DOCSLIB.ORG

Chapter 8 Bending Moment and Shear Force Diagrams

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Chapter 8 Bending Moment and Shear Force Diagrams CHAPTER 8 BENDING MOMENT AND SHEAR FORCE DIAGRAMS EXERCISE 51, Page 121 1. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. To calculate the reactions: Resolving vertically gives: RR12+= 3 But as the beam is symmetrically loaded, RR12= Hence, R12= R = 1.5kN Bending moment expressions: Range AC At x, M = Rx1 = 1.5 x i.e. a straight line Range CB At x, M = R1 x−− 3(x 1) = 1.5 x – 3x + 3 i.e. M = - 1.5 x + 3 i.e. a straight line Shearing Force expressions: Range AC SF = + 1.5 kN Range CB SF = + 1.5 – 3 = - 1.5 kN The bending moment and shearing force diagrams are shown below. 146 © John Bird & Carl Ross Published by Taylor and Francis 2. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Taking moments about B gives: R1 ×=× 3 41 4 Hence, R= = 1.333kN 1 3 Resolving vertically gives: RR12+= 4 Hence, R2 = 4−=− R1 4 1.333 = 2.667 kN Bending moment expressions: Range AC At x, M = Rx1 = 1.333 x i.e. a straight line Range CB At x, M = R1 x−− 4(x 2) = 1.333 x – 4x + 8 147 © John Bird & Carl Ross Published by Taylor and Francis i.e. M = - 2.667 x + 8 i.e. a straight line Shearing Force expressions: Range AC SF = R1 = + 1.333 kN Range CB SF = R1 - 4 = + 1.5 – 4 = - 2.667 kN The bending moment and shearing force diagrams are shown below. 3. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Taking moments about B gives: R1 ×=×+× 3 12 41 6 Hence, R= = 2kN 1 3 Resolving vertically gives: R12+=+ R 14 Hence, R2 = 52− = 3 kN 148 © John Bird & Carl Ross Published by Taylor and Francis Bending moment expressions: Range AC At x, M = Rx1 = 2 x i.e. a straight line Range CB At x, M = R1 x−− 1(x 1) = 2 x – x + 1 i.e. M = x + 1 i.e. a straight line Range DB At x, M = R1 x− 1(x −− 1) 4(x − 2) = 2 x – x + 1 – 4x + 8 i.e. M = - 3x + 9 i.e. a straight line Shearing Force expressions: Range AC SF = R1 = + 2 kN Range CD SF = R1 - 1 = 2 – 1 = 1 kN Range DB SF = R1 - 1 - 4 = 2 – 1 – 4 = - 3 kN The bending moment and shearing force diagrams are shown below. 4. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. 149 © John Bird & Carl Ross Published by Taylor and Francis Taking moments about D gives: R1 ×+×=× 24014 4 Hence, R= = 2kN 1 2 Resolving vertically gives: R12+=+ R 14 Hence, R2 = 52− = 3 kN Bending moment expressions: Range CA At x, M = −1x = - x i.e. a straight line Range AB At x, M = −+1x R1 (x − 2) = -1 x + 2 x - 4 i.e. M = x - 4 i.e. a straight line Range DB At x, M = −+1x R12 (x −+ 2) R (x −− 4) 4(x − 4) = -1 x + 2x - 4 + 3x - 12 – 4x + 16 i.e. M = 0 Shearing Force expressions: Range CA SF = - 1 kN Range AB SF = - 1 + R1 = - 1 + 2 = + 1 kN Range DB SF = - 1 + 2 – 4 + 3 = 0 kN The bending moment and shearing force diagrams are shown below. 150 © John Bird & Carl Ross Published by Taylor and Francis 5. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Taking moments about B gives: R1 ×+ 36 =× 42 86− Hence, R= = 0.667kN 1 3 Resolving vertically gives: RR12+= 4 Hence, R2 = 4− 0.667 = 3.333 kN Bending moment expressions: Range AC At x, M = Rx1 = 0.667 x i.e. a straight line Range CD At x, M = R1 x−− 4(x 1) = 0.667 x - 4 x + 4 i.e. M = - 3.333 x + 4 i.e. a straight line 151 © John Bird & Carl Ross Published by Taylor and Francis Range DB At x, M = R1 x− 4(x −+ 1) 6 = 0.667 x - 4x + 4 + 6 i.e. M = - 3.333 x + 10 i.e. a straight line Shearing Force expressions: Range AC SF = R1 = + 0.667 kN Range CD SF = R1 - 4 = 0.667 - 4 = - 3.333 kN Range DB SF = R1 – 4 = 0.667 – 4 = - 3.333 kN The bending moment and shearing force diagrams are shown below. 6. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Bending moment expressions: Range AC 152 © John Bird & Carl Ross Published by Taylor and Francis At x, M = - 4 x i.e. a straight line Range CB At x, M = - 4 x – 6(x – 2) = - 4 x - 6 x + 12 i.e. M = - 10 x + 12 i.e. a straight line Shearing Force expressions: Range AC SF = - 4 kN Range CB SF = - 4 - 6 = - 10 kN The bending moment and shearing force diagrams are shown below. 7. Determine expressions for the bending moment and shearing force distributions for the following simply supported beam; hence, or otherwise, plot the bending moment and shearing force diagrams. Bending moment expressions: Range AC At x, M = - 2 x i.e. a straight line Range CB At x, M = - 2 x + 6 i.e. a straight line 153 © John Bird & Carl Ross Published by Taylor and Francis Shearing Force expressions: Range AC SF = - 2 kN Range CB SF = - 2 kN The bending moment and shearing force diagrams are shown below. 8. A horizontal beam of negligible mass is of length 7 m. The beam is simply-supported at its ends and carries three vertical loads, pointing in a downward direction. The first load is of magnitude 3 kN and acts 2 m from the left end, the second load is of magnitude 2 kN and acts 4 m from the left end, and the third load is of magnitude 4 kN and acts 6 m from the left end. Calculate the bending moment and shearing force at the points of discontinuity, working from the left support to the right support. To determine the reactions R A and R B : Taking moments about B in the diagram below gives: Clockwise moment = anticlockwise moment i.e. R A × 7 = 3 × 5 + 2 × 3 + 4 × 1 i.e. R A × 7 = 25 25 from which, R = = 3.57 kN A 7 Resolving vertically, R A + R B = 3 + 2 + 4 = 9 from which, RB = 9 – 3.57 = 5.43 kN 154 © John Bird & Carl Ross Published by Taylor and Francis Range AC At x, M = R A × x = 3.57 x i.e. a straight line (1) SF = + R A = 3.57 kN (2) Range CD At x, M = R A × x – 3(x – 2) = 3.57 x – 3 x + 6 i.e. M = 0.57 x + 6 i.e. a straight line (3) SF = R A - 3 = 3.57 – 3 = 0.57 kN (4) Range DE At x, M = R A × x – 3(x – 2) – 2(x – 4) = 3.57 x – 3 x + 6 – 2x + 8 i.e. M = - 1.43 x + 14 i.e. a straight line (5) SF = R A - 3 – 2 = 3.57 – 5 = - 1.43 kN (6) Range EB At x, M = R A × x – 3(x – 2) – 2(x – 4) - 4(x – 6) 155 © John Bird & Carl Ross Published by Taylor and Francis = 3.57 x – 3 x + 6 – 2x + 8 – 4 x + 24 i.e. M = - 5.43 x + 38 i.e. a straight line (7) SF = R A - 3 – 2 – 4 = 3.57 – 5 – 4 = - 5.43 kN (8) Plotting equations (1) to (8) results in the bending moment and shearing force diagrams shown above. Summarising, Bending moments (kN m): 0, 7.14, 8.28, 5.42, 0 Shearing forces (kN): 3.57, 3.57/ 0.57, 0.57/ - 1.43, - 1.43/ - 5.43, - 5.43 156 © John Bird & Carl Ross Published by Taylor and Francis EXERCISE 52, Page 124 1. Determine expressions for the bending moment and shearing force distributions for the following simply supported beams; hence, plot the bending moment and shearing force diagrams. kN Total load = 6× 7m = 42 kN m By inspection, RR12= 42 Hence, R= R = = 21kN 122 Bending moment expression: x2 At x, M = Rx−× 6 1 2 i.e. M = 21 x - 3 x2 i.e. a parabola Shearing Force expression: At x, SF = R1 − 6x i.e. SF = 21 – 6x i.e. a straight line The bending moment and shearing force diagrams are shown below. 157 © John Bird & Carl Ross Published by Taylor and Francis 2. Determine expressions for the bending moment and shearing force distributions for the following simply supported beams; hence, plot the bending moment and shearing force diagrams. kN Total load = 5× 12m = 60 kN m By inspection, RR12= 60 Hence, R= R = = 30kN 122 Bending moment expression: x2 At x, M = Rx−× 5 1 2 i.e. M = 30 x – 2.5 x2 i.e. a parabola Shearing Force expression: At x, SF = R1 − 5x i.e.
