BFC 20903 Mechanics of Materials Chapter 2 Shear

BFC 20903 (Mechanics of Materials) Chapter 2: Shear Force and Bending Moment Shahrul Niza Mokhatar shahruln@uthm. edu. my

Chapter Learning Outcome 1. Draw the shear and moment diagrams. 2. Determine the largest shear and moment in a member and specify the location.

Introduction • Changes of shear force and bending moments in beams as reactions to several types of loading or combinations of loading using different support conditions. • Value of shear force and bending moment as well as the location of maximum values can be determined. • These maximum values are important because they will be used in the analysis of structural design.

Equilibrium system Brick / load Beam Roller • How to make this system to be balanced?

Types of beam • Beams can be defined as a structural component that has a smaller dimension of cross section than its length. – Statically determinate beam – Statically indeterminate beam • These classifications of beams depend on the unknown total reactions to the static equilibrium equation. • There are 3 static equilibrium equations: classified into two types based on their analyses

Types of beam • These classifications of beams depend on the unknown total reactions to the static equilibrium equation. There are 3 static equilibrium equations: • if unknown total reaction is more than 3, thus the beam is “Statically indeterminate beam”. Below is how the stability and capability of beams are being determined:

Type of support • Reaction can be determined according to its support. – Simply support • This beam is supported by rollers at one end and by pins at the other end. • This beam is a statically determinate beam.

– Cantilever beam • This beam has a fixed support at one end a free support at the other end. • The beam is a statically determinate beam

– Continuous Beam • This beam has several supports. • This beam is a statically indeterminate beam

Type of support

• Dead Load Types of loading – Loading is fixed and very close to the approximation values (very little change). Some examples are structure self weight, ceiling and fixed equipment such as the piping system services, lighting and air conditioning. For a sample of construction elements please refer to BS / EUCode. • Live Load/Imposed Load – It is a variable load on a structure for example; human weight, furniture, and other equipment based on structure purpose. Values for the loading can be referred to BS EN 1991. • Wind Load – The design of buildings must account for wind load especially in highrise buildings. However the effects of this wind load depend on the location, shape, building dimension, and wind speed in the particular region.

• Point load Types of loading – This load acts at one position only whether in vertical or inclined direction. The unit for this type of loading is N, k. N. • Distributed load – This load acts at a certain distance or a distributed load. This loading is measured in unit load per distance which is N/mm, k. N/m and so on. The value can be constant or it can increase and decrease along its reaction distance. • Moment load – Resulting load made by a pair of forces and this force will create moment on the point of loading The unit for this load is the unit of load multiple with the perpendicular distance (Nmm, k. Nm and so on).

Reactions on supports • Free Body Diagram – In order to calculate the reaction from any structure, first, draw the actual structure again in the form of a Free Body Diagram. – Free body diagram is the sketch of the actual structure in lines form that are free from supports as well as from loads acting on it. Draw all reactions at the supports & the directions of the reactions & their accurate dimensions. Draw all forces or loads acting on the structure elements.

Equilibrium equation • This consists of 3 equations, where by using these three equations, reactions of certain structures can be determined. The equation is: • To avoid a simultaneous equation, let’s use moment equation, ΣM = 0 first on any one support point, followed by other reactions on support point that can be determined using this equation ΣFx = 0 ; Σ Fy = 0. • If the solution of this simultaneous equation gives the negative value, it means the earlier assumption of reaction direction is incorrect and the actual direction is the opposite direction.

Example 1 • Determine the reaction of beam stressed with a load as shown in Figure

Example 1: Solution

Example 2 Determine the reaction on beam

Example 2: Solution

Example 3: Tutorial (Discuss in Group) Find all the reaction force. 4 k. N/m C D 3 m 2 k. N 4 m B E 3 m A 5 m

Example 3: Solution Answer: RAx = -2 k. N RAy = 8. 8 k. N RCy = 8. 8 k. N, MC = 6 k. Nm REy = 11. 2 k. N RDy = 11. 2 k. N, MD = 0

Shear Force Diagram (SFD) Bending Moment Diagram (BMD)? Why we need SFD & BMD? Before load act Moment (-ve) Moment (+ve)

Shear Force and Bending Moment – Shear force, V at a section is the total algebra force taken only at one side of the section. Any section can be taken to be totaled provided that both sections must have the same values and they follow the marking direction. – Bending moment, Mx is the total algebra of moments force acting on the section.

Shear Force and Bending Moment � To make sure that the beam is in a balanced vertical direction, there should exist a V force that balances the RA force and one moment M with an anti clock wise movement to prevent the beam from rotating. Both the force and moment must be balanced on the section that has been cut. The total force and moment at this section equals to zero. � The V force is known as a shear force, the force that causes the member to slide and separate into two sections. Meanwhile moment M is known as the bending moment, which is a reaction moment at a certain point for all forces as well as combinations reacting on the left or right side of the point.

Natural sign for shear force and bending moment • Common sign – Horizontal direction – Vertical direction – Moment direction

Shape of SFD and BMD

Procedure to draw SFD and BMD – Support reactions – determine all the reactive force & couple moments acting on the beam. – Shear & moment functions – specify separate coordinates x having an origin at the beam’s left & extending to regions of the beam between concentrated forces &/or couple moments. Section beam at each distance x, & draw the FBD of one of the segments. – Shear & moment diagram – Plot the shear diagram (V versus x) & moment diagram (M versus x). If the numerical values of the functions describing V & M are positive, the values are plotted above the x axis, whereas negative values are plotted below the axis. It is convenient to show the shear & moment diagram below the FBD of the beam.

Equation & Diagram of Shear Force & Bending Moment 1. Simply supported beam with point load acting. To get the V and M equations, make sections on each part between the point load, distributed load and support.

The above equation shows that shear force caused by point load is constant or fixed and the bending moment changes linearly with x.

Note: – SFD positive, +ve is drawn on top of the beam. – BMD positive, +ve is drawn on top of the beam. – To draw SFD & BMD, it is enough to determine V & M value on the important points only such as at the support, point under the point load & at both ends & in the middle of the distributed load.

2. Distributed uniform load acting on simply supported beam.

The above equation shows that shear force caused by the distributed uniform load will change linearly with x. SFD needs to drawn first to see the relation with BMD. From SFD above, there is one contra point, where the SFD line cuts across the beam axis (shear force = 0). The distance of the point from point A can be obtained as:

3. Uniform increase load acting on simply supported beam.

• To determine the value of gx which is the value of uniform increase load when the distance x is from A, the intensity of the load on the section needs to obtained. From the figure above;

• plot every point on SFD & BMD, thus SFD & BMD will appear as shown below. • When shear force, V = 0 ; bending moment is at the maximum and the point distance from A is: