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STREMA BOOK

Published by Ian Paul Gualberto, 2023-05-27 14:08:02

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page | 44 SAMPLE PROBLEM 5.1strength of materials: stresses on thin-walled cylinders— Given: Required: ● Outer diameter = 210 mm ● Maximum ● t = 2.1 mm normal stress ● p = 150 kPa within the wall Solution: The ball is in the shape of sphere thus the normal stress acting in this ball can be determined by using the equation below, σ = ������������ 4������ Note that if outer diameter is given in the problem, don’t forget to subtract the thickness to the given outside diameter before multiplying it to the internal pressure. σ= ������������−������ σ= (150×103������������)(207.9 ������������) 4������ 4(2.1 ������������) σ= (150×103������������)(210 ������������−2.1 ������������) σ = 3712500 ������������ 4(2.1 ������������) σ = 3. 7125 ������������������ ������������������


page | 45 TEST YOUR KNOWLEDGEstrength of materials: exercises A cylindrical tank has an internal pressure of 8.5 GPa. As shown in the figure below, determine its maximum diameter if the longitudinal joint (red line) and girth joint (blue line) are limited to 45 kN/mm2 and 28 kN/mm2 respectively. Answer: ������ = 10. 59 ������������, ������ = 13. 18 ������������ ������ ������


page | 47 SstrHengEthAofRmaAterNialDs: cMhaOptMer EsixN—T IN BEAMS (INC. MOVING LOADS) Introduction The primary problem in strength of materials is to determine the relations between the stresses and deformations caused by loads applied to structures. The study of bending loads is complicated considering the loading effects vary from section to section of the beam. These loading effects take the form of a shearing force and a bending moment which is referred to as shear and moment. This module encompasses the variation in shear and bending moment in beams subjected to various combinations of loadings under different conditions of support, particularly the determination of the maximum values of shear and moment. A. DEFINING A BEAM A beam is a bar that is subjected to forces or couplings in a plane that contains the bar's longitudinal section. A beam can be determinate or indeterminate depending on its determinacy. The internal force system in a beam consists of a shear force and a bending moment operating on the bar's cross section. The shear force and bending moment of a beam normally fluctuate continuously over its length. Internal forces create two types of stresses on a beam's transverse section: (1) normal stress from the bending moment, and (2) shear stress from the shear force. For the computation of stresses and deformations, knowing the distribution of shear force and bending moment in a beam is critical.


page | 48 Statically Determinate Beams VS Statically Indeterminate Beams Statically determinate beams are ones in which the reactions of the supports can be predicted using static equilibrium equations while the beam is considered to be statically indeterminate if the number of reactions applied to it exceeds the number of equations in static equilibrium. The static equations must be augmented by equations based on the beam's elastic deformations in order to solve the beam's reactions. Supports and Loads Beams are classified according to their supports. A simply supported beam is shown in Fig. 6.1 (a). The pin support prevents displacement but not rotation of the beams' ends. A pin connection is referred to as a roller support.is free to move parallel to the beam's axis; only the transverse displacement is suppressed by this sort of support. Figure 6.1. (a) Statically Determinate Beams As shown in Fig. 6.1 (b), a cantilever beam is formed into a stiff support at one end and is free at the other (b). The built-in support prevents the end of the beam from being displaced or rotated. P in Fig. 6.1 (a) is an example of a concentrated load. A dispersed load, on the other hand, is applied over a finite area. A line load can be used to approximate a distributed load that acts on a relatively small region..


page | 49 Figure 6.1. (b) Statically Determinate Beams An overhanging beam is supported by a pin and a roller support, with one or both ends of the beam extending beyond the supports, as shown in Fig. 6.1 (c). Figure 6.1. (c) Statically Determinate Beams The force per unit length (lb/ft, N/m, etc.) of this loading is written as w. The load distribution may be uniform, as shown in Fig. 6.1 (b), or it may vary with distance along the beam, as shown in Fig. 6.1 (c). The weight of the beam is an example of distributed loading, but it is usually insignificant in comparison to the loads imposed on the beam.