Load more

Recommended publications
  • Glossary: Definitions
    Appendix B Glossary: Definitions The definitions given here apply to the terminology used throughout this book. Some of the terms may be defined differently by other authors; when this is the case, alternative terminology is noted. When two or more terms with identical or similar meaning are in general acceptance, they are given in the order of preference of the current writers. Allowable stress (working stress): If a member is so designed that the maximum stress as calculated for the expected conditions of service is less than some limiting value, the member will have a proper margin of security against damage or failure. This limiting value is the allowable stress subject to the material and condition of service in question. The allowable stress is made less than the damaging stress because of uncertainty as to the conditions of service, nonuniformity of material, and inaccuracy of the stress analysis (see Ref. 1). The margin between the allowable stress and the damaging stress may be reduced in proportion to the certainty with which the conditions of the service are known, the intrinsic reliability of the material, the accuracy with which the stress produced by the loading can be calculated, and the degree to which failure is unattended by danger or loss. (Compare with Damaging stress; Factor of safety; Factor of utilization; Margin of safety. See Refs. l–3.) Apparent elastic limit (useful limit point): The stress at which the rate of change of strain with respect to stress is 50% greater than at zero stress. It is more definitely determinable from the stress–strain diagram than is the proportional limit, and is useful for comparing materials of the same general class.
    [Show full text]
  • Bending Stress
    Bending Stress Sign convention The positive shear force and bending moments are as shown in the figure. Figure 40: Sign convention followed. Centroid of an area Scanned by CamScanner If the area can be divided into n parts then the distance Y¯ of the centroid from a point can be calculated using n ¯ Âi=1 Aiy¯i Y = n Âi=1 Ai where Ai = area of the ith part, y¯i = distance of the centroid of the ith part from that point. Second moment of area, or moment of inertia of area, or area moment of inertia, or second area moment For a rectangular section, moments of inertia of the cross-sectional area about axes x and y are 1 I = bh3 x 12 Figure 41: A rectangular section. 1 I = hb3 y 12 Scanned by CamScanner Parallel axis theorem This theorem is useful for calculating the moment of inertia about an axis parallel to either x or y. For example, we can use this theorem to calculate . Ix0 = + 2 Ix0 Ix Ad Bending stress Bending stress at any point in the cross-section is My s = − I where y is the perpendicular distance to the point from the centroidal axis and it is assumed +ve above the axis and -ve below the axis. This will result in +ve sign for bending tensile (T) stress and -ve sign for bending compressive (C) stress. Largest normal stress Largest normal stress M c M s = | |max · = | |max m I S where S = section modulus for the beam. For a rectangular section, the moment of inertia of the cross- 1 3 1 2 sectional area I = 12 bh , c = h/2, and S = I/c = 6 bh .
