page | 94 point along the beam and drawing the M/EI diagram for each of the applied loads, including the support reactions, prior to the application of any of the theorems to determine what is required. Table 8.2. Areas and Centroid of Geometric Shapes Geometric Area Centroid Shape bh ������ ������ Rectangle 12 ������ ������ 22 Triangle ������ℎ ������ 2������ 23 3 ������ℎ ������ 3������ 34 4 Parabolic 2������ℎ 3������ 5������ Spandrel 38 8 ������ℎ ������ 4������ 45 5
page | 95 Cubic 3������ℎ 2������ 3������ Spandrel 45 5 General ������ℎ ������ ������(������+1) Spandrel ������+1 ������+2 ������+2 In most cases, tangential deviation t does not equal beam deflection. The tangent drawn to the elastic curve at the wall in cantilever beams, on the other hand, is horizontal and so coincides with the neutral axis of the beam. In this scenario, the tangential deviation is equal to the beam's deflection, as shown below. Figure 8.7. The deflection at B, indicated as δB, is equal to B's departure from a tangent line through A, denoted as tB/A in the diagram above. This is because the tangent line through A coincides with the beam's neutral axis. CONJUGATE-BEAM METHOD A conjugate beam is a fictitious beam having the same length as the actual beam but with a loading equal to the actual beam's bending moment divided by its flexural stiffness, EI. The conjugate beam approach takes advantage of the similarity of the relationships between load, shear force, and bending moment, as well as curvature, slope, and deflection. Properties of Conjugate Beam
page | 96 1. The length of a conjugate beam is always equal to the length of the actual beam. 2. The load on the conjugate beam is the M/EI diagram of the loads on the actual beam. 3. A simple support for the real beam remains simple support for the conjugate beam. 4. A fixed end for the real beam becomes a free end for the conjugate beam. 5. The point of zero shear for the conjugate beam corresponds to a point of zero slope for the real beam. 6. The point of maximum moment for the conjugate beam corresponds to a point of maximum deflection for the real beam. THEOREM I The slope at any point of a beam is equal to the shear force at that point of the conjugate beam loaded with the M/EI diagram of the real beam. ������ ������ = ������ ������/������ θ = ������/������ ������' ������ ������ ������ THEOREM II The deflection at any point of a beam is equal to the moment at the corresponding point of the conjugate beam loaded with the M/EI diagram of the real beam. ������������ = ������������/������ ������ − ������������/������ ∆ = ������������/������ ������ − ������������/������ ������ ������ Supports of Conjugate Beam Knowing that the slope on the real beam is equal to the shear on conjugate beam and the deflection on real beam is equal to the moment on conjugate beam, the shear and bending moment at any point on the conjugate beam must be consistent with the slope and deflection at that point of the real beam. Take for example a real beam with fixed support; at the point of fixed support there is neither slope nor deflection, thus, the shear and moment of the corresponding conjugate beam at that point must be zero. Therefore, the conjugate of fixed support is free end.
page | 97 Table 8.3. Support for Conjugate Beams.
page | 98 Table 8.4. Real Beams and their Conjugate. Rules of Sign For a positive curvature diagram, where there is a positive ordinate of the ME/I diagram, the load in the conjugate should point in the positive y direction (upward) and vice versa (see the figures below).
page | 99 Figure 8.8. Real beam diagram for positive moment Figure 8.9. Conjugate beam diagram for deflection and slope analysis If the convention stated for positive curvature diagrams is followed, then a positive shear force in the conjugate beam equals the positive slope in the real beam, and a positive moment in the conjugate beam equals a positive deflection (upward movement) of the real beam. This is shown in Figure 8.10. Figure 8.10. Procedure for Analysis by Conjugate Beam Method ● Draw the curvature diagram for the real beam. ● Draw the conjugate beam for the real beam. The conjugate beam has the same length as the real beam. A rotation at any point in the real beam corresponds to a shear force at the same point in the conjugate beam, and a displacement at any point in the real beam corresponds to a moment in the conjugate beam.
