lb is applied to the spliced nylon rod, what is the average shear stress in the glue going between the pipe and rods?

Figure:

I need help with this problem. I know that {average shear stress} = {force} / {area}. But I am very confused when you take into account this splicing! Thank you in advance!

 

[Chestermiller]

Problem statement:
Two 3/4-in.-diameter nylon rods are spliced together by gluing a 2-in. section of plastic pipe over the rod ends, as shown in the figure. If a tensile force of P = 500 lb is applied to the spliced nylon rod, what is the average shear stress in the glue going between the pipe and rods?

Figure:

I need help with this problem. I know that {average shear stress} = {force} / {area}. But I am very confused when you take into account this splicing! Thank you in advance!

Can you articulate in words what's happening here? For example, what forces are acting on each of the two rods (treated as free bodies)? What is the glue doing?

Chet

 

[kal854]

Can you articulate in words what's happening here? For example, what forces are acting on each of the two rods (treated as free bodies)? What is the glue doing?

Chet

I'm assuming there's some adhesive force acting on the rods, as well as P?? But I'm not sure...

 

[Borek]

Cross post, locked.

 

[HalfLostAndSearching]

I would approach this problem by first calculating the cross-sectional area of the spliced rod. Since the rod diameter is given in inches, I would convert it to feet to be consistent with the force unit of pounds. The cross-sectional area of a circle is calculated using the formula A = πr^2, where r is the radius of the circle. In this case, the radius would be 3/8 inches (3/4 inch diameter divided by 2) or 0.03125 feet. Therefore, the cross-sectional area of one rod would be approximately 0.000772 square feet (π x 0.03125^2). Since there are two rods spliced together, the total cross-sectional area would be double that, or 0.001544 square feet.

Next, I would calculate the average shear stress using the formula given in the problem statement: average shear stress = force / area. Plugging in the given force of 500 pounds and the calculated area of 0.001544 square feet, I would get an average shear stress of approximately 324.07 pounds per square foot (500 / 0.001544).

However, as a scientist, I would also take into account the type of glue used for the splice and its shear strength. Different types of glue have different shear strengths, so it is important to consider this in the calculation. If the shear strength of the glue is known, it can be used to calculate the actual average shear stress at the splice. If it is not known, experiments or tests on the specific type of glue can be conducted to determine its shear strength and therefore, a more accurate average shear stress value.

In conclusion, the average shear stress in the glue going between the pipe and rods can be calculated using the given formula, but it is important to also consider the type of glue and its shear strength for a more accurate calculation.