    [Show full text]
  • Bending Moment & Shear Force
    Strength of Materials Prof. M. S. Sivakumar Problem 1: Computation of Reactions Problem 2: Computation of Reactions Problem 3: Computation of Reactions Problem 4: Computation of forces and moments Problem 5: Bending Moment and Shear force Problem 6: Bending Moment Diagram Problem 7: Bending Moment and Shear force Problem 8: Bending Moment and Shear force Problem 9: Bending Moment and Shear force Problem 10: Bending Moment and Shear force Problem 11: Beams of Composite Cross Section Indian Institute of Technology Madras Strength of Materials Prof. M. S. Sivakumar Problem 1: Computation of Reactions Find the reactions at the supports for a simple beam as shown in the diagram. Weight of the beam is negligible. Figure: Concepts involved • Static Equilibrium equations Procedure Step 1: Draw the free body diagram for the beam. Step 2: Apply equilibrium equations In X direction ∑ FX = 0 ⇒ RAX = 0 In Y Direction ∑ FY = 0 Indian Institute of Technology Madras Strength of Materials Prof. M. S. Sivakumar ⇒ RAY+RBY – 100 –160 = 0 ⇒ RAY+RBY = 260 Moment about Z axis (Taking moment about axis pasing through A) ∑ MZ = 0 We get, ∑ MA = 0 ⇒ 0 + 250 N.m + 100*0.3 N.m + 120*0.4 N.m - RBY *0.5 N.m = 0 ⇒ RBY = 656 N (Upward) Substituting in Eq 5.1 we get ∑ MB = 0 ⇒ RAY * 0.5 + 250 - 100 * 0.2 – 120 * 0.1 = 0 ⇒ RAY = -436 (downwards) TOP Indian Institute of Technology Madras Strength of Materials Prof. M. S. Sivakumar Problem 2: Computation of Reactions Find the reactions for the partially loaded beam with a uniformly varying load shown in Figure.
    [Show full text]
  • Shear Force and Bending Moment in Beams and Frames CHAPTER2
    Shear Force and Bending Moment in Beams and Frames CHAPTER2 2.0 INTRODUCTION To bridge a gap in land on earth is most pressing problem for structural engineers. Slabs, beams or any combination of these are extensively used in bridging gaps. Construction of bridges, covering of door and window openings and covering of roof of enclosures are examples where beams and slabs are used. The space below a beam is available for other use. Structural actions of beams and slabs are slightly different. A beam collects fl oor load and transfers it longitudinally to the supports, which are usually provided either at both ends (beam action) or at one end only (cantilever action). Cantilevers are used in construction of balconies in homes and footpaths in bridges. A slab transfers fl oor load in both directions whereas a beam transfers fl oor load in only longitudinal direction. Therefore, a beam is a one-dimensional structural element and it can be represented by a line as shown in Fig. 2.1(b) in which arrow shows the direction of load transfer. Mechanical engineers extensively use cantilevers. The boom of a crane, the cutting blade of a bulldozer and wings of a fan are few examples of cantilevers. The span of beam or cantilever, which is defi ned as distance between adjacent supports, governs the complexity of its design. Shear force and bending moment diagrams quantify structural action of beams and cantilevers and are required for their rational design. Beams and cantilevers shown in Fig. 2.1 are known as fl exural elements.