page | 50 Note: Because the support reactions can be determined from the equilibrium equations, the three types of beams are statically determinate. Other sorts of beams are shown in Figure 4.2. These beams are over supported in that each one contains at least one additional reaction than is required for support. Because such beams are statically indeterminate, the inclusion of redundant supports necessitates the employment of extra equations derived from the beam's deformation. Chapter 9 will cover the study of statically indeterminate beams. Figure. 6.2. Statically Indeterminate Beams B. SHEAR-MOMENT EQUATION AND SHEAR-MOMENT DIAGRAM ★ The shear force V and the bending moment M at each cross section of the beam are determined by the beam analysis. ★ The couple M is called the resisting moment or moment and the force V is called the resisting shear or shear. ★ Draw a free-body diagram that exposes these forces and then compute the forces using equilibrium equations to determine the internal force system acting at a specific portion of a beam. ★ To calculate V and M in terms of the distance x measured along the beam. The shear force and bending moment diagrams for the beam are obtained by scaling these formulas.


page | 51 ★ The shear force and bending moment diagrams are useful visual references to internal forces in a beam; they identify the highest values of V and M in particular. a. Procedure for determining shear force and bending moment diagrams ▪ Compute the support reactions from the free-body diagram (FBD) of the entire beam. ▪ Divide the beam into segments so that the loading within each segment is continuous. Thus, the end-points of the segments are discontinuities of loading, including concentrated loads and couples. Perform the following steps for each segment of the beam: ▪ Introduce an imaginary cutting plane within the segment, located at a distance x from the left end of the beam, that cuts the beam into two parts. ▪ Draw a FBD for the part of the beam lying either to the left or to the right of the cutting plane, whichever is more convenient. At the cut section, show V and M acting in their positive directions. ▪ Determine the expressions for V and M from the equilibrium equations obtainable from the FBD. These expressions, which are usually functions of x, are the shear force and bending moment equations for the segment. ▪ Plot the expressions for V and M for the segment. It is visually desirable to draw the V-diagram below the FBD of the entire beam, and then draw the M- diagram below the V-diagram. The bending moment and shear force diagrams of the beam are composites of the V and M diagrams of the segments. These diagrams are usually discontinuous, or have discontinuous slopes. At the end-points of the segments due to discontinuities in loading. b. Properties of Shear and Moment Diagrams The following are some important properties of shear and moment diagrams: 1. The area of the shear diagram to the left or to the right of the section is equal to the moment at that section.


page | 52 2. The slope of the moment diagram at a given point is the shear at that point. 3. The slope of the shear diagram at a given point equals the load at that point. 4. The maximum moment occurs at the point of zero shears. This is in reference to property number 2, that when the shear (also the slope of the moment diagram) is zero, the tangent drawn to the moment diagram is horizontal. 5. When the shear diagram is increasing, the moment diagram is concave upward. 6. When the shear diagram is decreasing, the moment diagram is concave downward. c. Sign Convention The customary sign conventions for shearing force and bending moment are represented by the figures below. A force that tends to bend the beam downward is said to produce a positive bending moment. A force that tends to shear the left portion of the beam upward with respect to the right portion is said to produce a positive shearing force. Figure 6.3. Sign Convention for External Loads; Shear Force and Bending Moment An easier way of determining the sign of the bending moment at any section is that upward forces always cause positive bending moments regardless of whether they act to the left or to the right of the exploratory section.


page | 53 C. Reference while Solving for Shear-Moment Equation and Shear-Moment Diagram Figure. 6.4.1 Reference for Uniformly Distributed Load Figure. 6.4.2 Reference for Uniform Load Partially Distributed


page | 54 Figure. 6.4.3 Reference for Uniform Load Partially Distributed at One End Figure. 6.4.4. Reference for Uniform Load Partially Distributed at Each End Figure. 6.4.5. Reference for Load Increasing Uniformly at One End


page | 55 Figure. 6.4.6. Reference for Load Increasing Uniformly to Center Figure. 6.4.7. Reference for Concentrated Load at Center Figure. 6.4.8. Reference for Concentrated Load at Any Point


page | 56 Figure. 6.4.9. Reference for Two Equal Concentrated Loads Symmetrically Placed Figure. 6.4.10. Reference for Two Equal Concentrated Loads Unsymmetrical Placed Figure. 6.4.11. Reference for Two Unequal Concentrated Loads Unsymmetrical Placed


page | 57 Figure. 6.4.12. Reference for Uniformly Distributed Load Figure. 6.4.13. Reference for Concentrated Load at Free End Figure. 6.4.14. Reference for Concentrated Load at Any Point


page | 58 Fig. 6.4.15. Reference for Beam Fixed at One End,Supported at Other – Uniformly Distributed Load Figure. 6.4.16. Reference for Beam Fixed at One End, Supported at Other – Concentrated Load at Center Figure. 6.4.17. Reference for Beam Fixed at One End, Supported at Other – Concentrated Load at Any Point


page | 59 Figure. 6.4.18. Beam Overhanging One Support – Uniformly Distributed Load Figure. 6.4.19. Beam Overhanging One Support – Uniformly Distributed Load on Overhang Figure. 6.4.20. Beam Overhanging One Support – Concentrated Load at End of Overhang


page | 60 Figure. 6.4.21. Beam Overhanging One Support – Concentrated Load at Any Point between Supports Figure. 6.4.22. Beam Overhanging Both Support – Unequal Overhangs – Uniformly Distributed Load Figure. 6.4.23. Beam Fixed at Both Ends – Uniformly Distributed Load


page | 61 Figure. 6.4.24. Beam Fixed at Both Ends – Concentrated Load at Center Figure. 6.4.25. Beam Fixed at Both Ends – Concentrated Load at Any Point Figure. 6.4.26. Continuous Beam – Two Equal Spans – Uniform Load on One Span


page | 62 Figure. 6.4.27. Continuous Beam – Two Equal Spans – Concentrated Load at Center of One Span Figure. 6.4.28. Continuous Beam – Two Equal Spans – Concentrated Load at Any Point Figure. 6.4.29. Continuous Beam – Two Equal Spans – Uniformly Distributed Load


page | 63 Figure. 6.4.30. Continuous Beam – Two Equal Spans – Two Equal Concentrated Loads Symmetrically Placed Figure. 6.4.31. Continuous Beam – Two Unequal Spans – Uniformly Distributed Load Figure. 6.4.32. Continuous Beam – Two Unequal Spans – Concentrated Load on Each Span Symmetrically Placed


page | 64 SAMPLE PROBLEM 6.1strength of materials: shear and moment in beams (inc. moving loads) — STEP 1. Determine what is being asked in the problem. REQUIRED: The shear and moment equations and create the diagram STEP 2. Proceed with the solution. SOLUTION: The computation for the support reactions must be done first; Σ������ = 0 Σ������ = 0 ������ ������ -������ (7) + 7(5) + 14(2) = 0 -������ (7) + 14(5) + 7(2) = 0 ������ ������ ������������ = 9 kN ������������ = 12 kN After computing for the support reactions, create the load diagram which includes all the loads present in the beam. Also include the support reactions. Additionally, divide the beam into segments with cutting planes.


page | 65 In each segment, draw a free body diagram for the part of the beam and determine the equations for V and M from the equilibrium equations from the free body diagram. Segment 1 (0 ������ < ������ < 2 ������) Σ������������ = 0 Σ������ = 0 9 - ������ = 0 9(������) - ������ = 0 1 1 ������ = 9 kN ������ = 9������ kN·m 1 1 Segment 2 (2 ������ < ������ < 5������) Σ������������ = 0 Σ������ = 0 9 - 7 - ������ = 0 9(������) - 7(������ − 2) − ������ = 0 2 2 ������2 = 2 kN ������2 = 9������ - 7������ + 14 ������ = 2������ + 14 kN·m 2


page | 66 Segment 3 (5 ������ < ������ < 7 ������) Σ������������ = 0 Σ������ = 0 9 - 7 - 14 - ������3 = 0 9(������)- 7(������ − 2) − 14(������ − 5) − ������3 = ������ = -12 kN 0 ������ = 9������ - 7������ + 14 - 14������ + 70 3 3 ������ = -12������ + 84 kN·m 3 STEP 3. Create the shear and moment diagram. After obtaining the shear and moment equations, plot the expressions V and M for the segment to obtain the diagram.


page | 67 SAMPLE PROBLEM 6.2strength of materials: shear and moment in beams (inc. moving loads) — Step 1: Solve for the value of reactions ∑������ = 0 ∑������ = 0 ������ ������ ������ = 24 ������������ ������ (7) − 24(5) − 28(2) = 0 ������ ������ ������ (7) − 28 (5) − 14 (2) = 0 ������ ������ = 18 ������������ ������ Step 2: Make FBD with cuts.


