**SIMPLY SUPPORTED BEAMS WITH UNIFORM DISTRIBUTED LOAD (UDL)**

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*Authors of this Blog are Atharva Dandapur, Sonal Dangare, Tejas Dani, Arjun Desai.*

**WHAT ARE BEAMS?**

Beams are important types of structural elements that play a key role in creating a safe load path to transfer the weight and forces on a structure to the foundations and into the ground. They are horizontal members which carry perpendicular loads in their longitudinal direction. They are used to support the weight of the floor, roofs of the building and transfer the load to the vertical load bearing member of the structure.

Following are the loads acting on a BEAM:

a. Uniform

b. Varied by length

c. Single point

d. Combination

**TYPES OF BEAMS: -**

Beams are classified according to various features such as:

According to the End Support Conditions:

· Simply Supported Beam

· Cantilever beam

· Continuous beam

· Fixed end beam

· Overhanging beam

· Double overhanging beam

Ø According to the Cross Section:

· I-Beam

· T-Beam

· C-Beam

According to Geometry

· Straight beam

· Curved Beam

· Tapered Beam

According to Casting Condition

· In-situ casting beam

· Precast concrete beam

· Pre-stressed concrete beam

**SIMPLY SUPPORTED BEAMS: -**

The simply supported beam is one of the simplest structures. It features only two supports, one at each end. One is a pinned support and the other is a roller support. With this configuration, the beam is inhibited from any vertical movement at both ends whereas it is allowed to rotate freely. Due to the roller support it is also allowed to expand or contract axially, though free horizontal movement is prevented by the other support. Removing any of the supports inserting an internal hinge, would render the simply supported beam to a mechanism, that is body the moves without restriction in one or more directions. Obviously this is unwanted for a load carrying structure. Therefore, the simply supported beam offers no redundancy in terms of supports, and if a local failure occurs the whole structure would collapse. These type of structures that offer no redundancy are called critical or determinant structures. To the contrary, a structure that features more supports than required to restrict its free movements is called redundant or indeterminate structure.

**FINITE DIFFERENCE METHOD: -**

In numerical analysis, finite-difference methods (FDM) are a class of numerical techniques for solving differential equations by approximating derivatives with finite differences. Here , spatial domain and time interval are discretized, or broken into a finite number of steps, and the value of the solution at these discrete points is approximated by solving algebraic equations containing finite differences and values from nearby points. Finite difference methods convert ordinary differential equations (ODE) or partial differential equations (PDE), which may be solved by matrix algebra techniques. Today, FDM are one of the most common approaches to the numerical solution of PDE, along with finite element methods.

We solve second-order ordinary differential equations of the form:

**Basic equation:** The differential equation that governs the deflection of a simply supported beam under uniformly distributed load is:

Where,

· x = Location along the beam (in)

· E = Young’s modulus of elasticity of beam (psi)

· I = Second moment of area (in⁴)

· Q = Uniform loading intensity (lb/in)

· L = Length of beam (in)

Steps:

1. Discretization

2. Solution of equation

3. Conversion of equation to Matrix

4. Applying boundary conditions

5. Find Error

**PROBLEM: -**

The deflection ‘y’ in a simply supported beam with a uniform load ‘q’ and a tensile axial load T is given by:

Where,

· x = location along the beam (in)

· T = tension applied (lbs)

· E = young’s modulus of elasticity of beam (psi)

· I = second moment of area (in⁴)

· q = uniform load intensity (lb/in)

· L = length of beam (in)

**Given,**

T = 7200lbs, q = 5400lbs/in, L = 75in, E = 30Msi , I = 120 in⁴

**Find,**

1. Find the deflection of beam at x=50”. Use a step size of △x = 25” and the approximate the derivatives by central divided difference approximation.

2. Find the relative true error in the calculation of y (50).

**SOLUTION: -**

d²y ∕ dx²-Ty / El = qx(L -X)/2EL

d²y ∕ dx² -7200y/(30*10⁶)*(120) = (5400)*x*(75-x) / 30*10⁶*2*120

d²y ∕ dx² -2*10⁶y = 7.5*10^-2x*(75-x)

Approximating the derivative at node i by CENTRAL DIVIDED DIFFERENCE approximation

d² y/dx² =y(i+1) -2yi+ y(i-1) 1/(∆x)²

Since △x=25, we have 4 nodes

1/((∆x)² ) y(i+1) -2yi+y(i-1) -2∗10⁶ yi=7.5∗10^(-7) xi∗(75-xi)

Equations at nodes:

Node 1- y1=0 …..(from simply supported boundary condition at x=0)

Node 2 — (y3 — 2y2+ y1)/(25)² -2∗10^(-6) y2=7.5∗10^(-7) x2∗(75-x2 ) — 0.0016y1–0.003202y2+0.0016y3=9.375∗10^(-4)

Node 3 — (y4 — 2y3+ y2)/(25)² -2∗10^(-6) y3=7.5∗10^(-7) x3∗(75-x3 )- 0.0016y2–0.003202y3+0.0016y4=9.375∗10^(-4)

Node 4 — y4=0 …..(from simply supported boundary condition at x=75)

Solving the equations we get,

y(50)= y(x2) ≈ y2=-0.58521

yh= k1 e⁰.0014142x+ k2 e^(-0.0014142x) ……homogeneous part

yp=Ax²+Bx+C …..particular part

(d² yp)/dx² -2∗10⁶ yp=7.5∗10^(-7) x∗(75-x)

(d² (Ax²+Bx+C ))/dx² -2∗10⁶ (Ax²+Bx+C )=7.5∗10^(-7) x∗(75-x)

Equating the terms we get,

Solving the equations,

Applying the following boundary conditions,

To calculate the RELATIVE TRUE ERROR,

The true error is given by,

Et = Exact Value — Approximate Value ; Et = 0.05320

The Relative Error is given by,

**MATLAB CODE: -**

**OUTPUTS: -**

**APPLICATIONS: -**

A common usage is for things like solving Differential Equations numerically, and approximating derivatives for root finding and numerical optimization schemes.

They are generally used to solve Partial Differential Equation (PDE) like- heat, diffusion, flow, and electromagnetics.

**CONCLUSION: -**

From the graph we can see that as the nodes increases the value of error decreases. As the number of nodes increases the deflection between the true value and the approximate value decreases. Finite difference method is easy, simple but errors are more, thus to reduce these errors we need to increase the step size. Beam Deflection problems are accurately solved by Finite Difference Method.