    [Show full text]
  • Glossary of Notations
    108 GLOSSARY OF NOTATIONS A = Earthquake peak ground acceleration. IρM = Soil influence coefficient for moment. = A0 Cross-sectional area of the stream. K1, K2, = aB Barge bow damage depth. K3, and K4 = Scour coefficients that account for the nose AF = Annual failure rate. shape of the pier, the angle between the direction b = River channel width. of the flow and the direction of the pier, the BR = Vehicular braking force. streambed conditions, and the bed material size. = BRa Aberrancy base rate. Kp = Rankine coefficient. = = bx Bias of ¯x x/xn. KR = Pile flexibility factor, which gives the relative c = Wind analysis constant. stiffness of the pile and soil. C′=Response spectrum modeling parameter. L = Foundation depth. = CE Vehicular centrifugal force. Le = Effective depth of foundation (distance from = CF Cost of failure. ground level to point of fixity). = CH Hydrodynamic coefficient that accounts for the effect LL = Vehicular live load. of surrounding water on vessel collision forces. LOA = Overall length of vessel. = CI Initial cost for building bridge structure. LS = Live load surcharge. = Cp Wind pressure coefficient. max(x) = Maximum of all possible x values. = CR Creep. M = Moment capacity. = cap CT Expected total cost of building bridge structure. M = Moment capacity of column. = col CT Vehicular collision force. M = Design moment. = design CV Vessel collision force. n = Manning roughness coefficient. = D Diameter of pile or column. N = Number of vessels (or flotillas) of type i. = i DC Dead load of structural components and nonstructural PA = Probability of aberrancy. attachments. P = Nominal design force for ship collisions. DD = Downdrag. B P = Base wind pressure.
    [Show full text]
  • Truss Action  Consider a Loaded Beam of Rectangular Cross Section As Shown on the Next Page
    TRUSSES Church Truss Vermont Timber Works Truss Action Consider a loaded beam of rectangular cross section as shown on the next page. When such a beam is loaded, it is subjected to three internal actions: an axial force (A), a shear force (V), and an internal bending moment (M) The internal bending moment is a direct result of the induced shear force. If this shear force is eliminated, the bending moment is also eliminated. The end result is a tensile or compressive axial force Since this loaded beam does deflect (bend), the top of the beam will be subjected to compressive forces while the bottom of the beam will be subjected to tensile forces The magnitudes of these forces and moments and their internal distribution depends upon several factors. The details of this are beyond the scope of a course in statics but will be developed in a Strength of Materials course. For now, simply accept these phenomena at face value 2 Truss Action F (a) A simply supported beam of rectangular cross section is shown with a point load at midspan. Consider a section of the beam between the two dotted lines F (b) Force 'F' induces an internal shear (V) and axial (A) force. The shear force causes an internal bending moment (M) M A V V A M 3 Truss Action But if one replaces the beam with a truss as shown below, all shear forces within the beam are translated to pure compressive and tensile forces in the truss members. Since there is no shear, there is no bending moment.
    [Show full text]
  • Explicit Analysis of Large Transformation of a Timoshenko Beam : Post-Buckling Solution, Bifurcation and Catastrophes Marwan Hariz, Loïc Le Marrec, Jean Lerbet
    Explicit analysis of large transformation of a Timoshenko beam : Post-buckling solution, bifurcation and catastrophes Marwan Hariz, Loïc Le Marrec, Jean Lerbet To cite this version: Marwan Hariz, Loïc Le Marrec, Jean Lerbet. Explicit analysis of large transformation of a Timoshenko beam : Post-buckling solution, bifurcation and catastrophes. Acta Mechanica, Springer Verlag, 2021, 10.1007/s00707-021-02993-8. hal-03306350 HAL Id: hal-03306350 https://hal.archives-ouvertes.fr/hal-03306350 Submitted on 29 Jul 2021 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Explicit analysis of large transformation of a Timoshenko beam : Post-buckling solution, bifurcation and catastrophes Marwan Hariza,1, Lo¨ıc Le Marreca,2, Jean Lerbetb,3 aUniv Rennes, CNRS, IRMAR - UMR 6625, F-35000 Rennes, France bUniversit´eParis-Saclay, CNRS, Univ Evry, Laboratoire de Math´ematiques et Mod´elisation d’Evry, 91037, Evry-Courcouronnes, France Abstract This paper exposes full analytical solutions of a plane, quasi-static but large transformation of a Timoshenko beam. The problem isfirst re-formulated in the form of a Cauchy initial value problem where load (force and moment) is prescribed at one-end and kinematics (translation, rotation) at the other one.
