Mechanical Principles and Applications

Reactions and LoadsApplied to Beams

Dr Andrew KimmancePhD; MSc, BSc-Eng, MCIOB, MAPM, MAHEA

Learning Summary & Outcome What is a Beam? (Beam Analysis) What is a Load ? (Type &

Distribution) What is a Force ? (Supports) What is Equilibrium ? (reaction to

forces) What is a Bending Moments ?

(diagrams) Reactions Applied to Beams Calculations and Examples How to Check Reaction and

Answers

Learning Overview

Before starting to calculate the reactions to loads applied to a beam we first need to understand the mechanisms involved ; e.g., how, when, where, who, why, and what we are going to calculate

Learning Summary

Why is this important?

In Civil Engineering and Construction, calculations of simple loads applied to a beam can become critical and require special attention, as well as accurate implementation of the results. A simple yet effective way of calculating the reactions

to load (weight, mass, force) applied to a beam is presented here through examples, derived formulas and expressions.

Example: note the Reactions (Ra & Rb) at the end supports with the load at the centre of the beam, is expressed as F/2:

Learning Outcome

Civil Engineering

F = 56 NRa = 28 NRb = 28 N

F = 56N (point load P)

F/2 F/2concentrated load (P)

Sample (1)

RbRa

Reaction FormulaR = WL / 2W = Load Applied (UDL)L - Length of Beam

What is a Beam?

A structural member (e.g., timber, metal) that is “loaded” laterally (sideways) with forces to its length; basically, a load applied diagonally or vertically to the length of a beam (girder) to support a roof or floor.

Applied loads produce bending in a beam, bending is considered the most severe way of stressing (tension or compression) a beam.

Applied loads can be point load (concentrated ) KN or distributed KN/m² (uniform loads - UDL), also varying loads diagonally applied or both .

Beams

Applied Loads F Beam

bendingRa Rb

Beams are sized appropriately to support the loads a structure will carry.

Beams are primarily subjected to bending and shear, hence designed to carry the Shear, Bending Moment, also the Deflection caused by the design load.

The Reactions (Ra & Rb ) of the load (forces) applied to a beam can be calculated, and also the shear, bending, deflection and deformation.

Beam Analysis

Why?Safely

0M = 0

material cost, failure, injury, reputation

Ra RbStatic Equilibrium

Applied Load

Considerations

Beams come in many shapes and sizes and can be classified according to how they are supported.1.A simple or overhanging beam are supported by a pinned and a roller support.

2.Continuous beams are supported over several supports

3.Cantilever beams have a moment fixed at one end

4.Fixed beams are supported by fixed connections at both end and have end moments

5.Propped- Fixed at one end supported at other

Typical Beam Types

Types

Beams

Failure

A Load (structural) or action is a Force, deformation, or acceleration applied to a structure or its components.

Loads cause stresses, deformations, and displacements, and excess loads or overloading may cause structural failure.

Types of Load include, dead and live loads (gravity & lateral):– Dead loads are static forces that are relatively constant for a

lengthy time, such as weight of the structure and roof.– Live (dynamic) loads or imposed loads, usually unstable or

moving loads, such as temporary, transition or impact.What other loads do we need to consider?

What is a Load?

Lateral, wind, show

Loads Explained

The loads (dead, live, wind, snow) are initially applied to a building surface (floor or roof)

Transferred through walls, floors and beams which then transfer the load to other building component.

The beams in a building transfer the load to the structural components that supports its ends which may be a girder as indicated by the arrows.

How are Load Distributed?

Girder

Beam Girder

Beam

A Force is considered the action of one body on another and is usually characterised by three vector (trajectory) quantities:

1) Point of application (point load)2) Magnitude (size of load)3) Direction (bearing or path of load)

In order to design a beam, the magnitude of the reaction forces (Ra + Rb) need to be calculated to achieve static equilibrium.

Reaction forces are linear or rotational.– A linear reaction is often called a shear reaction (F or R)– A rotational reaction is often called a moment reaction (M)

Reaction forces indicate type and magnitude of the forces that are transferred and “must balance” the applied forces ∑ = 0.𝐹

What is a Force?

Forces Explained

At a roller support, beams can rotate at the support and, in this case, move horizontally and can only react with a single linear reaction force in the direction of the support.

A pin connection will hold the beam stationary both vertically and horizontally but will allow the beam to rotate. This results in horizontal (Fx) and vertical (y) linear reaction forces from the support, but no rotational reaction.

A fixed support will not allow the beam to move in any way. The resulting reaction forces include a horizontal force (Fx), a vertical force (Fy), and a moment reaction, M.

Important Note: No matter what support is used, the resulting forces must be remain static (stationary) and be in equilibrium.

Forces and Supports

Supports Explained

Forces and Supports

The state of a beam or object in which the load, weight or force counteract each other so that the beam remains stationary.

A bean can move when a load or force is applied, thus must be in static equilibrium to successfully carry loads.

In order to be static, a beam must be in equilibrium, all of the forces or loads acting on the beam must counteract and balance each other.

What is Equilibrium?

Static Equilibrium

Note: What come down or across must be equal to what goes up and across

Applied Forces

Applied Loads

zero

The loads applied to the beam (from a roof or floor) must be resisted by forces from the beam supports to be static.

These resisting forces are called “Reaction Forces”.

Reaction Forces in Equilibrium

Reaction Forces

0

50KNApplied Point Load

Reaction Force

Reaction Force

R1

R2

25KN 25KNStatic Equilibrium

Length of beam (m)

Reaction FormulaR = WL / 2W = Load Applied (UDL)L - Length of Beam

The sum of all vertical forces (F) acting on a body (beam) must equal zero.