page | 100 ● Apply the curvature diagram of the real beam as a distributed load on the conjugate beam. ● Using the equations of static equilibrium, determine the reactions at the supports of the conjugate beam. ● Determine the shear force and moment at the sections of interest in the conjugate beam. These shear forces and moments are equal to the slope and deflection, respectively, in the real beam. Positive shear in the conjugate beam implies a counter-clockwise slope in the real beam, while a positive moment denotes an upward deflection in the real beam. SAMPLE PROBLEM 8.1strength of materials: deflection in beam — Solution:
page | 101 ( )������ =1 • ������ ������/������������������ ������������������������ ������ ������������ ������ = 1 ⎣⎡ 1 (6)(5400)(6) − 1 (8)(6400)(6)⎦⎤ • 123 (1.5������106)(60) 2 3 ������/������ ������������/������ =− 0. 09984 ������������. (������������������������������������) SAMPLE PROBLEM 8.2strength of materials: deflection in beam — Solution: Σ������ = 0 Σ������ = 0 ������2 ������1 6������ = 400 + 1000(2) 1 6������ + 400 = 1000(2) 2 ������ = 400������ ������ = 600������ 1 2 ( )������������������������ ������ ������������ ������ = 1 (2)(800)( 4 ) + 1 (4)(2400)( 10 ) − 1 (2)(2000) 2 3 2 3 2
page | 102 = 11733. 33������ • ������3 (������������������������������������) ( )������������������������ ������ ������������ ������ = 1 (2)(800)( 14 ) + 1 (4)(2400)( 8 ) − 1 (2)(2000) 2 3 2 3 2 = 9866. 67������ • ������3 (������������������������������������) TEST YOUR KNOWLEDGEstrength of materials: exercises For the beam loaded as shown in the figure below, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. (Hint: Draw the moment diagram by parts from right to left.)
page | 103 GLOSSARY Beam - a bar that is subjected to forces or couplings in a plane that contains the bar's longitudinal section. Bearing stress - a measure of the tendency for the applied force to crush the supporting member. Cantilever beam - formed into a stiff support at one end and is free at the other Circumferential stress - acts perpendicular to the axis of a cylinder. Compressive stress - defined in the same way as tensile stress, but with negative numbers to show compression because delta L is the opposite. Deflection - the degree to which a particular structural element can be displaced with the help of a considerable amount of load. Elastic potential energy - stored in any deformed system that obeys Hooke’s law and has a displacement x from equilibrium and a force constant k. Equilibrium Position - if it experiences neither linear acceleration nor angular acceleration. Flexural strength - defined as a material's ability to resist deformation under load. Flexural/Bending Stresses - stresses caused by the bending moment. Modulus of elasticity - dependent on the material of the beam. Moment of inertia - the linear deflection of the beam at position x (∆ ). ������ Normal stress - takes place when an applied force is perpendicular to the resisting area. Oscillation - a back and forth motion of an object between two points of deformation Overhanging beam - supported by a pin and a roller support, with one or both ends of the beam extending beyond the supports. Restoring force - force in the opposite direction, Shearing stress - an action that tends to slide on an adjacent section. Slope - obtained from the first integration Sphere - regarded as the most structurally efficient shape for closed pressure vessels because internal pressure acts uniformly in all directions, resulting in uniform wall stresses.
page | 104 Statically determinate beams - ones in which the reactions of the supports can be predicted using static equilibrium equations Static indeterminacy - simply means that the solution cannot be obtained from the equilibrium equations alone. Stress - a measure of the magnitude of the forces that cause deformation. Tensile stress - happens whenever a load leads to the material stretching in the course of the load. Thermal strain - refer to the strain that a change in temperature causes in a body. Thermal stress - refer to the tension that a change in temperature causes in a body. Thin-walled pressure vessel - a versatile and dependable solution for a variety of industrial applications, from storing compressed gasses to transferring high-pressure liquids. Torque - the result of the turning force and the distance between the force's application point and the shaft's axis. Torsion - the twisting of the shaft as a result of this torsion. Transverse bending test - most frequently employed, in which a rod specimen is bent until fracture using a three point flexural test technique.