page | 68 Step 3: Determine V and M for each beam segment (AB, BC, and CD) SEGMENT AB SEGMENT BC ↑ + ∑������ = 0 ↑ + ∑������ = 0 ������ ������ 18 – V = 0 18 – 14 – V = 0 V = 18 kN V = 4 kN ↻ + ∑������ = 0 ↻ + ∑������ = 0 ������ ������ Fig. 5.3 -18x + 14(x-2) + M= 0 M = 4x + 28 kN•m -18x + M = 0 M = 18x kN•m SEGMENT CD ↑ + ∑������ = 0 ������ 18 – 14 – 28 – V = 0 V = -24 kN ↻ + ∑������ = 0 ������ -18x – 14(x-2) + 28 (x-5) + M = 0 M = -24x + 168 kN•m Step 4: Compute for the values that will be plotted in the diagram. when x = 0 and x = 2 when x = 2 and x = 5 V1 = 18 kN ; V1 = 18 kN V2 = 4 kN ; V2 = 4 kN M1 = 0 ; M1 = 36 kN•m M2 = 36 kN•m ; M2 = 48 kN•m


page | 69 when x = 5 and x = 7 V3 = -24 kN ; V3 = -24 kN M3 = 48 kN•m ; M3 = 0 Step 5: Make V and M Diagram


page | 70 GRAPHICAL METHOD FOR DRAWING SHEAR- MOMENT DIAGRAMS Relations between the loading, shear force, and bending moment can be derived from the equilibrium equations. These relations allow us to plot the shear force diagram directly from the load diagram, and then create the bending moment diagram from the shear force diagram. This technique, called graphical method, permits us to draw the shear force and bending moment diagrams without deriving equations for V and M. STEPS TO DETERMINE SHEAR FORCE AND BENDING MOMENT DIAGRAMS USING THE GRAPHICAL METHOD The following steps delineates the procedure for determining shear force and bending moment diagrams by the graphical method: 1. Compute for the support reactions from the free body diagram of the entire beam. 2. Create the load diagram of the beam showing the values of the loads, also include the support reactions. Refer to the sign conventions shown in Table 1 to determine the appropriate sign of each load. 3. Work from left to right, create the V and M diagrams for each segment of the beam. 4. When the right end of the beam is reached, check to see whether the computed values of V and M are consistent with the end conditions. If they are not, an error is made in the computations. The use of a graphical method may appear to be more inconvenient than plotting the shear force and bending moment equations. But, with thorough understanding you will find this method not only faster but also more accurate to avoid numerical errors since it has a self-checking nature of the computations.


page | 71 SAMPLE PROBLEM 6.3strength of materials: shear and moment in beams (inc. moving loads) — STEP 1. Determine what is being asked in the problem. REQUIRED: Draw the shear and moment diagrams STEP 2. Proceed with the solution. SOLUTION: The computation for the support reactions must be done first; Σ������ = 0 Σ������������ = 0 Cv + ������ - 100 - 60(5) = 0 ������ ������ -������ (5) + 100(7) + 60(5)( 5 ) = 0 2 Cv + 290 - 100 - 60(5) = 0 ������ Cv = 110 lb ������ = 290 lb ������


page | 72 After computing for the support reactions, create the load diagram which includes all the loads present in the beam. Also include the support reactions. Work from left to right, create the V and M diagrams for each segment of the beam. Shear Diagram ������ = 0 lb ������������ ������ = 0 - 100 = -100 lb ������������


page | 73 ������ = -100 lb ������������ ������������������ = -100 + 290 = 190 lb ������ = 190 - 60(5) = -110 lb ������������ ������ = -110 + 110 = 0 lb ������������ Moment Diagram ������ = 0 lb·ft ������ ������ = 0 + (-100)(2) = -200 lb·ft ������ To solve for x: 190 = 110 ������ 5−������ x = 3.1667 ������ = -200 + 1 (3.1667)(190) = 100.8365 lb·ft 2 ������ ������ = 100.8365 - 1 (5-3.1667)(110) = 0 lb·ft 2 ������ Moving Loads A moving vehicle across a beam or girder constitutes a system of concentrated loads at fixed distances from one another. For beams with concentrated loads, the maximum bending moment occurs under one of the loads. Thus, the problem here is to determine the bending moment under each load when each load is in a position that can cause a maximum moment to occur. We know that: ● The maximum moment occurs at a point of zero shears. ● For beams loaded with concentrated loads, the point of zero shears usually occurs under a concentrated load and so the maximum moment. ● Beams and girders such as in a bridge or an overhead crane are subject to moving concentrated loads, which are at fixed distance with each other. ● The problem here is to determine the moment under each load when each load is in a position to cause a maximum moment. The largest value of these moments governs the design of the beam.


page | 74 Shown in figure , ������ , ������ , ������ , and ������ represent a system of loads at fixed distances ������, 123 4 ������, and ������ from one another. Additionally, the loads move as a unit across the simply supported beam with span ������. Let us locate the position of ������ when the bending moment 2 under this load is maximum. Denoting the resultant of the loads on the span by ������ and its position from ������ by ������, the value of the left reaction is 2 ������ = ������ (������ − ������ − ������) ������ 1 The bending moment under ������ will be 2 [������ = (Σ������) ] ������ = ������ (������ − ������ − ������)(������) - ������ ������ ������ ������ 2 1 Set the derivative of ������ with respect to ������ equal to zero in order to compute for the 2 value ������ which will give maximum ������ : 2 ������������ = ������ (������ − ������ − 2������) = 0 ������ 2 ������������ from which ������ = ������ - ������ (equation 1) 2 2 Since the derivative of all terms of the form ������ ������ with respect to ������ will be zero he 1 value of ������ is independent of the number of loads to the left of ������ . 2 The equation 1 can be expressed in terms of the following rule: The bending moment that is under a particular load is a maximum if the center of the beam is midway between that load and the resultant of all loads then on the span. Applying this rule, we locate the position of each load when the moment at that load is a maximum and compute the value of each such maximum moment. The maximum shearing force occurs at and is equal to the maximum reaction.


page | 75 SAMPLE PROBLEM 6.3strength of materials: shear and moment in beams (inc. moving loads) — STEP 1. Determine what is being asked in the problem. REQUIRED: The maximum moment and maximum shearing force STEP 2. Proceed with the solution. SOLUTION: To be able to analyze the problem properly, draw a free body diagram. And solve for the resultant of the loads and x. ������ = 15 + 25 = 40 ������������ x������ = 4(15) x(40) = 4(15) x = 1.5 m After that, the maximum moment under each wheel will be computed. Maximum moment under 15 kN wheel:


page | 76 Σ������ = 0 ������ 2 ������ (8) = 2.75(40) 1 ������ = 13.75 ������������ 1 ������ = 2.75������ ������������ ������ℎ������ ������������������������ ������������ 15 ������������ 1 ������ = 2.75(13.75) ������������ ������ℎ������ ������������������������ ������������ 15 ������������ ������ = 37.8125 ������������·������ ������������ ������ℎ������ ������������������������ ������������ 15 ������������ Maximum moment under 25 kN wheel: Σ������ = 0 ������ 1 ������ (8) = 3.25(40) 2 ������ = 16.25 ������������ 2 ������ = 3.25������ ������������ ������ℎ������ ������������������ℎ������ ������������ 25 ������������ 2 ������ = 3.25(16.25) ������������ ������ℎ������ ������������������ℎ������ ������������ 25 ������������ ������ = 52.8125 ������������·������ ������������ ������ℎ������ ������������������ℎ������ ������������ 25 ������������


page | 77 After computing for the maximum moment under each wheel, choose between the obtained values that will be considered as the maximum moment. Next is to compute the maximum shear force that will occur in the beam. The maximum shear will occur when the 25 kN is over a support: Σ������ = 0 ������ 1 ������ (8) = 6.5(40) 2 ������ = 32.5 ������������ 2 Thus, the maximum moment is 52.8125 ������������·������ and the maximum shearing force is 32.5 ������������.


