Upcoming SlideShare
×

# Centroids moments of inertia

79,663 views

Published on

19 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

Views
Total views
79,663
On SlideShare
0
From Embeds
0
Number of Embeds
149
Actions
Shares
0
1,308
0
Likes
19
Embeds 0
No embeds

No notes for slide

### Centroids moments of inertia

1. 1. University of ManchesterSchool of Mechanical, Aerospace and Civil EngineeringMechanics of Solids and StructuresDr D.A. BondPariser Bldg. C/21e-mail: d.bond@umist.ac.ukTel: 0161 200 8733 UNIVERSITY OF MANCHESTER 1st YEAR LECTURE NOTES MECHANICS OF SOLIDS AND STRUCTURES SEMESTER 2 § 11: CENTROIDS AND MOMENTS OF AREA § 12: BEAM SUPPORTS AND EQUILIBRIUM § 13: BEAM SHEAR FORCES & BENDING MOMENTS § 14: BENDING THEORY 1
2. 2. Centroids and Moments of Area11. CENTROIDS AND MOMENTS OF AREAS11.1 Centroid and First Moment of Area11.1.1 DefinitionsThe Centroid is the geometric centre of an area. Here the area can be said to be concentrated, analogous tothe centre of gravity of a body and its mass. In engineering use the areas that tend to be of interest are crosssectional areas. As the z axis shall be considered as being along the length of a structure the cross-sectionalarea will be defined by the x and y axes.An axis through the centroid is called the centroidal axis. The centroidal axes define axes along which thefirst moment of area is zero.The First Moment of Area is analogous to a moment created by a Force multiplied by a distance except thisis a moment created by an area multiplied by a distance. The formal definition for the first moment of areawith respect to the x axis (QX):Q X = ∫ ydA (11.1)Similarly for the first moment of area with respect to the y axis (QY) is:QY = ∫ xdA (11.2)where x, y and dA are as defined as shown in Figure 11.1. Y x dA C y y x X Total Area = A Figure 11.1The X and Y subscripts are added to indicate the axes about which the moments of area are considered.11.1.2 Co-ordinates of the CentroidThe centroid of the area A is defined as the point C of co-ordinates x and y which are related to the firstmoments of area by:Q X = ∫ ydA = yAQY = ∫ xdA = x A (11.3)An area with an axis of symmetry will find its first moment of area with respect to that axis is equal to zeroi.e. the centroid is located somewhere along that axis. Where an area has two axes of symmetry the centroidis located at the intersection of these two axes 2
3. 3. Centroids and Moments of Area11.1.3 Example: Centroid of a TriangleDetermine the location of the centroid of a triangle of base b and height h. Y dA x dx ^ Y x C h ^ X y b X Figure 11.2Ans: x = 2b/3 and y = h/311.1.4 Centroid of a Composite AreaWhere an area is of more complex shape a simple method of determining the location of the centroid may beused which divides the complex shape into smaller simple geometric shapes for which the centroidallocations may easily be determined. Consider Figure 11.3 which shows a complex shape of Area A madefrom three more simple rectangular shapes of Areas A1, A2 and A3. As the centroids of the rectangular shapesare easily determined from symmetry the locations of their respective sub-area centroids are used to calculatethe location of the centroid of the composite shape. A1 Y C1 x C A2 C2 y C3 A3 X Figure 11.3Q Y = ∫ xdA = ∫ xdA + ∫ xdA + ∫ xdA A1 A2 A3 = x 1 A1 + x 2 A 2 + x 3 A 3 = xAThe same method can be used to calculate the y-wise location of the centroid of the composite area. 3
4. 4. Centroids and Moments of Area11.1.5 Example: Centroid of a L section A1 x= (b + ht 2 ) 2(b + h ) Y h 2 + t (b + 2h ) y= x 2(b + h ) h C C1 y t C2 A2 X t b Figure 11.411.2 Second Moment of Area11.2.1 DefinitionsThe second moment of the area about the x axis (IX) is defined as:I X = ∫ y 2 dA (11.4)and the second moment of the area about the y axis (IY) similarly as:I Y = ∫ x 2 dA (11.5)Some texts refer to second moment of Area as Moment of Inertia. This is not technically correct and SecondMoment of Area should be preferred.11.2.2 Example: Rectangle (of dimensions b × h)Derive an expression for the second moments of Area for a rectangle with respect to its centroidal axes (Usethe symbol ^ to indicate centroidal axes and properties with respect to these axes).The centroid is easily located by using intersecting axes of symmetry. ˆ dy Y dA h/2 y C ˆ X h/2 bh 3 Ans: I X = b ˆ 12 Figure 11.5 4
5. 5. Centroids and Moments of AreaThe solution for second moment of area for a rectangle is frequently used as many composite shapes arebroken into rectangular sections to determine their composite second moments of area. The rule is oftenrecalled as: The second moment of area of a rectangle about its horizontal centroidal axis is equal to one-twelfth its base (b) multiplied by its height (h) cubed.Similarly I Y may be determined to be equal to b3h/12. ˆ11.2.3 Relationship to Polar Second Moment of AreaThe Rectangular Second Moments of Area IX and IY are able to be related to the Polar Second Moment ofArea about the z axis (J) which was introduced in the section on Torsion.I X + I Y = ∫ y 2 dA + ∫ x 2 dA ( ) = ∫ y 2 + x 2 dA = ∫ r 2dAI X + IY = J Z (11.6)11.2.4 Radii of GyrationThe radius of gyration of an Area A with respect to an axis is defined as the length (or radius r) for which:I X = ∫ y 2 dA = rX A 2I Y = ∫ x 2 dA = rY2 A (11.7)J Z = ∫ r dA = r A 2 Z 2As for the First Moment of Area, the X, Y and Z subscripts are added to indicate the axes about which thesecond moments of area or radii of gyration are considered.11.2.5 Parallel Axis TheoremIf the axes system chosen are the centroidal axes the Second Moments of Area calculated are known as theSecond Moments of Area about the centroidal axes. Such axes are often annotated differently to otheraxes indicating that they are centroidal axes. In this course the symbol ^ shall be used. If the second momentof area about another set of axes is required then the Parallel Axis Theorem may be used rather than havingto recalculate the Second Moments of Area. Y dA ˆ Y C ˆ y ˆ X Total Area = A y d X Figure 11.6 5