    [Show full text]
  • 7.4 the Elementary Beam Theory
    Section 7.4 7.4 The Elementary Beam Theory In this section, problems involving long and slender beams are addressed. As with pressure vessels, the geometry of the beam, and the specific type of loading which will be considered, allows for approximations to be made to the full three-dimensional linear elastic stress-strain relations. The beam theory is used in the design and analysis of a wide range of structures, from buildings to bridges to the load-bearing bones of the human body. 7.4.1 The Beam The term beam has a very specific meaning in engineering mechanics: it is a component that is designed to support transverse loads, that is, loads that act perpendicular to the longitudinal axis of the beam, Fig. 7.4.1. The beam supports the load by bending only. Other mechanisms, for example twisting of the beam, are not allowed for in this theory. applied force applied pressure cross section roller support pin support Figure 7.4.1: A supported beam loaded by a force and a distribution of pressure It is convenient to show a two-dimensional cross-section of the three-dimensional beam together with the beam cross section, as in Fig. 7.4.1. The beam can be supported in various ways, for example by roller supports or pin supports (see section 2.3.3). The cross section of this beam happens to be rectangular but it can be any of many possible shapes. It will assumed that the beam has a longitudinal plane of symmetry, with the cross section symmetric about this plane, as shown in Fig.
    [Show full text]
  • Basic Mechanics
    93 Chapter 6 Basic mechanics BASIC PRINCIPLES OF STATICS All objects on earth tend to accelerate toward the Statics is the branch of mechanics that deals with the centre of the earth due to gravitational attraction; hence equilibrium of stationary bodies under the action of the force of gravitation acting on a body with the mass forces. The other main branch – dynamics – deals with (m) is the product of the mass and the acceleration due moving bodies, such as parts of machines. to gravity (g), which has a magnitude of 9.81 m/s2: Static equilibrium F = mg = vρg A planar structural system is in a state of static equilibrium when the resultant of all forces and all where: moments is equal to zero, i.e. F = force (N) m = mass (kg) g = acceleration due to gravity (9.81m/s2) y ∑Fx = 0 ∑Fx = 0 ∑Fy = 0 ∑Ma = 0 v = volume (m³) ∑Fy = 0 or ∑Ma = 0 or ∑Ma = 0 or ∑Mb = 0 ρ = density (kg/m³) x ∑Ma = 0 ∑Mb = 0 ∑Mb = 0 ∑Mc = 0 Vector where F refers to forces and M refers to moments of Most forces have magnitude and direction and can be forces. shown as a vector. The point of application must also be specified. A vector is illustrated by a line, the length of Static determinacy which is proportional to the magnitude on a given scale, If a body is in equilibrium under the action of coplanar and an arrow that shows the direction of the force. forces, the statics equations above must apply.
    [Show full text]
  • Minimizing Deflection and Bending Moment in a Beam with End Supports
    MINIMIZING DEFLECTION AND BENDING MOMENT IN A BEAM WITH END SUPPORTS Samir V. Amiouny John J. Bartholdi, III John H. Vande Vate School of Industrial and Systems Engineering Georgia Institute of Technology Atlanta, Georgia 30332, USA December 3, 1991; revised April 3, 2003 Abstract We give heuristics to sequence blocks on a beam, like books on a bookshelf, to minimize simultaneously the maximum deflection and the maximum bending moment of the beam. For a beam with simple supports at the ends, one heuristic places the blocks so that the maximum deflection √ is no more than 16/9 3 ≈ 1.027 times the theoretical minimum and the maximum bending moment is within 4 times the minimum. Another heuristic allows maximum deflection up to 2.054 times the theoretical minimum but restricts the maximum bending moment to within 2 times the minimum. Similar results hold for beams with fixed supports at the ends. Key words: combinatorial mechanics, heuristics, sequencing, beam, de- flection, bending moment 1 1 Introduction The limiting factor in the design of beams loaded by weights is often either the permissible deflection [6] or the bending moment [8]. For given loads, the positions at which they are placed determine the deflection and bending moment along the beam; therefore deflection and bending moment can be controlled to some extent by judicious sequencing of the loads. Unfortunately, as we shall show, it can be computationally difficult to determine the best sequence of loads on the beam; however we give fast heuristics that position the loads so that both the maximum deflection and the maximum bending moment are guaranteed to be not “too much” larger than the minimum possible.
