# Simply Supported Beam: 9 Important Facts

## Simply Supported Beam Definition

A simply supported beam is a beam, with one end normally hinged, and other-end is having support of roller. So because of hinged support’s, restriction of displacement in (x, y) will be and because of roller support’s will be prevented the end-displacement in the y-direction and will be free to move parallel to the axis of the Beam.

## Simply Supported Beam free body diagram.

The free-body diagram for the Beam is given below in which with point load acting at a distance ‘p’ from the left end of the Beam.

## Simply Supported Beam boundary conditions and Formula

Evaluating Reaction forces acting on the Beam by using Equilibrium conditions

Fx + Fy = 0

For vertical Equilibrium,

Fy = RA +RB – W = 0

Taking Moment about A equals to 0 with standard notations.

Rb = Wp/L

From above equation,

RA + Wp/L = W

Let X-X be the intersection at ‘a’ distance of x from the end point denoted by A.

Considering standard Sign-convention, we can compute the Shear force at the point A as described in figure.

Shear force at A,

Va = Ra = wq/L

Shear force at region X-X is

Vx = RA – W = Wq/L – W

Shear Force at B is

Vb = -Wp/L

This proves that the Shear Force remains constant between points of application of Point Loads.

Applying standard rules of Bending Moment, Clockwise Bending Moment from the Left end of the Beam is taken as +ve and Counter Clockwise Bending moment is considered as -ve respectively.

• B.M at the point A = 0.
• B.M at the point C = -RA p   ………………………… [since the moment is counter-clockwise, Bending Moment is coming out as negative]
• B.M at the point C is as follows
• B.M = -Wpq/L
• B.M at the point B = 0.

## Simply supported Beam Bending moment for uniformly distributed Loading as a function of x.

Given below is a simply-supported beam with uniformly distributed Loading applied across the complete span,

Region X-X be any region at a distance x from A.

The resultant equivalent load acting on the Beam Due to Uniform Loading case can be elaborated by

F = L * f

F=fL

Equivalent Point Load fL acting at the mid-span. i.e., at L/2

Evaluating Reaction forces acting on the Beam by using Equilibrium conditions

Fx = 0 = Fy = 0

For vertical Equilibrium,

Fy = 0

Ra + Rb = fL

taking standard sign conventions, we can write

L/2 – R = 0

From above equation,

RA + fl/2

Following the standard Sign convention, shearforce at A will be.

Va = Ra = FL/2

Shear Force at C

Vc = Ra – fL/2

Shear force at region X-X is

Vx = RA – fx = fL/2 – fx

Shear Force at B

Vb = -fL/2

For Bending Moment Diagram, we can find that by taking standard notation.

• B.M at the point A = 0.
• B.M at the point X is
• B.Mx = MA – Fx/2 = -fx/2
• B.M at the point B = 0.

Thus, the bending moment can be written as as follows

B.Mx = fx/2

## Case I: For Simply supported Beam with a concentrated load F acting at the center of the Beam

Below is a free body diagram for a simply supported steel beam carrying a concentrated load (F) = 90 kN acting at the Point C. Now compute slope at the point A and maximum deflection. if I = 922 centimer4, E = 210 GigaPascal, L =10 meter.

Solutions:

The F.B.D. Given an example is given below,

The slope at the end of the Beam is,

dy/dx = FL/16E

For a simply supported steel beam carrying a concentrated load at the centre, Maximum Deflection is,

Ymax = FL/48 EI

Ymax = 90 x 10 x 3 = 1.01m

## Case II: For Simply supported Beam having load at ‘a’ distance from support A.

For this case acting load(F) = 90 kN at the Point C. Then compute slope at the point A and B and the maximum deflection, if I = 922 cm4, E = 210 GigaPascal, L =10 meter, a = 7 meter, b = 3 meter.

So,

The slope at the end support A of the Beam,

θ = Fb(L2 – b2) = 0.211

Slope at the end support B of the Beam,

θ = Fb (l2 – B2 ) (6 LE) = 0.276 rad

The equation gives maximum Deflection,

Ymax = Fb (3L – 4b) 48EI

## Slope and Deflection in Simply supported Beam with uniformly distributed Loadingcase

Let weight W1 acting at a distance a from End A and W2 acting at a distance b from end A.