6. 6. Centroids and Moments of Area ˆFigure 11.6 shows an area with a centroid at C (with a centroidal x axis shown as X ) and for which theSecond Moment of Area with respect to the X axis is required. The X axis is a distance d away from thecentroidal axis.I X = ∫ y 2dA = ∫ (y + d ) dA { as y = y + d} 2 ˆ ˆ = ∫ y 2 dA + 2d ∫ y dA + ∫ d 2dA ˆ ˆ = ∫ y 2 dA + ∫ d 2dA ˆ {as ∫ y dA = 0 about centroidal axes} ˆI X = I X + Ad 2 ˆ (11.8)This demonstrates that if the Second Moment of Area is known around an area’s centroidal axis the SecondMoment of Area of that area about another axis distance d from the centroidal axis is simply the sum of theCentroidal Second Moment of Area and the product Area × d2.This theorem applies to the Second Moments of Area IX and IY as well as to the Polar Second Moment ofArea provided the appropriate centroidal values are used.11.2.6 Example: Second moment of Area of a Rectangle about its base axisFor the rectangle shown; determine its second moment of area about its base and left edge axes (X and Y). ˆ Y Y h/2 C ˆ X h/2 X b Figure 11.7 bh3 hb3Ans: I X = , IY = 3 3 6
7. 7. Centroids and Moments of Area11.2.7 Second Moment of Area for a Composite SectionConsider the composite area shown in Figure 11.8. To determine the Second Moment of Area of such acomplex structure a similar approach to that used for calculating the centroids of complex areas is used.Follow these steps:i. Determine the Centroids of the sub-Areasii. Calculate the Second Moments of Area of the sub-Areas about their centroidal axesiii. Use the Parallel axis theorem to move sub-Area Second Moments of Area to axis of interestiv. Sum the contributions of each sub-Area to the overall Second Moment of Area. A1 Y d1 C1 A2 d2 C2 d3 A3 C3 X Figure 11.8The validity of the above approach can be seen below for determining IY of the area in Figure 11.8:IY = ∫ x 2 dA = ∫ x 2 dA + ∫ x 2 dA + ∫ x 2 dA A1 A2 A3 = ∫ (ˆ + d ) dA + ∫ (ˆ + d 2 ) dA + ∫ (ˆ 3 + d 3 ) dA 2 2 2 x 1 1 x 2 x A1 A2 A3 = ∫ (ˆ x 2 1 x )x2 ˆ ( 2 ˆ2 ) 2 + 2 ˆ 1d1 + d12 dA + ∫ ˆ 2 + 2 x2 d 2 + d 2 dA + ∫ x3 + 2 x3d 3 + d 32 dA ˆ ( ) 2 A1 A2 A3 ( ˆ 1 ˆ 1 1 )( = I Y A + A d + I Y A + A2 d + I Y A + A d 2 ˆ 2 2 2 )( 3 2 3 3 ) as ∫ (2 xd )dA = 2d ∫ (ˆ )dA = 0 ˆ x A3 A3This method is often well suited to a tabular layout or a spreadsheet.11.3 Tabulated Centroids and Second Moments of AreaMany text books list the locations of standard area centroids and provide the Second Moment of Area aroundthese centroids. The departmental databook has such a table and will be allowed for use in examstherefore students should become familiar with the use of this table.11.4 UnitsFirst Moment of Area has units of Length3.Second Moment of Area has units of Length4. 7
8. 8. Centroids and Moments of Area11.4.1 Example: A Regular I sectionDerive an expression for the second moment of Area for a regular I beam with respect to its centroidal x axis. Y ˆ Y t ˆ X a t t X b Figure 11.9The I section may be represented as being comprised of a rectangle of dimensions b×(2t+a) from which twosmaller rectangles of dimensions ½(b-t)×a have been taken out all of which have the same x-wise centroidalaxis. The total second moment of area is then simply the sum of all the contributions (with the missing areasbeing subtracted). b(a + 2t ) 3 1 (b − t )a 3IX = ˆ − 2. 2 12 12 b(a + 2t ) − (b − t )a 3 3 = 12This solution could also have been derived by considering the three rectangles separately and using parallelaxis theorem although there would have required significantly more work. 8
9. 9. Centroids and Moments of Area11.4.2 An Unsymmetrical I sectionConsider an unsymmetrical section shown below. The section is symmetrical about the vertical centroidalŶ( ) axis only. The y-wise position of centroid is to be found so that the second moment of area about its x-wise centroidal axis can be determined. In examples such as this where the component is constructed from"regular sub-areas" it is best to follow a tabular method as shown below. 5cm Y y3 3 1.5cm y2 2 6cm y ^ 1cm X y1 1 1cm X 10cm Figure 11.10The section above is divided into three rectangular areas, (1), (2), (3). The bottom x-axis is used as datum.The tabular method of finding the centroid and the second moment of area are demonstrated in the followingTable. Section Area (Ai) yi (Ay )i d = y - yi Ad2 IX i ˆ I X i + A i d i2 ˆ i (cm2) (cm) (cm3) (cm) (cm4) (cm4) (cm4) 1 10 0.5 5 3.208 102.913 0.833 103.746 2 6 4 24 -0.292 0.512 18 18.512 3 7.5 7.75 58.125 -4.042 122.533 1.406 123.939 Totals 23.5 87.125 246.197y= ∑ (Ay) i ∑A i= 3.708cmI X = ∑ I X i + ∑ A i d i2 ˆ ˆ= 246.197 cm 4 2Use first three columns to find y before proceeding to calculate d, Ad etc. 9
10. 10. Beam Supports and Equilibrium12. BEAM SUPPORTS AND EQUILIBRIUM IN BENDING12.1 Introduction12.1.1 What is beam bending?Tension, compression and shear are caused directly by forces. Twisting and bending are due to moments(couples) caused by the forces. Tension F F ⇒ F F Compression F F ⇒ F F ⇒ V Shear V V V Torsion T T ⇒ T T Bending M M ⇒ M M Figure 12.1 Structural DeformationsBending loads cause a straight bar (beam) to become bent (or curved). Any slender structural member onwhich the loading is not axial gets bent. Any structure or component that supports the applied forces(externally applied or those due to self weight) by resisting to bending is called a beam.12.1.2 Eraser ExperimentWhat is the basic effect of bending? Mark an eraser on the thickness face with a longitudinal line along thecentre and several equi-spaced transverse lines. Bend it. The centre line has become a curve.Question:• What happens to the spacing of the transverse lines?Bending causes compression on one side and extension on the other. By inference there is a section whichdoes not extend or shrink. This is called the Neutral Plane. On the eraser this will be the central longitudinalline. Consistent with extension and compression, bending must cause tensile (pulling) stresses on one sideand compressive (pushing) stresses on the other side of the neutral plane.Bending is predominantly caused by forces (or components of forces) that are act perpendicular to the axis ofthe beam or by moments acting around an axis perpendicular to the beam axis. 10
11. 11. Beam Supports and Equilibrium12.2 Representation of a beam and its loading Uniformly Concentrated or Distributed Loads Beam Point Load (UDL) Non-uniform Distributed Load Moment y WC x wDE -wFG MH B J A C D E F G H z RBy RJy Reaction Loads Figure 12.2: Typical representation of a beamFor a schematic diagram (suitable for a FBD), normally only a longitudinal view along the centre line (thelocus of the centroids of all the transverse sections, called the Centroidal Axis) is used to represent the beam(see Figure 12.2). Vertical (y-direction) forces acting on the beam will be assumed to act at the centre line,but normal to it. Concentrated forces that act at specific points, such as W at C, are shown as arrows.Distributed loads are shown as an area (or sometimes as a squiggly line) to represent a load distributed over agiven length of the beam. Distributed loads have dimensions of force per unit length. Moments arerepresented by a curved arrow.Question:• What is the most common form of distributed load?For reference, a Cartesian co-ordinate system (xyz) consistent with the right hand screw rule is always used.The origin can be located at any convenient point (usually an end or the centre of the beam). The z-axis isaligned along the axis of the beam, the y-axis in the direction of the depth of the beam and the x-axis in thedirection of the beam width (into the page). When representing a beam on paper the y and z planes arenormally drawn in the plane of the page and the x axis is perpendicular to the page. The bending forces andmoments considered in this 1st year course will only act in the yz-plane (i.e. the plane of the page).Become accustomed to this axis system as it is common to most analyses in future years.12.3 Supports for a beam and their schematic representation12.3.1 IntroductionA beam must be supported and the reactions provided by the supports must balance the applied forces tomaintain equilibrium. Types of support and their symbolic representations are given in the followingsections.12.3.2 Simple supportA simple support will only produce a reaction force perpendicular to the plane on which it is mounted (seeFigure 12.3). Simple supports may move in the plane on which they are mounted but prevent any motionperpendicular to this plane. Simple supports do not produce forces in the plane on which they are mountedand moments are not restrained by a simple support. So for a simply supported beam the axial displacementsand rotations (which cause slope changes) at the supports are unrestrained (i.e. in Figure 12.3 the beam is 11
12. 12. Beam Supports and Equilibriumfree to move along the z-axis and rotate about the x axis). Imagine these supports as being similar to thesupporting wheel of a wheelbarrow. y y Beam A z z RAy Support y-direction Reaction Load Figure 12.3: Simple support reaction loads12.3.3 Hinged or Pinned end supportHinged or pinned supports provide similar support to simple supports with the addition of support in theplane on which they are mounted i.e displacements in the axial direction are prevented (in Figure 12.4 thebeam is only free to rotate about the x axis). Imagine these supports as being the same as the connection atthe top of a grandfather clock pendulum. y y Beam A RAz z Support z-direction z Reaction Load RAy Support y-direction Reaction Load Figure 12.4: Hinged/Pinned support reaction loads12.3.4 Fixed or built-in end supportFixed end supports (also called encastre) support moments in addition to lateral and axial forces. No axial,lateral or rotational movements are possible at a built in end (i.e in Figure 12.5 the end is not able to move ineither the y or z direction nor can it rotate about the x axis). Imagine these supports as being like theconnection of a balcony onto a building. y MA Support Reaction y Moment Beam A RAz z Support z-direction z Reaction Load RAy Support y-direction Reaction Load Figure 12.5: Fixed support reaction loads and moments12.4 Distributed loadsTo simplify the analysis of a distributed load it is usually easier to replace the distributed load with a pointload acting at an appropriate location. As the units of distributed loads are load per unit length the equivalentpoint load may be determined by statics. 12
13. 13. Beam Supports and Equilibrium Area under w(z) = Aw w(z) We y y e z z L L dAw = w(z).dz w(z) y dz z z Figure 12.6: Replacing a distributed load with an equivalent point loadFor the two cases to be equivalent the sum of the forces in the y and z directions have to be the same and thesum of the moments about any point have to be the same. Considering forces in the y direction first: L∑ Fy = ∫ w( z ) dz = Aw = We 0That is, the equivalent point force of a distributed load (We) is equal to the area under the w(z)function (Aw).Now consider moments about the origin: L L∑ M o = ∫ w( z ).z dz = ∫ z dAw = We .e 0 0Note the similarity between this equation and Equation (11.3) for the first moment of area which allows theprevious expressions to be re-written as:L∫ z dA0 w = z AwWe .e = Aw .e∴e = zThat is, the point along the beam at which the effective force (We) must act is at the centroid of the area(Aw) under the distributed load curve w(z). 13
14. 14. Beam Supports and Equilibrium12.5 Equilibrium considerations for a beamConsider a beam carrying loads as shown in the figure below. The right hand support at B is a simple supportand can only carry vertical forces. All the horizontal force components have to be supported by the left hand(hinged) support, at A. y c W M w A B C z l1 l2 l3 l4 L Figure 12.7 A beam hinged at the left and simply supported on the right, loads as shownConsider a point C where the left and right hand parts of the beam are to be separated into two free bodydiagrams. To maintain equilibrium in the separated sections additional forces and moments must be appliedat the new ends to keep both sections of the beam in the same geometry as when the beam was intact. Theseforces and moments are known as the axial and shear forces and bending moments at position C. Theseforces and moments determine how a beam deforms under loading. To determine these forces and momentsthe support reactions must first be obtained from the conditions of equilibrium of forces and moment for thewhole beam. Then the forces at the point C (shear force and bending moment) may be obtained from forceand moment equilibrium of the part of the beam to the left or to the right of the section. The left and righthand parts and all the possible forces acting the two new ends are shown in the free body diagrams below.12.5.1 The support reactionsThe support forces are obtained from the conditions of equilibrium of forces and moments on the wholebeam so DRAW a FBD of the beam and apply equilibrium conditions. W M y w B RAz A z We is equivalent RAy point load to RBy distributed load w We Figure 12.8: FBD of entire beam used to calculate support loads and momentsEquilibrium conditions require:∑F z = 0, ∑F y = 0, ∑M = 0First with the condition MA = 0, we get the vertical support reaction at B. ∑ 14
15. 15. Beam Supports and Equilibrium ∑M A =0 ( = −W .l1 − w l3 − l 2 ) (l 3 + l2 2 ) − M + RBy .L W .l1 + 2 ( w 2 2 l3 − l2 + M ) ⇒ RBy = L Equilibrium of forces in the vertical direction, Fy = 0, gives the vertical support force at A: ∑ ∑F y =0 = −W − w(l3 − l 2 ) + RBy + RAy ⇒ RAy = W + w(l3 − l 2 ) − RBy∑ Fz = 0 provides the axial support force at A. ∑F z =0 = R Az Note: If any of the forces calculated are negative then they act in the opposite sense to that assumed in the FBD. 12.6 Sign Conventions The forces and moments that act on a beam at point C (MC, FC and VC) are assigned positive or negative signs depending on the face that they act on. If the face they act on has a normal in the positive z direction then positive forces and moments are in the positive y or z directions or as defined by the right hand rule. If the face has a normal in the negative z direction then a positive force or moment is in the opposite direction. This sign convention is shown below. AXIAL FORCES F +ve F F -ve F SHEAR FORCES V +ve V V -ve V BENDING MOMENTS M M M M +ve -ve w w DISTRIBUTED LOADS +ve -ve Figure 12.9 Sign convention for Axial Forces, Shear Forces, Bending Moments and distributed loads 12.6.1 The forces at point C If the beam is cut at point C (at a distance c from A) then for equilibrium we require equal and opposite forces FC and VC as well as equal and opposite moments MC, acting at the severed sections of the two parts of the beam. MC, FC and VC are provided in the complete beam by the internal stresses in the material of the beam. • MC = Sum of moments due to all forces to one side of the point C (including support forces) is called the BENDING MOMENT acting on the vertical face of the beam at position C. 15
16. 16. Beam Supports and Equilibrium• FC = Sum of all x-direction (axial) forces to one side of the point C is called the AXIAL FORCE acting on the vertical face of the beam at position C.• VC = Sum of all the y-direction (transverse/shear) forces to one side of the point C is called the SHEAR FORCE acting on the vertical face of the beam at position C. y c y W w M VC A VC B MC MC RAz FC PC z z l2-c RBy RAy l3-c c-l1 C l4-c L-c C Figure 12.10: Left Hand and Right Hand FBDs for beam sectioned at CThus, in the above problem, MC, FC and VC can be found by considering the LH end FBD to be:M C = − R Ay .c + W ( c − l1 ) = − w(l3 − l2 )c + Wl1c wc 2 2 Mc L + L l3 − l 2 + L − Wl1 ( ) VC = − R Ay + W = − w(l3 − l2 ) + 1 + Wl L 2L w 2 2 M l3 − l2 + L ( ) FC = − R Az = 0or by using the RH end FBD to be:  (l + l ) M C = w(l3 − l2 ) 3 2 − c  + M − RBy (L − c ) = − w(l3 − l2 )c + 1 + Wl c wc 2 2 Mc l3 − l 2 + − Wl1 ( )  2  L L LVC = − w(l 3 − l2 ) + RBy = − w(l3 − l2 ) + 1 + Wl L 2L w 2 2 M l3 − l 2 + L ( )FC = 0Note that the numerical quantities of bending moment, axial force and shear forces must be the same inmagnitude and sense (sign) at a section irrespective of whether they are calculated considering the free bodydiagrams of the beam to the left or to the right of the section.The importance of MC, FC and VC are that the beam’s performance at position C is directly related to theseforces i.e., the stress, the deflection and the local rotation (angle) of the beam are all determined by thesemoments and forces. 16
17. 17. Beam Supports and Equilibrium12.6.2 Example (easy): Beam forces at mid span for a cantilever beamDetermine the beam bending moment, shear and axial forces at mid span (C) for the beam and loadingshown in Figure 12.11. 2kN/m 3kN y 1m A B C 2m 2kN/m y VC MCRAz A FC MA RAy C z Figure 12.11: A cantilever beam with concentrated and distributed loadsAnswers: RAy = 5 kN, MA = 7 kN.m, FC = 0kN, MC = 3 kN.m, VC = -3 kN12.6.3 Example (more difficult): Beam forces at mid span for a simply supported beamDetermine the beam bending moment, shear and axial forces at mid span (C) for the beam and loadingshown in Figure 12.12. 2kN 2kN/m y 1/3m 1/3m A B C 2m 2kN y 1kN/m VC MCRAz A FC RAy C z Figure 12.12: A simply supported beam with concentrated and distributed loadsAnswers: RAy = 59/27 kN, RBy = 31/27 kN, FC = 0kN, MC = 7/9 kN.m, VC = -2/27 kN 17
18. 18. Beam Supports and Equilibrium12.7 Relationships between M, V and Distributed loads12.7.1 Relationship between M and VThe bending moment and the shear force at a given section are not independent of each other. The mutualrelationship between these quantities is derived by considering the equilibrium of a small length of the beambetween z and z+ z. Assume that the Bending Moment M and Shear Force V vary along the length of the δbeam such that at z+ z the Bending Moment is M+ M and the Shear Force is V+ V. This element is shown δ δ δin Figure 12.13. Note that the shear forces and bending moments as shown are all positive. M M+ M y z V V+ V z z Figure 12.13 FBD of beam element to related M and VConsider the moment equilibrium about the left hand end of the element∑ M = (V + δV ).δz + M − (M + δM ) = 0 ⇒ δM = Vδz + δVδz ≈ VδzIn the limit when z approaches zero this reduces to: δdM =V dzor (12.1)M = ∫ VdzNote that the moments are taken about the left hand end of the element and that the second order quantity,the product of V and z, is neglected because it will be negligibly small. δ δEquation (12.1) states that the variation of bending moment with z will have a slope/gradient equal to thevalue of the shear force. 18
19. 19. Beam Supports and Equilibrium12.7.2 Relationship between V and a Distributed LoadConsider the equilibrium of a small length of the beam between z and z+ z upon which a Distributed Load δ(w) is applied. Assume that the Shear Force V varies along the length of the beam such that at z+ z the Shear δForce is V+ V. This element is shown in Figure 12.14. δ w w+ w y z V V+ V z z Figure 12.14: FBD of a beam element to relate Q and a UDLEquilibrium of forces in the y direction gives: = −V + (V + δV ) + w.δz + δwδz = 0 1∑F y 2 1 ⇒ δV = − wδz − δwδz ≈ − wδz 2In the limit when z approaches zero this reduces to: δdV = −w dzor (12.2)V = ∫ − wdzNote that the second order quantity, the product of w and z, is neglected because it will be negligibly δ δsmall.Therefore if a small element of a beam has no distributed load on it then w = 0 and the shear force in thesection must be constant. If w is non-uniform (i.e. w = w(z)), this analysis assumes that dz is so small that thedistributed load across dz may be considered uniform at a level defined by w(z). Thus this equations is validwhether the distributed load is uniform or a function of z (i.e. w may equal w(z)).Equations (12.1) and (12.2) can be used to check the consistency of predicted shear force and bendingmoment variation along a beam.Combining the previous two equations gives the key relationship: dV d 2 M−w= = (12.3) dz dz 2 19
20. 20. Beam Supports and Equilibrium12.7.3 Example: Simply supported beam with a Uniformly Distributed LoadDetermine the shear force and bending moment at mid-span of the beam shown in Figure 12.15 using:a. The Free Body Diagram approach of section 12.6.1, andb. Equation (12.3) y 5kN/m A B 2m Figure 12.15 Simply supported beam with a Uniformly Distributed LoadAnswers: Vmid-span = 0kN, Mmid-span = -2.5kN/m 20
21. 21. Shear Forces and Bending Moments13. SHEAR FORCE & BENDING MOMENT DIAGRAMS13.1 IntroductionBending moments cause normal tensile and compressive stresses simultaneously in different parts of a beamsection. Shear forces cause shear stresses that try to cut the beam. The magnitudes of bending moment andshear forces generally vary from one section to another in a beam. As shown in the previous section, the twoquantities are dependent on each other. Graphs showing the variation of M and V along the length of thebeam are called Bending Moment (BM) and Shear Force (SF) diagrams. The BM and SF diagrams help toidentify the critical sections in beams where bending moments and shear forces are highest.13.2 Examples of SF and BM diagrams13.2.1 Example: Cantilever beam with a concentrated loadConsider beam AB fixed at A, carrying a concentrated load (-W) at any position C (distance z = c from A). W W y c y c MA A B z z C RAz L RAy Figure 13.1 Cantilever beam with a concentrated load and its FBD• To draw the SF and BM diagrams first calculate the support reactions by considering force and moment equilibrium conditions. DRAW the FBD of the beam (see Figure 13.1). ∑F ⇒z RAz = 0 ∑F ⇒ y RAy = W ∑M ⇒ A M A = Wc• Determine the shear force and bending moment relationships for each section of the beam - where sub-length boundaries are defined by point loads, moments and the start/finish of Distributed Loads. Again use a FBD for each section (see Figure 13.2). W y z y z-c c M M MA MA z z V z V RAy RAy Figure 13.2: Sub-lengths (A ≤ z ≤ C) and (C ≤ z ≤ B) FBDs In length (A ≤ z ≤ C): In section (C ≤ z ≤ B): V (z ) = − R Ay = −W V (z ) = R Ay -W = 0 M (z ) = M A − R Ay z = Wc − Wz = W (c − z ) M (z ) = M A − R Ay z + W (z-c ) = Wc − Wz + W (x-c ) = 0 21
22. 22. Beam Bending Theory• Plot the shear force (V) and Bending Moment (M) diagrams along the length of the beam. M Wc V A C B A C B z z -W Figure 13.3: Shear Force and Bending Moment Diagrams for cantilever beam with concentrated load• Check consistency of shear force and bending moment expressions between sections and at locations where concentrated loads/moments are applied or where UDLs start or finish.Hints & Checks for BM and SF diagrams• Notice that the magnitudes of the shear force and bending moments at A (i.e. V(0) & M(0)) are equal to the magnitudes of the end reaction load and moment• Note that the Shear Force (V) is equal to the negative value of the end reaction load (V(0) = -RAy) as by our definition for shear forces, RAy is acting in the –ve direction. That is, in the +ve y direction but on a face with a normal in the –ve z direction.• Whenever there is a concentrated load or bending moment applied to a beam the corresponding Shear Force and Bending Moment diagrams should show a step of the same magnitude. In this example the steps at A due to the point load and moment RA and MA are from zero to -W and Wc respectively.• Notice that the shear force and bending moment at a free end are zero.• Note that the bending moment varies linearly from Wc to 0 over the distance of z = c in the region 0 ≤ z ≤ c; i.e., it has a constant gradient of -W as expected due to V being equal to -W throughout that section of the beam (see equation (12.1)). 22