    [Show full text]
  • Course 3 Structural Action: Trusses and Beams
    Basis of Structural Design Course 3 Structural action: trusses and beams Course notes are available for download at https://www.ct.upt.ro/studenti/cursuri/stratan/bsd.htm Arch Truss rafter tie Linear arch supporting a Relieving of support concentrated force: large spreading: adding a tie spreading reactions at supports between the supports Truss forces . Truss members connected by pins: axial forces (direct stresses) only . Supports: – one pinned, allowing free rotations due to slight change of truss shape due to loading – one roller bearing support ("simple - (C) - (C) support") - allowing free rotations and lateral movement due to + (T) loading and change in temperature . Forces in the truss: – tie is in tension (+) – rafters are in compression (-) Truss forces . If more forces are present within the length of the rafter bending stresses - - . To avoid bending stresses, + diagonal members and vertical - - - - posts can be added + + . More diagonals and posts can be added for larger spans in order to avoid bending stresses Alternative shape of a truss . For a given loading find out the shape of a linear arch (parabolic shape) . Add a tie to relieve spreading of supports . Highly unstable shape Alternative shape of a truss . Add web bracing (diagonals and struts) in order to provide stability for the pinned upper chord members . If the shape of the truss corresponds to a linear arch web members are unstressed, but they are essential for stability of the truss . Reverse bowstring arches: – advantage: longer members are in tension – disadvantage: limited headroom underneath Truss shapes . Curved shape of the arch: difficult to fabricate trusses with parallel chords .
    [Show full text]
  • Failure Theories Why Do Mechanical Components Fail? Mechanical Components Fail Because the Applied Stresses Exceeds the Material’S Strength (Too Simple)
    Failure Theories Why do mechanical components fail? Mechanical components fail because the applied stresses exceeds the material’s strength (Too simple). What kind of stresses cause failure? Under any load combination, there is always a combination of normal and shearing stresses in the material. What is the definition of Failure? Obviously fracture but in some components yielding can also be considered as failure, if yielding distorts the material in such a way that it no longer functions properly Which stress causes the material to fail? Usually ductile materials are limited by their shear strengths. While brittle materials (ductility < 5%) are limited by their tensile strengths. Stress at which point? Stress at which point? Failure Theories Load type Material Property Application of Stress Uniaxial Ductile Static Biaxial Brittle Dynamic Pure Shear Static Loading Maximum Normal Stress Dynamic Loading Modified Mohr Goodman Yield strength Gerber Maximum shear stress Soderberg Distortion energy Static Failure Theories The idea behind the various classical failure theories is that whatever is responsible for failure in the standard tensile test will also be responsible for failure under all other conditions of static loading. Ductile Material Brittle Material Characteristic Yield Stress Ultimate Stress Failure Stress Important 1. Maximum Shear Stress 1. Maximum Normal Theories 2. Maximum Octahedral Stress Shear Stress 2. Modified Mohr. Maximum Ductile Materials Shear Stress Failure occurs when the maximum shear stress in Theory the part exceeds the shear stress in a tensile test specimen (of the same material) at yield. Hence in a tensile test, S τ = y max 2 For a general state of stresses σ− σ S τ = 1 3 = y max 2 2 This leads to an hexagonal failure envelop.
    [Show full text]