page | 78 TEST YOUR KNOWLEDGEstrength of materials: exercises 1. The overhanging beam ABC carries a concentrated load of 250 lb and a uniformly distributed load 0f 150 lb/ft. (a) Derive the shear force and bending moment equations; and (b) draw the shear force and bending moment diagrams. Neglect the weight of the beam. Figure 2 Answer: ������ = − 250 ������������, ������ = − 250������ ������������ · ������������ 11 ������ = 1925 − 120������ ������������, ������ = − 75������2 + 1925������ − 4950 ������������ · ������������ ] 22 Answer for the shear-moment diagram:


page | 79 2. A four-wheeler truck was about to cross a 35-m span bridge. Its axle loads of 2500lb, 9000 lb, 16500 lb and 16500 lbs are separated by a distance of 5 ft, 9 ft and 15 ft respectively. Determine the maximum moment and maximum shear developed in the span. Answer: a.) ������ = 29192 ������������ ������������������ b.) ������ = 275491. 01 ������������ − ������������ ] ������������������


page | 81 FLEXURAL STRESSstrength of materials: chapter seven— Introduction Flexural strength, also known as modulus of rupture, bend strength, or fracture strength, a mechanical parameter for brittle material, is defined as a material's ability to resist deformation under load. The transverse bending test is most frequently employed, in which a rod specimen having either a circular or rectangular cross-section is bent until fracture using a three point flexural test technique. The flexural strength represents the highest stress experienced within the material at its moment of rupture. It is measured in terms of stress, here given the symbol σ. Stresses caused by the bending moment are known as flexural or bending stresses. Consider a beam to be loaded as shown: Consider a fiber at a distance from the neutral axis, because of the beam's curvature, as the effect of bending moment, the fiber is stretched by an amount of. Since the curvature of the beam is very small, bcd and Oba are considered as similar triangles. The strain on this fiber is ε= ������������ = ������ ������������ ρ


page | 82 By Hooke's law, ε = σ then, ������ σ = ������ ; σ = ������ ������ ������ ρ ρ Which means that the stress is proportional to the distance from the neutral axis. Considering a differential area at a distance from N.A., the force acting over the area is ������������ = ������ ������������ = ������ ������������������ = ������ ������������������ ρ ρ ������ The resultant of all the elemental moments about N.A. must be equal to the bending moment on the section. ������ = ∫ ������������ = ∫ ������������������ = ∫ ������( ������ ������������������) ρ ������ = ������ ∫ ������2������������ ρ ������������������ ∫ ������2������������ = ������, ������ℎ������������ ������ = ������������ ������������ ρ = ������������ ρ ������ ������������������������������������������������������������������������ ρ = ������������ ������ ������ ������������ = ������������ ������������ ������ ������ℎ������������ ������ = ������������ ������ ������


page | 83 ������������������ (������ ) = ������������ ������ ������ ������������������ The bending stress in beam due to curvature is ������ = ������������ = ������������ ������ ������ ρ ������ ������ ������ = ������������ ρ ������ The beam curvature is: ������ = 1 ρ Where ρ is the radius of curvature of the beam in mm (in), is the bending moment in N·mm (lb·in), ������ is the flexural stress in MPa (psi), I is the centroidal moment of inertia in mm4 ������ (in4 ), and c is the distance from the neutral axis to the outermost fiber in mm (in). ������������������������ = ������������ = ������ ������ ������/������ The ratio ������ is called the section modulus and is usually denoted by units of mm3 (in3). The ������ maximum bending stress may then be written as (������ ) = ������ ������ ������ ������������������ This form is convenient because the values are available in handbooks for a wide range of standard structural shapes.


page | 84 SAMPLE PROBLEM 7.1strength of materials: shear and moment in beams (inc. moving loads) — Solution: Σ������ = 0 ������2 12������ = 9(2000) 1 ������ = 1500 ������������ 1 Σ������ = 0 ������1 12������ = 3(2000) => ������ = 500 ������������ 22 Maximum fiber stress: (������ ) = ������������ = 4500(12)(2) ������ ������ ������������������ 2(43) 12 (������ ) = 10, 125 ������������������ (Answer) ������ ������������������ Stress in a fiber located 0.5 in from the top of the beam at midspan: ������ = 4500 ������ = 3000(12)(1.5) 9 ������ ������ 2(43) 6 12 ������ = 3000 ������������. ������������ ������ = 5, 602. 50 ������������������ ������ ������ ������ = ������������ (Answer) ������ ������


page | 85 TEST YOUR KNOWLEDGEstrength of materials: exercises A high strength steel band saw, 20 mm wide by 0.80 mm thick, runs over pulleys 600 mm in diameter. What maximum flexural stress is developed? What minimum diameter pulleys can be used without exceeding a flexural stress of 400 MPa? Assume E = 200 GPa. Answer: d= 400 m


page | 87 DEFLECTION IN BEAMstrength of materials: chapter eight— Introduction Deflection is the degree to which a particular structural element can be displaced with the help of a considerable amount of load. It can also be referred to as the angle or distance. The distance of deflection of a member under a load is directly related to the slope of the deflected shape of the body under that load. It can be computed by integrating the function which is used to describe the slope of the member under that load. Beam is a long piece of a body that is capable of holding the load by resisting the bending. The beam can be bent or moved away from its original position. This distance at each point along the member is the representation of the deflection. The deflection of the beam in a particular direction when force is applied to it is known as Beam deflection. Excessive beam deflection could cause damage to the other parts of the structure, becomes unsafe to the user, and will fail to meet its intended function, even if stresses in the beam are low and risk of failure is not present. As a result, building rules restrict a beam's maximum deflection to around 1/360th of its span. The deflection of a beam from its original unloaded location is used to describe its deformation. The deflection is calculated by comparing the original neutral surface to the distorted beam's neutral surface. The elastic curve of the beam is the configuration assumed by the deformed neutral surface. Figure 8.1 shows the beam in its original undeformed state and Figure 8.2 shows the beam in its deformed state. Figure 8.1. Undeformed State of Beam Figure 8.2. Deformed State of Beam


page | 88 The deflection of the beam is specified by the displacement y. It will frequently be necessary to calculate the deflection y for each x value along the beam. The elastic curve or deflection curve of the beam is this relationship. It can be calculated using a variety of analytical approaches. The differential equation that relates the deflection to the bending moment is their common ground. The solution to this equation is difficult because the bending moment is frequently a discontinuous function, requiring piecewise integration of the equations. Deflection analysis methods are essentially different approaches to solving this differential problem. In this chapter, we look at four such methods: DOUBLE-INTEGRATION METHOD Deflection by double integration is also known as deflection by the direct or constant integration method. The deflection of a beam is calculated by integrating the differential equation of the elastic curve of a beam twice and determining the constants of integration using boundary conditions. The slope is obtained from the first integration, while the deflection is obtained from the second integration. When the applied loading is consistent, this strategy works well. Consider an undeflected beam (Fig. 8.3). The beam will deflect (Fig. 8.4) due to bending forces if loads are applied perpendicularly. Let y represent the vertical linear deflection of the beam in relation to the longitudinal axis, and x represent the distance from the origin. One feature of the deflected shape of the beam is that it is nearly flat; thus, the slope of the curve at (x, y) is quite modest. We can express the first derivative as follows by definition: ������������������θ = ������������ ������������ The tangent of a tiny angle is almost equal to the angle itself since the slope is modest. θ= ������������ ������������ Let's get the derivative of this expression: (1) ������θ = ������2������ ������������ ������������2


page | 89 We've established a relationship between x, y, and θ. We also have to account for the length of the beam changing due to bending. Consider the following differential arc ds with a central angle dθ. If ρ is the radius, then the arc length formula is as follows: ������������ = ������������θ We can assert that ds = dx because ds is so small. ������������ = ������������θ (2) 1 = ������ ������ ������������ We have obtained the following from the derivation of the flexure formula: (3) 1 = ������ ������ ������������ The internal moment of the beam at location x is M, the modulus of elasticity is E, and the plane section's moment of inertia is I. It's worth noting that (2) and (3) can be combined to form: ������θ = ������ ������������ ������������ Let's simplify the equation by equating (1) and (4): ������2������ = ������ ������������2 ������������ The equation below is what we use to express the deflection of the beam with respect to position x. ( )������������������2������ ������������2 = ������ where: ● E is the modulus of elasticity which is dependent on the material of the beam ● I is the moment of inertia of the beam plane section at position x y is the linear deflection of the beam at position x (∆ ) ������ ● x is the position of the section from the origin