page | 106 ANSWER KEYSstrength of materials: exercises Topic 1: Simple Stress 1. A 110-foot steel rod inside a lofty tower carries a 200-pound weight at its lower end. Determine the maximum normal stress in a circular rod with a diameter of 1/4 inch and weight of the rod. Given: P = 200 lb L = 110 ft D = ¼ in. Required: σ =? ������������������ Formula: σ= ������+������ = ������������ ������ ������ ������ ������������������ Solution: ������������ = (490 ������������ )(110)( 1 ������������2 ) = 374. 3 ������������������ ������������3 144 ������������2 ������ = 200 ������������ = 4074 ������������������ ������ π (0.25)2 4 σ = 374. 3 ������������������ + 4074 ������������������ ������������������ σ = 4448. 3 ������������������ ������������������
page | 107 Topic 2: Deformation: Hooke’s Law, Statically Indeterminate Members A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of a cast iron 5 mm thick. Compute the load that will compress the combined bar a total of 0.8 mm in the length of 2 m. For steel, E = 200 GPa, and for cast iron, E = 100 GPa. Solution: δ = ������������ ������������ δ =δ ������������������������ ������������������������ = δ = 0. 80 ������������ ������������������������������ ������ (2000) δ= ������������������������ ������������������������ = 0. 80 ������������������������ ������������������������ [ 1 π(602−502)](100000) 4 ������������������������������ ������������������������ = 11000π ������ ������ (2000) δ= ������������������������������ = 0. 80 ������������������������������ [ 1 π(502)](200000) 4 ������ = 50000π ������ ������������������������������ Σ������������ = 0 ������ = ������ ������������������������ ������������������������ + ������ ������������������������������ ������ = 11000π + 50000π ������ = 61000π ������ ������ = 191. 64 ������������ (Answer)
page | 108 Topic 3: Thermal Stresses, Statically Indeterminate Members A bronze bar 3 m long with a cross sectional area of 320 ������������2 is placed between two rigid walls as shown in figure below. At a temperature of -20°C, the gap Δ = 2.5 mm. Find the temperature at which the compressive stress in the bar will be 35 MPa. Use α = 18. 0 × 10−6m/(m·°C) and E = 80 GPa. Given: ● cross sectional area of 320 ������������2 ● temperature of -20°C ● gap Δ = 2.5 mm ● α = 18. 0 × 10−6m/(m·°C) ● E = 80 GPa. Required: ● Temperature at which the compressive stress in the bar will be 35 MPa Solution: δ =δ+∆ ������ α������∆������ = σ������ + 2. 5 ������ 18. 0 × 10−6(3000)(∆������) = 35(3000) + 2. 5 80000 ∆������ = 70. 6 °������ ������ = (70. 6 − 20) °������ ������ = 50. 6 °������
page | 109 Topic 4: Torsion A steel propeller shaft is to transmit 4.5 MW at 3 Hz without exceeding a shearing stress of 50 MPa or twisting through more than 1° in a length of 26 diameters. Compute the proper diameter if G = 83 GPa. Given: ● 4.5 MW at 3 Hz ● 50 MPa or twisting through more than 1° ● length of 26 diameters ● G = 83 GPa Required: ● proper diameter Solution: ������ = ������ 2π������ ������ = 4.5(1000000) 2π(3) ������ = 238, 732. 41 ������ − ������ Based on maximum allowable shearing stress: τ= 16������ π������3 ������������������ 50 = 16(238732.410)(1000) π������3 ������ = 289. 71 ������������ Based on maximum allowable angle of twist: θ= ������������ ������������ 1°( π ) = (238732.41)(26������)(1000) 180° 1 π������4(83000) 32 ������ = 352. 08 ������������ Therefore, using the larger diameter, d = 352 mm.
