# Beam Deflection Formula

Beam Deflection is defined as a phenomenon in which some load is used to deflect a body. In other words, it is the deflection of a beam in one direction when a force is applied to it. It is evaluated by integrating the function that describes the slope of the member under that load. It is directly proportional to the force applied and beam length but changes inversely with Young’s modulus and the moment of inertia of the object. It is denoted by the symbol D. Its unit of measurement meters (m), and the dimensional formula is given by [M^{0}L^{1}T^{0}].

**Beam Deflection Formula**

D = WL^{3}/3EIWhere,

- W is the applied force,
- L is the length of beam,
- E is the Young’s Modulus,
- I is the moment of inertia.

### Sample Problems

**Problem 1: Calculate the beam deflection for a length of 5 m if a force of 250 N is applied on an object whose Young’s modulus is 40 N/m ^{2} and moment of inertia is 50 kg m^{2}.**

**Solution:**

We have,

W = 250

L = 5

E = 40

I = 50

Using the formula we have,

D = WL

^{3}/3EI= (250 × 5

^{3})/(3 × 40 × 50)= 31250/6000

= 5.2 m

**Problem 2: Calculate the beam deflection for a length of 4 m if a force of 200 N is applied on an object whose Young’s modulus is 60 N/m ^{2} and moment of inertia is 60 kg m^{2}.**

**Solution:**

We have,

W = 200

L = 4

E = 60

I = 60

Using the formula we have,

D = WL

^{3}/3EI= (200 × 4

^{3})/(3 × 60 × 60)= 12800/10800

= 1.18 m

**Problem 3: Calculate the force applied if beam deflection for a length of 8 m is 0.78 m. The value of Young’s modulus is 20 N/m ^{2}, and the moment of inertia is 45 kg m^{2}.**

**Solution:**

We have,

D = 0.78

L = 8

E = 20

I = 45

Using the formula we have,

D = WL

^{3}/3EI=> 0.78 = (W × 8

^{3})/(3 × 20 × 45)=> W = 2106/512

=> W = 4.11 N

**Problem 4: Calculate the force applied if beam deflection for a length of 2 m is 0.05 m. The value of Young’s modulus is 30 N/m ^{2}, and the moment of inertia is 25 kg m^{2}.**

**Solution:**

We have,

D = 0.05

L = 2

E = 30

I = 25

Using the formula we have,

D = WL

^{3}/3EI=> 0.05 = (W × 2

^{3})/(3 × 30 × 25)=> W = 112.5/8

=> W = 14.06 N

**Problem 5: Calculate Young’s modulus if beam deflection for a length of 2 m is 1.5 m when a force of 15 N is applied. The moment of inertia is 30 kg m ^{2}.**

**Solution:**

We have,

D = 1.5

W = 15

L = 2

I = 30

Using the formula we have,

D = WL

^{3}/3EI=> E = WL

^{3}/3ID= (15 × 2

^{3})/(3 × 30 × 1.5)= 120/135

= 0.88 Nm

^{-2}

**Problem 6: Calculate Young’s modulus if beam deflection for a length of 6 m is 3 m when a force of 30 N is applied. The moment of inertia is 20 kg m ^{2}.**

**Solution:**

We have,

D = 3

W = 30

L = 6

I = 20

Using the formula we have,

D = WL

^{3}/3EI=> E = WL

^{3}/3ID= (30 × 6

^{3})/(3 × 20 × 3)= 6480/180

= 36 Nm

^{-2}

**Problem 7: Calculate the moment of inertia if beam deflection for a length of 8 m is 5 m when a force of 60 N is applied. The value of Young’s modulus is 12 N/m ^{2}.**

**Solution:**

We have,

D = 5

W = 60

L = 8

E = 12

Using the formula we have,

D = WL

^{3}/3EI=> I = WL

^{3}/3ED= (60 × 8

^{3})/(3 × 12 × 5)= 30720/180

= 170.67 kg m

^{2}