page | 90 We refer to EI as flexural rigidity in structural analysis. As you may have guessed, this equation is a differential equations issue. We use integration to solve for the linear and angular deflection at any point on the beam. If we integrate the equation once, we will get the following: ������������ ������������ = ∫ ������ + ������ = ������������θ ������������ 1 ������ We may use this expression to get the rotation (θ) of any point in the beam because dy/dx is the slope. To put it another way, integrating once will give you an expression for the angular deflection of the beam at location x. We can reach the following result by reintegrating the equation: ������������������ = ∫∫ ������ + ������ ������ + ������ = ������������∆ 12 ������ You can get an expression for the linear deflection (Δ) of the beam at location x by integrating twice from the second-order differential equation. This is why this method of determining deflections is known as the double integration method. It's worth noting that the solution is completely analytical. It's worth noting that this method only works if the deflections are minor; if the deflections are somewhat significant, the procedure may fail due to the assumptions we made during the derivation. MOMENT DIAGRAM BY PARTS The moment-area method of finding the deflection of a beam will demand the accurate computation of the area of a moment diagram, as well as the moment of such an area about any axis. To pave its way, this section will deal on how to draw moment diagrams by parts and to calculate the moment of such diagrams about a specified axis. Basic Principles 1. The bending moment caused by all forces to the left or to the right of any section is equal to the respective algebraic sum of the bending moments at that section caused by each load acting separately. ������ = (Σ������) = (Σ������) ������ ������ 2. The moment of a load about a specified axis is always defined by the equation of a spandrel


page | 91 ������ = ������������������ where n is the degree of power of x. The graph of the above equation is as shown below in figure 8.6, Figure 8.6. and the area and location of the centroid are defined as follows. ������ = 1 ������ℎ ������ = 1 ������ ������+1 ������+2 ������ Cantilever Loadings A = area of moment diagram ������ = moment about a section of distance x ������ ������ = location of centroid Degree = degree power of the moment diagram Table 8.1. Load Diagram A ������ ������ Degree − ������������ zero Couple or Moment ������ 1 ������ Load 2 − ������ Concentrated Load − 1 ������������2 − ������������ 1 ������ first 2 3


page | 92 Uniformly − 1 ������ ������3 − 1 ������ ������3 1 ������ second Distributed Load 6 2 4 ������ ������ Uniformly Varying Load − 1 ������ ������3 − ������ ������2 1 ������ third 24 5 ������ ������ 6������ AREA-MOMENT METHOD Another method of determining the slopes and deflections in beams is the area-moment method, which involves the area of the moment diagram. The 'moment area theorems' or 'Mohr's theorems' are the foundations of the moment area method. The first is concerned with the change in slope between any two places on the beam, while the second is concerned with the deflection at a specific position. After the following graphic, which will be used as a reference, the two theorems will be stated. It shows a simple beam deflected by random loading and the bending moment diagram that goes with it. Figure 8.6. Moment Diagram


page | 93 THEOREM I The change in slope between the tangents drawn to the elastic curve at any two points A and B is equal to the product of 1/EI multiplied by the area of the moment diagram between these two points. ( )θ = 1 ������������ ������������ ������������������������ ������������ THEOREM II The deviation of any point B relative to the tangent drawn to the elastic curve at any other point A, in a direction perpendicular to the original position of the beam, is equal to the product of 1/EI multiplied by the moment of an area about B of that part of the moment diagram between points A and B. ( )������ =1 ( )������ =1 ������/������������������ ������/������������������ ������������������������ • ������ ������������������������ • ������ ������������ ������ ������������ ������ Rules of Sign Procedure for Analysis by Area-Moment Method ● Sketch the free-body diagram of the beam. ● Draw the M/EI diagram of the beam. This will look like the conventional bending moment diagram of the beam if the beam is prismatic (i.e. of the same cross section for its entire length). ● To determine the slope at any point, find the angle between a tangent passing the point and a tangent passing through another point on the deflected curve, divide the M/EI diagram into simple geometric shapes, and then apply the first moment-area theorem. To determine the deflection or a tangential deviation of any point along the beam, apply the second moment-area theorem. ● In cases where the configuration of the M/EI diagram is such that it cannot be divided into simple shapes with known areas and centroids, it is preferable to draw the M/EI diagram by parts. This entails introducing a fixed support at any convenient


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