The U.D.L. applied over the complete Beam doesn’t require any special treatment associated with Macaulay’s brackets or Macaulay’s terms. Keep in mind that Macaulay’s terms are integrated with respect to themselves. For the above case (x-a), if it comes out negative, it must be ignored. Substituting the end conditions will yield constants’ values of integration conventionally and hence the required slopes and deflection value.

In This case, the U.D.L. starts at point B, the bending moment equation is modified, and the uniformly distributed load term becomes Macaulay’s Bracket terms.

The Bending Moment equation for the above case is given below.

EI (dy/dx) = Rax – w(x-a) – W1 (x-a) – W2 (x-b)

Integrating we get,

EI (dy/dx) = Ra (x2/2) – frac w(x-a) (6) – W1 (x-a) – W1 (x-b)

Given below is the Simply-supported Beam of span L subjected to Triangular Loading and derived the equation of slope and Bending moment utilizing the Double-integration methodology is as follows.

For the symmetrical Loading, every support reaction bears half of the total load and the reaction at support is wL/4 and considering moment at the point which is at a distance x from Support A is calculated as.

M = wL/4x – wx/L – x/3 = w (12L) (3L – 4x)

Using the diffn-equation of the curve.

by the double Integrating we can find as.

EI (dy/dx) = w/12L (3L x 2x 2) (-x ) + C1

putting x = 0, y = 0 in equation [2],

C2 = 0

Thus, slope = 0 at x = L/2,

0 = w/12L (3L x L2 – L4 +C1)

Substituting the constants values of C2 and C1 we get,

EI (dy/dx) = w 12L (3L) (2) – 5wl/192

The highest deflection is found at the center of the beam. i.e., at L/2.

Ely = w/12L (3L x 2L x 3) (2 x 8) / l5(5 x 32) (192)

Evaluating slope at L = 7 m and deflection from given data: I = 922 cm4 , E = 220 GPa, L =10 m, w = 15 N-m

From the above equations: at x = 7 m,

EI (dy/dx) = w (12L)(3L x 2x x 2) – x4 – 5wl/192

using equation [4]

Ely = – wl/120

220 x 10 x 922 = 6.16 x 10-4 m

Negative sign represents downward deflection.

Given below is an example of a simply supported steel beam carrying a point load and the Supports in this beam are pin supported on one end, and another is roller support. This Beam has the following given material, and loading data

loading shown in the Figure below has F=80 kN. L = 10 m, E = 210 GPa, I = 972 cm4, d = 80 mm

Evaluating Reaction forces acting on the Beam by using Equilibrium conditions

Fx = 0 ; Fy = 0

For vertical Equilibrium,

Fy = 0 (Ra + Rb – 80000 = 0)

Taking Moment about A, Clock wise Moment +ve, and anticlockwise moment is taken as -ve, we can calculated as.

80000 x 4 – Rb x 10 = 0

Rb = 32000N

Putting the value of RB in equation [1].

RA + 32000 = 80000

Ra = 48000

Let, X-X be the section of interesting at the distance of x from the endpoint A, so Shear force at A will be.

VA = RA = 48000 N

Shear force at region X-X is

Vx = RA – F = Fb/L – F

Shear Force at B is

Vb = -Fa/L = -32000

This proves that the Shear Force remains constant between points of application of Point Loads.

Applying standard rules of Bending Moment, Clockwise Bending Moment from the Left end of the Beam is taken as positive. Counter Clockwise Bending moment is taken as Negative.

• Bending Moment at A = 0
• Bending Moment at C = -RA a   ………………………… [since the moment is counter-clockwise, Bending Moment is coming out as negative]
• Bending Moment at C is
• B.M = -80000 x 4 x 6/4 = -192000 Nm
• Bending Moment at B = 0

Euler-Bernoulli’s Equation for Bending Moment is given by

M/I = σy = E/R

M = Applied B.M over the crosssection of the Beam.

I = 2nd area moment of Inertia.

σ = Bending Stress-induced.

y = normal distance between the neutral axis of the Beam and the desired element.

E = Young’s Modulus in MPa

R = Radius of Curvature in mm

Thus, the bending Stress in the Beam

σb = Mmax / y = 7.90

To know about Deflection of beam and Cantilever beam other article click below.

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