1.a concentrated load p is applied at the end of cantilever.the crosssection of the beam is a square of side "a" and with a hole of diameter "a/2" then deflection at the tip of beam is?lengh of beam is L.

Question

1.a concentrated load p is applied at the end of cantilever.the crosssection of the beam is a square of side "a" and with a hole of diameter "a/2" then deflection at the tip of beam is?lengh of beam is L.

Answer 1

The deflection of a cantilever beam with a point load P at the end is given by

δx=PL³/(3EI)

P=point load
L=length
E=modulus of elasticity
I=area moment of inertia
all in compatible units.

The area moment of inertia for the beam in question is equal to that of the square less that of the hole.

Area moment of inertia:
I=bh³/12) (b=width, h=height)
and for a circle of radius r
Ic=πr⁴/4

For other shapes, see:
http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia

So for the beam in question,
I=a⁴/12-πa⁴/1024

and the deflection can be calculated accordingly.

Answer 2

To calculate the deflection at the tip of the beam, we can use the formula for the deflection of a cantilever beam applied with a concentrated load at the free end.

The formula for the deflection of a cantilever beam under a concentrated load is given by:

δ = (P * L^3) / (3 * E * I)

Where:
- δ is the deflection at the tip of the beam.
- P is the applied load.
- L is the length of the beam.
- E is the modulus of elasticity of the material.
- I is the moment of inertia of the beam's cross-section.

In this case, the cross-section of the beam is a square with side "a" and a hole of diameter "a/2".

The moment of inertia of a square cross-section about its centroid is given by:

I = (a^4) / 12

However, since there is a hole in the center, the moment of inertia will be different. We need to calculate the moment of inertia taking into account the hole:

I_total = I_full - I_hole

For a square cross-section, the moment of inertia excluding the hole is given by:

I_full = (a^4) / 12

For a circular hole with diameter "a/2", the moment of inertia is given by:

I_hole = (π * (a/4)^4) / 4

Therefore, the total moment of inertia is:

I_total = (a^4) / 12 - (π * (a/4)^4) / 4

Now we can substitute the values into the formula for deflection:

δ = (P * L^3) / (3 * E * I_total)

Please provide the values for P, L, a, and E so that we can calculate the deflection at the tip of the beam.

Answer 3

To find the deflection at the tip of the beam, we can use the formula for deflection of a cantilever beam under a concentrated load.

Given:
- Concentrated load, P
- Side of square cross-section, a
- Diameter of the hole, a/2
- Length of the beam, L

To solve this problem, we'll use the following steps:

Step 1: Calculate the area moment of inertia (I) for the square cross-section
The formula for calculating the area moment of inertia for a square cross-section is: I = (a^4) / 12

Step 2: Calculate the effective area moment of inertia (Ieff) for the cross-section with a hole
Since the hole has a diameter of a/2, it creates a circular void in the square cross-section. The effective area moment of inertia can be calculated using the parallel axis theorem:
Ieff = I - (π / 4) * ((a / 2)^4)

Step 3: Calculate the maximum bending moment (M) at the fixed end of the cantilever
The maximum bending moment occurs at the fixed end of the cantilever and is given by:
M = P * L

Step 4: Calculate the deflection at the tip of the beam (δ)
The deflection of a cantilever under a concentrated load can be calculated using the formula:
δ = (P * L^3) / (3 * E * Ieff)

Where E is the Young's modulus of the material.

By following these steps and substituting the given values, you can calculate the deflection at the tip of the beam.