page | 110 Topic 5: Stresses on Thin-walled cylinders 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 and girth joint are limited to 45 kN/mm2 and 28 kN/mm2 respectively. Given: p = 8.5 GPa longitudinal joint = 45 kN/mm2 girth joint = 28 kN/mm2 Required: Dmax= ? Solution: ������������ σ= ������ ������ 2������ =45������������/������������2( 1000 ������ ) (8.5×103������������������)������ 1 ������������ ������ ������ 2������ 45000������/������������2 = (8.5×103������������������)������ ������ ������ 2������ ������ = 10. 59 ������������ ������ ������������ σ= ������ ������ 4������ =28������������/������������2( 1000 ������ ) (8.5×103������������������)������ 1 ������������ ������ ������ 4������ 28000������/������������2 = (8.5×103������������������)������ ������ ������ 4������ ������ = 13. 88 ������������ ������ Therefore, the maximum allowable diameter is Dc = 10.59 mm
page | 111 Topic 6: Shear and Moment in Beams (inc. moving loads) 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. Required: a. The shear force and bending moment equations b. Draw the shear force and bending moment diagrams Free Body Diagram: Solution: A. As seen in the free body diagram, the uniformly distributed load has been replaced by its resultant, which is the force 150(12) = 1800 lb (area under the loading diagram) acting at the centroid of the loading diagram. The reactions at the supports B and C were also computed. Note that we consider the clockwise rotation and upward forces as positive. Σ������ = 0 ↻+ ������ (12) − 250(18) − 1800(6) = 0 ������ ������
������ = 1275 ������������ page | 112 ������ 1275 − 250 − 1800 + ������ = 0 Σ������������ = 0 ↑+ ������ @ Segment AB (0<x<6): ������������ = 775 ������������ Σ������ = 0 ↑+ Σ������ = 0 ↻+ ������ ������ − 250 − ������1 = 0 − ������1 − 250(������) = 0 ������ = − 250 ������������ ������ = − 250������ ������������ · ������������ 1 1 @ Segment BC (6<x<18): Σ������ = 0 ↑+ ������ − 250 − ������ + 1275 − 150(������ − 6) = 0 2 − 250 − ������ + 1275 − 150������ + 900 = 0 2 ������ = 1925 − 120������ ������������ 2 Σ������ = 0 ↻+ ������ − ������ − 250(������) + 1275(������ − 6) − 150(������ − 6)( ������−6 ) = 0 2 2
page | 113 − ������ − 250������ + 1275������ − 7650 − 75(������ − 6)2 = 0 2 − ������ − 250������ + 1275������ − 7650 − 75(������2 − 12������ + 36) = 0 2 − ������2 − 250������ + 1275������ − 7650 − 75������2 + 900������ + 2700 = 0 ������ = − 75������2 + 1925������ − 4950 ������������ · ������������ 2 B. Shear and Moment Diagram: Final Answer: ������ = − 250 ������������, ������ = − 250������ ������������ · ������������ 11 ������ = 1925 − 120������ ������������, 2 ������ = − 75������2 + 1925������ − 4950 ������������ · ������������ ] 2 Answer for the shear-moment diagram:
page | 114 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. Figure 6 Required: a.) maximum moment and b.) maximum shear Solution: ������ = 2500 + 9000 + 16500 + 16500 = 44500 ������������ +↑ Σ������ = 0 2500 44500(������) = 9000(5) + 16500(14) + 16500(29) ������ = 754500 44500 ������ = 16. 95505618 ≈ 16. 96 ������������ Maximum Shear: At ������ since 12.04 ft < 16.96 ft ������ +↑ Σ������ = 0 ������ ������ (35) = 44500(22. 96) ������ ������ = 1021720 35 ������ ������������ = ������������������������ = 29192 ������������ Maximum moment under 2500 wheel load;
page | 115 Since the 16500 lb load is outside the span it is disregarded form the equation. ������ = 2500 + 9000 + 16500 = 28000 ������������ +↑ Σ������ = 0 2500 28000(������) = 9000(5) + 16500(14) ������ = 276000 28000 ������ = 9. 86 ������������ +↑ Σ������������ = 0 ������ (35) = 28000(12. 57) ������ ������ = 351960 35 ������ ������ = 10056 ������������ ������ +↑ Σ������ = 0 2500
page | 116 ������ = 10056(12. 57) 2500 ������2500 ≈ 126403. 92 ������������ − ������������ Maximum moment under 9000 wheel load; Since the 16500 lb load is outside the span it is disregarded form the equation. ������ = 2500 + 9000 + 16500 = 28000 ������������ +↑ Σ������ = 0 2500 28000(������) = 9000(5) + 16500(14) ������ = 276000 28000 ������ = 9. 86 ������������
page | 117 +↑ Σ������ = 0 ������ ������ (35) = 28000(15. 07) ������ ������ = 421960 35 ������ ������ = 12056 ������������ ������ +↑ Σ������ = 0 9000 ������9000 = 12056(15. 07) − 2500(5) ������ ≈ 169183. 92 ������������ − ������������ 9000 Maximum moment under 16500 (3rd wheel from left) wheel load; ������ = 2500 + 9000 + 16500 + 16500 = 44500 ������������ +↑ Σ������ = 0 2500 44500(������) = 9000(5) + 16500(14) + 16500(29) ������ = 754500 44500 ������ = 16. 95505618 ≈ 16. 96 ������������ +↑ Σ������������ = 0
page | 118 ������ (35) = 44500(16. 02) ������ ������ = 712890 35 ������ ������������ = 20368. 28571 ≈ 20368. 29 ������������ +↑ Σ������16500 (3������������ ������ℎ������������������ ������������������������ ������������������������) =0 ������ = 20368. 29(16. 02) − 2500(14) − 9000(9) 16500 (3������������ ������ℎ������������������ ������������������������ ������������������������) ������ ≈ 210300. 01 ������������ − ������������ 16500 (3������������ ������ℎ������������������ ������������������������ ������������������������) Maximum moment under 16500 (4th wheel from left) wheel load; Since the 2500 lb and 9000 lb loads are outside the span it is disregarded from the equation. ������ = 16500 + 16500 = 33000 ������������ +↑ Σ������16500 (3������������ ������ℎ������������������ ������������������������ ������������������������) =0
page | 119 33000(������) = 16500(15) ������ = 247500 33000 ������ = 7. 5 ������������ +↑ Σ������ = 0 ������ ������ (35) = 33000(21. 25) ������ ������ = 701250 35 ������ ������������ = 20035. 71 ������������ +↓ Σ������ 16500 (2������������ ������ℎ������������������ ������������������������ ������������������������) =0 ������ = 20035. 71 (13. 75) 16500 (2������������ ������ℎ������������������ ������������������������ ������������������������) ������ ≈ 275491. 01 ������������ − ������������ 16500 (2������������ ������ℎ������������������ ������������������������ ������������������������) Final Answer a.) ������ = 29192 ������������ ������������������ b.) ������ = 275491. 01 ������������ − ������������ ������������������
page | 120 Topic 7: Flexural Stress 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. Solution: Flexural test developed: ������ = ������������ ρ ������ = ������������ = (������������/ρ)������ ������ ������ ������ ������ = ������������ = 200000(0.80/2) ρ 300 ������ ������ = 266. 67 ������������������ (Answer) ������ Minimum diameter of pulley: ������������ = ������������ ρ 400 = 200000(0.80/2) ρ ρ = 200 ������������ ������������������������������������������������, ������ = 400 ������������ (Answer)
page | 121 Topic 8: Deflection (in Beams) For the beam loaded as shown in the figure below, compute the moment of area of the M diagrams between the reactions about both the left and the right reaction. Solution: Σ������ = 0 Σ������ = 0 ������1 ������2 4������ = 400(3)(1. 5) + 500(2) 4������ = 400(3)(2. 5) + 500(2) 2 1 ������ = 700������ ������ = 1000������ 2 1 ( )������������������������ ������ ������������ ������ = 1 (4)(2800)( 4 ) − 1 (2)(1000)( 2 ) − 1 (3)(1800)( 3 ) 2 3 2 3 3 4 = 5450������ • ������3 (������������������������������������) ( )������������������������ ������ ������������ ������ = 1 (4)(2800)( 8 ) − 1 (2)(1000)( 10 ) − 1 (3)(1800)( 13 ) 2 3 2 3 3 4 = 5750������ • ������3 (������������������������������������)
page | 122 This book entitled CHAIR UP!! Elevating Engineering Prowess Through In-Depth Scrutiny and Easier Approach prepared by GUALBERTO, IAN PAUL R., LEGASPI, CRISTINE JOY L., MALABANAN, RENZ MATHEW S., MAULION, ELOISA JANE B., and ZAGALA, JAMES RYU C., in partial fulfillment of the requirements for the subject CE–402 Strength of Materials has been evaluated and recommended for acceptance and approval. ENGR. HILARION AQUINO III Strength of Materials Instructor Checked by: Leslie Jhoyce Evangelista BS Civil Engineering Graduate University of Batangas Daniel Joshua Taligatos 4th year Civil Engineering Student Batangas State University-Alangilan Prince Jeric C. Gomez 4th year Civil Engineering Student Batangas State University-Alangilan