23. 23. Beam Bending Theory13.2.2 Example: Cantilever beam with a Uniformly Distributed Load.Consider beam AB fixed at A, carrying a uniformly distributed load w positioned between C and D as shownin Figure 13.4. a c y y w MA w z z A C D B RAz RAy L Figure 13.4 Cantilever beam with a UDL and its FBD• To draw the SF and BM diagrams first calculate the support reactions by considering force and moment equilibrium conditions. DRAW the FBD of the beam (see Figure 13.4). ∑F z ⇒ R Az = 0 ∑F y ⇒ RAy = − wc { = Area under Distributed Load curve}  c ∑M A ⇒ M A = − wc  a+  { = Moment created by effective force acting at centroid}  2• Determine shear force and bending moment relationships for each sub-length of the beam. In this example sub-lengths are defined by the reaction loads/beam ends and the start and finish of the UDL. Again use a FBD for each sub-length (see Figure 13.5). z y y z y z-a z z-a z-a-c a a M w M w M MA MA MA z x z V V V RAy RAy RAy Figure 13.5: FBDs for the sections (A ≤ z ≤ C), (C ≤ z ≤ D) and (D ≤ z ≤ B) respectively. In section (A ≤ z ≤ C): V (z ) = − RAy = wc  c  c M (z ) = M A − RAy z = -wc a +  + wcz = wc  z-a −   2  2 Again M(0) = MA as MA is a concentrated moment input to the beam at the left hand end. This is one good check to see that derived expression for M is correct. 23
24. 24. Beam Bending Theory In section (C ≤ z ≤ D): V (z ) = − RAy − w( z − a ) = w(− z + a + c ) w(z − a ) ( ) ( ) 2 w w w M (z ) = M A + RAy z − = 2 zc-2ac − c 2 − z 2 − 2az + a 2 = − (− z + a + c ) 2 2 2 2 2 Note: that M(z) could have been more easily derived using RHS FBD as it would not have included MA In section (D ≤ z ≤ C): V (z ) = − RAy − wc = 0  c M (z ) = M A − RAy z − wc z − a −  = 0  2• Plot the shear force (V) and Bending Moment (M) diagrams along the length of the beam. V M wc A C D B z z A C D B -(wc2)/2 -wc(a+c/2) Figure 13.6: Shear Force and Bending Moment diagrams for a cantilever beam with a UDL• To check consistency of results note that in section C to D, V varies linearly with slope –w as expected from equation (12.2) and in all sections dM/dx = V.• By varying a and c any particular case of a cantilever beam with a uniformly distributed load can be solved. 24
25. 25. Beam Bending Theory13.2.3 Example: Non-uniform distributed load on a cantilever wo y z L Figure 13.7: Non-uniform distributed load on a cantileverDerive an expression for the variation of shear force and bending moment in a cantilever beam loaded by anon-uniform distributed load as shown in Figure 13.7. Try this using both the FBD approach and by usingequations (11.1) and (11.2).Answers:V (z ) = 2L ( ) ( z − L2 and M ( z ) = o z 3 − 3 L2 z + 2 L3 wo 2 w 6L ) 25
26. 26. Beam Bending Theory13.2.4 Example: Simply supported beam AB of length L.Consider a beam, pin supported at one end and simply supported at the other (this combination is oftenreferred to as simply supported). A concentrated load (-W) acts at distance c from A. W y c A B C L Figure 13.8: A simply supported beam with concentrated load• Determine support reactions using the FBD of the entire beam: ∑F x ⇒ RAz=0 Wc ∑M A ⇒ RBy= L Wc W (L-c ) ∑F y ⇒ RAy= W-RBy = W − L = L• Derive expressions for Shear Force and Bending Moment in each section: W y z y z-c c M M z z V V RAy RAy Figure 13.9 Simply supported beam with a concentrated load FBDs In section (A ≤ z ≤ C): In section (C ≤ z ≤ B): − W (L − c ) Wc V (z ) = − RAy = V (z ) = − RAy + W = − RBy = L L − W (L-c )z − W (L-c )z − Wc(L-z ) M (z ) = − RAy z = M (z ) = − RAy z + W ( z-c ) = + W (z-c ) = L L L• Plot the shear force (V) and Bending Moment (M) diagrams along the length of the beam. V M Wc L A C B A B z z -W (L − c ) - Wc (L − c ) L C L Figure 13.10: Shear force and Bending Moment diagrams for a simply supported beam with a concentrated load 26
27. 27. Beam Bending Theory13.2.5 Example: A simply supported beam with a UDL. a c y w A B C D L Figure 13.11 Simply supported beam with a UDL and the beam FBD• Determine support reactions using the FBD of the entire beam: ∑F z ⇒ RAz = 0 wc  c ∑M A ⇒ RBy = L  a +  2 wc   c  ∑M B ⇒ RAy = L   L −  a +    2  • Derive expressions for Shear Force and Bending Moment in each section: z z z-a y y y z a M w M z z z M V V V RAy RAy RBy Figure 13.12: FBDs for beam sections (A ≤ z ≤ C), (C ≤ z ≤ D) and (D ≤ z ≤ B) respectively. In section (A ≤ z ≤ C): In section (C ≤ z ≤ D): In section (D ≤ z ≤ C): − wc   c  − wc   c  wc  c V (z ) =  L − a +  V (z ) =  L −  a +   + w( z − a )  V (z ) = a +  L   2   L   2   L  2 − wc  c  − wc  c  w( z − a ) M (z ) = − wc  c  a + ( L − z ) 2   M (z ) =  L − a + z  M (z ) =  L −  a +  z + L   2   L   2   2 L  2• Plot the shear force (V) and Bending Moment (M) diagrams along the length of the beam. V M ( wc a + c 2 ) L C A C E D B A D B z z - wc L ( ( L− a+c 2 )) Mmax Figure 13.13: Shear force and Bending Moment diagrams for a simply supported beam with a UDL• How might Mmax be determined? 27