The sum of all horizontal forces (F) acting on a body (beam) must equal zero.

The sum of all moments (about any point) acting on a body (beam) must equal zero.

Fundamental Principles of Equilibrium

yF 0

xF 0

pM 0

Zero: where

Note: A moment is created when a force tends to rotate an object. It is equal to the force times the perpendicular distance to the force.

The measure (magnitude) of bending effect due to forces acting on a beam, measured in terms of force times the distance.

A moment is a rotational force created when a force is applied perpendicularly to a point (axis) at a distance away from that point. – Calculated as the perpendicular force times the distance from the point.

Two things to remember when calculating bending moments: 1) Standard units are Nm/KN(Newton), and 2) Clockwise bending is generally taken as negative (–) (sagging –

hogging)

What is a Bending Moment?

∑M = 0

Explanation

Mx My

Positive

MomentMyMx

Negative

Moment

Bending Moment Diagram

To calculate the reaction to load applied to a beam or element; first, draw the structure in the form of a Free Body Diagram; for instance: Sketch the structure in line form free from supports and loads

acting on it. Draw all the reactions at the supports and the directions of the

reactions, and also their the precise dimensions. Draw forces or loads acting on the beam or structure elementExamples

Reactions on Supports

Free Body Diagram

Concentrated Load

Distributed Load7m

Free Body Diagram

The beam and free body diagram for a simple beam are shown here, the applied load can be concentrated, uniform or both.

To calculate the reaction to load applied to a beam; for ease, we will primarily deal with simple beams which posses a single span, one pinned connection, and one roller connection.

FREE BODY DIAGRAM

Applied Load

Note: When there is no applied horizontal load, you can ignore the horizontal reaction at the pinned end

Applied Load

BEAM DIAGRAM

Examples

Finding the reactions Ra & Rb is the first step in the analysis of a beam. If the problem is statically determinate the reactions can be found from free body

diagrams using the equations of equilibrium ∑ = 0.𝐹 Consider a simple example of a 4m beam with pin support at A and roller support

at B. Free body diagram shown below where Ra +Rb are the vertical reactions at the supports

(1) Firstly, consider the sum of moments about point B and let ∑𝑀B equal zero. ∑MB = 0 = 20(2) – RA(4) or ∑MA = 0 = -20(2) + RB (4) RA = 10KN RB = 10KN

NOTE: sign convention chosen is counter-clockwise moments are positive and clockwise moments are negative; i.e., upward forces (our reactions) as positive and downward forces (the point load) is negative, must kept to the same convention.

Calculating ReactionsExample

0 0

Next we need to solve another equation in order to find Rb (vertical reaction force at support B)(2) Let the sum of vertical forces equal 0 (ΣF = 0)Remember to include all forces including reactions and normal loads such as point loads. So if we sum all the vertical forces we get the following equation:

∑F = 0 = Ra + Rb – 20KN and since we know Ra = 10KN0 = 10KN + Rb – 20KN Rb = 10KN

We have used the two above equations (sum of moments equals zero and sum of vertical forces equals zero) and calculated that the reaction at support A is 10kN and reaction at support B 10kN. This makes sense as the point load is right in the middle of the beam, meaning both supports should have the same vertical forces (i.e. it is symmetric).

Calculating the Reactions

A simply supported 1m beam with a pin support at A and a roller support at B, calculate the reactions (Ra + Rb) to the load applied to the beam, include method for checks.

Simple Reaction Calculations

Example Problem

Point Load

20KN

RbRa

o.6m

0.4m

(1) We know ∑ = 0 then from the Equilibrium Equation 𝐹 (Ra + Rb) = 20(2) Assume clockwise direction as +ve and start by taking reaction moment

at point Ra

(3) This will be: -Rb x 1 + 20 x 0.4 = 0 …. hence, (4) Rb = 8KN now take moments about the reaction on the right:(5) Ra x 1 + 20 x 0.6 = 0 (6) Ra = 12KN ….. check these answers by balancing vertical forces ∑F = Ra + Rb- 20 = 0(7) ∑F = 0 then ∑F = 12 + 8 - 20 = 0 check

Free body DiagramA B

Calculate the reactions to loads applied to the simple supported beam below:

SolutionTaking moments about the left-hand support yields(1) ∑ML= 0 = 12kN x 3m + 18KN x 5m – RR x 6m = 0

6 + 90 – RR x 6 = 0 Hence RR = 21 kN

Note that the moment of RL about the left=hand support is zero, because the force passes through the point there. Remember also that convention clockwise moments are taken to be positive and counter clockwise moments negative. Now, summation of vertical forces yields:(2) ∑F = RL – 12kN – 1 8KN + RR x = 0 But we know RR = 21 kN

Therefore, R L – 12kN – 18KN + 21kN = 0 Hence RL = 9 kN

Beam Calculations - Example

12kN

18kN

2m3m 1m

RL RR

Let (RL + RR) stand for the reactions forces at the left-hand and right-hand support respectively

Solution continuedAlternatively, RL could be found also by taking moments about the right-hand support as follows:(3) ∑MR = – 12kN x 3m – 1 8KN x 1m + RL x 6m = 0

Hence RL = 9 kNIn this case, you can now check all the reaction answers be resolving or balancing the vertical forces together, as a result the following can be resolved:(4) ∑F = FL – 12kN – 1 8KN + RR = 9kN – 12kn – 18kN + 21kN = 0 (zero) vertical forces balanced and checked

Beam Example Cont….12kN

18kN

2m3m 1m

RL RR

Understand of Beams and Applied Loads Understanding of Reactions and Moments How to calculate the reaction to loads applied to a beam Questions Thank you

Question on Leaning Outcome