page | 123 REFERENCES ★ Simple Stress Goodno, B. J., & Gere, J. M. (2017). Mechanics of Materials (9th ed.). Cengage Learning. Stress Analysis Manual. (1986, October). Air Force Flight Dynamics Laboratory. ★ Deformation: Hooke’s Law, Statically Indeterminate Members Libretexts. (2022). 16.1: Hooke’s Law - Stress and Strain Revisited. Physics LibreTexts. https://phys.libretexts.org/Bookshelves/College_Physics/Book%3A_College_Physic s_1e_(OpenStax)/16%3A_Oscillatory_Motion_and_Waves/16.01%3A_Hookes_Law _-_Stress_and_Strain_Revisited Solution to Problem 233 Statically Indeterminate | Strength of Materials Review at MATHalino. (n.d.). https://mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-pro blem-233-statically-indeterminate ★ Thermal Stresses, Statically Indeterminate Members Thermal Stress | Strength of Materials Review at MATHalino. (n.d.). Mathalino.com. https://mathalino.com/reviewer/mechanics-and-strength-of-materials/thermal-stress AYUSH. (2021, August 26). STRENGTH OF MATERIALS- Thermal Stresses and Strains. EDUCATIONAL STUFFS. https://www.educationalstuffs.in/thermal-stresses-and-strains/ Mishra, P. (2016, October 2). What is Thermal Stress in Strength of Materials? Mechanical Booster. https://www.mechanicalbooster.com/2016/10/what-is-thermal-stress.html Statically Indeterminate Members - NUEVA VIZCAYA STATE UNIVERSITY Bambang, Nueva Vizcaya - Studocu. (n.d.). Studocu. Retrieved May 13, 2023, from https://www.studocu.com/ph/document/nueva-vizcaya-state-university/mechanical-e ngineering/statically-indeterminate-members/35132400 ★ Torsion AMIT. (2021, October 4). STRENGTH OF MATERIALS- Torsion. EDUCATIONAL STUFFS. https://www.educationalstuffs.in/torsion/
page | 124 Torsion | Strength of Materials Review at MATHalino. (n.d.). Mathalino.com. https://mathalino.com/reviewer/mechanics-and-strength-of-materials/torsion Holmes, D. (2019). Mechanics of Materials: Torsion» Mechanics of Slender Structures | Boston University. Bu.edu. https://www.bu.edu/moss/mechanics-of-materials-torsion/ ★ Stresses on Thin-walled cylinders A. (2022, January 14). Stresses in thin-walled Pressure Vessels. Arveng Training & Engineering. https://arvengtraining.com/en/stresses-in-thin-walled-pressure-vessels/#:~:text=Thin %2Dwalled%20pressure%20vessels%20are,one%20shape%20or%20the%20other. Thin walled pressure vessel - skyscrapers - priodeep's home. (n.d.). Retrieved March 7, 2023, from https://priodeep.weebly.com/uploads/6/5/4/9/65495087/thin_walled_pressure_vessel. pdf Ibrahim, A., Ryu, Y., & Saidpour, M. (2015). Stress Analysis of Thin-Walled Pressure Vessels. Modern Mechanical Engineering, 05(01), 1–9. https://doi.org/10.4236/mme.2015.51001 ★ Shear and Moment in Beams (inc. moving loads) linsgroup. (n.d.). BEAM FORMULAS WITH SHEAR AND MOM. Retrieved May 15, 2022, from https://www.linsgroup.com/MECHANICAL_DESIGN/Beam/beam_formula.htm Pytel, A. & Kiusalaas, J. Mechanics of Materials (2nd Edition). Cengage Learning. Pytel, A. & Singer, F.L. Strength of Material (3rd edition). Gere, J. Mechanics of Materials (6th Edition). Nash, W. Schaum’s Outlines Series Strength of Materials (5th Edition). ★ Flexural Stress David Rusenko, Dan Veltri, Chris Fanini. (2007). Flexural stresses in beams. Weebly. Retrieved May 12, 2023, from https://www.google.com/url?sa=t&source=web&rct=j&url=https://priodeep.weebly.c om/uploads/6/5/4/9/65495087/flexural_stresses_in_beams.pdf&ved=2ahUKEwjmtc7 uze_-AhVDe94KHV9lDmcQFnoECEwQAQ&usg=AOvVaw2jLXz-P-PeldOTXkpp v8sR
page | 125 DE-12: Lesson 20. FLEXURAL STRESS. (n.d.). http://ecoursesonline.iasri.res.in/mod/page/view.php?id=3666 ★ Deflection (in Beams) Beam Deflections | Strength of Materials Review at MATHalino. (n.d.). https://mathalino.com/reviewer/mechanics-and-strength-of-materials/chapter-6-beam -deflections Alderliesten, R. (2018). Introduction to Aerospace Structures and Materials. Pytel, A. & Kiusalaas, J. (2011). Mechanics of Materials: Second Edition.