28. 28. Beam Bending Theory13.3 Principle of Superposition13.3.1 TheoryThe support reactions and fixing moments as well as shear forces and bending moments (and all othermechanical entities such as stresses and displacements) at a given section (or point) due to the individualloads can be calculated separately and summed up algebraically to obtain the total effect of all the loadsacting simultaneously. This is applicable to conservative (i.e. linearly elastic) systems only.13.3.2 Example: Cantilever beam with a concentrated load and UDLConsider a cantilever beam with a concentrated load (N) and a UDL (w) as shown in Figure 13.14. This canbe separated into two more simple to analyse cantilever loading cases: a single concentrated load and a singleUDL. The separate results for these two loading cases may be added together to obtain the results for thecomplete beam (provided the beam remains in the linear elastic region of its stress-strain curve). This cangreatly simplify analysis as separate, simple expressions for SF and BM may be obtained for each of theloading cases and then be added together to obtain the SF and BM expressions for the overall beam. d e y N c w z A C D E B L d e y y N c w z z A C D E B A C D E B Concentrated Load Case (CL) UDL Case (UDL) Figure 13.14: Example of Principle of SuperpositionThe results of sections 13.2.1 and 13.2.2 may be used to quickly write the SF and BM expressions for thecombined load case: In section (A ≤ z ≤ C): V (z ) = V( CL ) + V( UDL ) = N + we  e M (z ) = M ( CL ) + M ( UDL ) = N ( z − c ) + we z-d −   2 In section (C ≤ z ≤ D): V (z ) = V( CL ) + V( UDL ) = we  e M (z ) = M ( CL ) + M ( UDL ) = we z-d −   2 28
29. 29. Beam Bending Theory In section (D ≤ z ≤ E): V (z ) = V( CL ) + V( UDL ) = w(− z + d + e ) M (z ) = M ( CL ) + M ( UDL ) = (− z + d + e )2 w 2 In section (E ≤ z ≤ B): V (z ) = V( CL ) + V( UDL ) = 0 M (z ) = M ( CL ) + M ( UDL ) = 0Note that there are more sections in the overall beam than in either of the individual beams. Simply add theappropriate expressions from each separated load case to determine the overall expression for each section.This method may also be used to determine the support loads and moments for the beam. In the exampleabove:R Ay = R Ay( CL ) + R Ay( UDL ) = − N − we  eM A = M A( CL ) + M A( UDL ) = − Nc − we d +   2Check these results using FBDs for the complete beam.13.4 Summary of Procedure for drawing SF and BM diagrams:i. Find support reactions by considering force and moment equilibrium conditions.ii. Determine shear force and bending moment expressions.iii. Plot shear force (V) vs z and bending moment (M) vs z to appropriate scales.iv. Check for consistency of sign convention and agreement of values at the supports and at the ends of the beam and that equations (11.1) and (11.2) are valid along the length of the beam.13.5 Macauley1 Notation13.5.1 IntroductionFrequently it is beneficial to use a single expression for the shear force and bending moment distributionsalong a beam rather than the collection of sub-length expressions.Consider a cantilever beam on which two concentrated forces and a UDL are acting. Y W1 W2 W1 W2 w w a MA z RAz A B b RAy L Figure 13.15: Beam for which Macauley expression is to be derived1 Macauley W.H. Note on the deflection of beams, Messenger of Mathematics, Vol 48 pp 129-130. 1919. 29
30. 30. Beam Bending TheoryThere are three lengths of the beam in which the bending moment is different:0≤ z≤a M = M A − R Ay za≤ z≤b M = M A − R Ay z − W1 ( z − a ) w(z − b ) 2b ≤ z ≤ L M = M A − R Ay z − W1 (z − a ) − W2 (z − b ) − 2Clearly the expression for the length b ≤ z ≤ L contains the terms in the other two lengths. To reduce thetedium of working with three separate equations and sub-lengths, the following notation (due to Macauley)may be used: w{z − b} 2M = M A − R Ay z − W1 {z − a} − W2 {z − b} − 2In this version of the bending moment equation the terms within { } should be set to zero if the valuewithin these brackets become -ve.In the general case:  0 for z < a{z − a}n =  (z − a ) for z ≥ a nThus the last two terms become zero if z < b and the last three terms are zero if z < a.13.5.2 Derivation of Macauley Shear Force and Bending Moment expressionsUsing Macauley parentheses the distributed loading on the beam may be written as:w = w{z − b} 0The term in the parentheses is equal to zero if z < b and if z ≥ b the term in the parentheses raised to thepower of zero equals 1. This is essentially an on switch for the distributed load that says that when z < bthere is no distributed load and when z ≥ b the distributed load equals w.Integrating2 the Macauley expression for distributed load gives an expression for the shear force along thebeam (from Equation 11.3):∫ w = ∫ w{z − b} = w{z − b} + C1 = −V 0 1 where C1 is a constant of integrationThe constant however, has to be evaluated for each sub-length of the beam. This is made fairly straightforwardby recalling that the shear force only has a step change when a point load is applied and that the step change isequal to the value of the point load. So for the example of Figure 13.15 the constant C1 represents a Macauleyexpression for all point loads along the beam:C1 = RAy + W1 {z − a} + W2 {z − b} 0 0hence the shear force may be written as:V = − w{z − b}− C1 = − w{z − b}− R Ay − W1 {z − a} − W2 {z − b} 0 02 When integrating a Macauley expression, the whole term within the brackets should be treated as a variable. This isjustified by recognising that only the integration constant varies if the term is expanded out, for example: {x − a} 2 x2 a2Treating { x − a } as a variable gives ∫ {x − a}dx = 2 + c1 = 2 − ax + + c1 2 2 ∫ {x − a}dx = ∫ xdx − ∫ adx = 2 − ax + c2 x Expanding { x − a } gives a2 ∴ c2 = c1 + but both expressions are constant 2 30
31. 31. Beam Bending TheoryIntegrating this term again to obtain an expression for bending moments (from Equation 12.3) gives:M = ∫ Vdz = − w {z − b}2 − R Ay z − W1 {z − a} 1 − W2 {z − b}1 + C2 2Similar to the relationship between the shear force constant and point loads, the constant C2 represents aMacauley expression for all point moments along the length of the beam. Note: that anti-clockwise momentsapplied to the beam are considered positive as they introduce a +ve step in the variation of M along the beam.C2 = M ACombining these last two expressions gives the full Macauley expression for bending moment along thebeam as: w{z − b} 2M = M A − R Ay z − W1 {z − a}− W2 {z − b}− 213.5.3 Example: Macauley Expression for a beamDerive the Macauley expressions for shear force and bending moment for the beam and loading shown inFigure 13.16.Then use the expressions derived to determine the value and location of the maximum bending moment inthe beam. a a y w B A 4a Figure 13.16 Simply supported beam with a UDLAnswers: V = w{z − a} + {z − 2a}0 {z − a}2 + 3wa z − 15 wa {z − 2a}1 3 wa 15 wa w − M = 4 4 2 4 4 31
32. 32. Beam Bending Theory13.5.4 Use of superposition to simplify Macauley expressionsIn the previous examples the bending moment equations have been easy to develop in Macauley notationbecause the distributed loads have been open ended i.e., they cease at the end of the bar. There are howeversome loading situations that are less easily expressed in this way, particularly closed end distributed loads. Aclosed end distributed load is one where the DL ceases or step-changes at some location along the length ofthe beam. In this situation superposition may be used to develop a bending moment expression that hasmultiple open-ended (i.e. ceasing at the end of the beam) DLs. Y Y Y w w w z z z A L/2 B L/2 C L/2 L/2 L/2 L/2 Figure 13.17: A cantilever beam with a UDL over part of the beamFor the cantilever beam shown in Figure 13.17 the closed end distributed load may be replaced by two openended distributed loads. The net combination of these two new load cases is equivalent to the original loadcase.To derive the Macauley expression simply add the expressions for both cases together. w( z ) = wUDL1 + wUDL 2 = − w{z − 0} + w{z − L } 0 0 2 ∫ w(z )dz = − w{z − 0} + w{z − L } + C1 1 1 2 ( ) C1 = R Ay UDL1 ( ) + R Ay UDL 2 = wL 2 V ( z ) = − ∫ w( z )dz = w{z − 0} − w{z − L } − 1 1 wL 2 2∫V (z ) dz = 2 {z − 0} w w {z − L }2 − wL z + C2 2 − 2 2 2 wL2 C 2 = (M A )UDL1 + (M A )UDL 2 = 8 wL2 M = ∫ V ( z ) dz = {z − 0} − {z − L } − w w wL z+ 2 2 2 2 2 2 8 32
33. 33. Beam Bending Theory14. BENDING THEORY14.1 IntroductionBending causes tensile and compressive stresses in different parts of the same cross-section of a beam. Thesestresses vary from a maximum tension on one surface to a maximum compression on the other passingthrough a point where the stress is zero (known as the neutral point). The maximum stresses areproportional to the bending moment at the cross-section. As the magnitude of the maximum stress dictatesthe load bearing capacity of the beam (i.e. for most engineering applications the stresses should be keptbelow yield), it is important to find out how the stresses the bending moments are related. The relationshipbetween stresses and bending moments will be developed in this section. The analysis is restricted by theassumptions stated in section 14.2. The assumptions can be relaxed and improved analysis can be made butthis is beyond the scope of the first year course and will be covered in future years.14.2 Assumptions• The beam is made of linear-elastic material.• The cross section of the beam is symmetrical about the plane in which the forces and moments act (i.e. the YZ plane).• A transverse section of the beam which is plane before bending remains plane after bending.• Youngs Modulus is same in tension and compression.• The lateral surface stresses (in the y-direction) are negligible. The lateral stress within the beam and the shear stresses between adjacent "layers" throughout the depth of the beam are ignored (until next year).It is possible to do the analysis without these assumptions. But the algebra becomes very complicated.14.3 The beam bending equation14.3.1 Location of the neutral axisBending moment generally varies along the length of the beam. However it is reasonable to assume thatbending moment is constant over a very small (infinitesimally small) length of the beam. So the case of sucha small length of a beam subjected to a constant bending moment along its length (known as pure bending)is analysed below.A small element of a beam is schematically shown in Figure 14.1. An initially straight beam element abcdis bent to a radius R at point z by bending moments M, to abcd. The layers above line ef (i.e. on the convexside) lengthen and those below ef (i.e. on the concave side) shorten. The line ef (and ef) is therefore thelayer within the beam that neither lengthens nor shortens i.e. the NEUTRAL AXIS.3 The initial length ofef is also equal to the arc length of ef, i.e:ef = dz = ef = RθConsider the line gh at a distance y from the neutral axis. The original length of gh was the same as for allother layers within the element i.e dz. The new length of gh may is related to its bend radius (R+y) and bendangle ( ) so that the new length may be written as (R+y)θ. Therefore as strain is the ratio of change in length θto original length and stress ( ) = strain ( ) × modulus of elasticity (E): σ εStrain in layer gh = z direction strain at distance y from neutral axis (ε z ( y )) gh − g h (R + y ) θ − dz (R + y )θ − Rθ yθ y ⇒ ε z( y ) = = = = = g h dz Rθ Rθ Rand3 Note: A prime () is used to indicate points in the undeformed condition. 33
34. 34. Beam Bending Theory Stress in layer gh = z direction stress at distance y from neutral axis (σ z ( y )) Ey ⇒ σ z ( y ) = Eε z = R y z dz a d y1 a d y g h z e f y2 e f b c M b c M R θ Figure 14.1 Element of a beam subject to pure bendingTherefore as E and R are constants for a given position z and bending moment M the variation of stress througha beam is linear as shown in Figure 14.2. Note in Figure 14.2 positive stress is defined using the convention forthe right hand end of beam. y z σz = Ey1/R y1 σz z y2 M σz = Ey2/R M Figure 14.2: Axial stress distribution at z 34