Determine the bending strain energy in the simply supported beam. Solve the problem two ways, (a) Apply Eq. 14–17. (b) The load w dx acting on the segment dx of the beam is displaced a distance y, where y = w(−x4 + 2Lx3 − L3x)/(24EI), the equation of the elastic curve. Hence the internal strain energy in the differential segment dx of the beam is equal to the external work, i.e.,
Prob. 14–24
Want to see the full answer?
Check out a sample textbook solutionChapter 14 Solutions
Mechanics of Materials (10th Edition)
Additional Engineering Textbook Solutions
Thinking Like an Engineer: An Active Learning Approach (4th Edition)
Management Information Systems: Managing The Digital Firm (16th Edition)
Electric Circuits. (11th Edition)
Vector Mechanics for Engineers: Statics and Dynamics
Database Concepts (8th Edition)
Elementary Surveying: An Introduction To Geomatics (15th Edition)
- The Figure shows a loading set up similar to that of the 'Bending in Beams' laboratory experiment. E = 200 GNm-2 and I = 2 x 10-9 m4. The bending moment and the radius of curvature at any point in the beam between the support points B and Care respectively: 1 kN 1 kN B A l0.125 m 1.25 m O a. (-)12.5 kNm, 2.8 m O b.(-)125 Nm, 3.2 m OC (-)2.5 kNm, 2.4 m O d.(-)125 Nm, 1.2 marrow_forwardThe Figure shows a loading set up similar to that of the 'Bending in Beams' laboratory experiment. E = 200 GNm"2 and I = 2 x 109 m“. The bending moment and the radius of curvature at any point in the beam between the support points B and C are respectively: 1 kN 1 kN B 0.125 m 1.25 m O a. (-)125 Nm, 1.2 m O b.(-)12.5 kNm, 2.8 m O C. (-)2.5 kNm, 2.4 m O d.(-)125 Nm, 3.2 marrow_forwardGive the expression for the shear force, V = V(x), and the bending moment,M = M(x), as a function of the distance, x, measured from point A.Hint: Find the expressions of shear force and bending moment in each section:AB (0< x <4), BC (4 <x <7) and CD (7< x <10)arrow_forward
- M = 2. A beam of length 10 m and of uniform rectangular section is supported at its ends and carries uniformly distributed load over the entire length. Calculate the depth of the section if the maximum permissible bending stress is 16 N/mm² and central deflection is not to exceed 10 mm. Take the value of E = 1.2 × 104 N/mm². Use the following g equations: w.L² 8 ус W.L 161 8 d = 2 W.L 8 == WL³ 5 384 ΕΙ (: W = w.L)arrow_forwardDetermine the reaction forces R1 and R2 and show the shear and bending moment diagram. Data: Point load (left-most) = 90.126 N Uniformly distributed load (80mm to 260mm) = 0.14444 N/mm Uniformly distributed load (160mm to 180mm) = 6.5 N/mmarrow_forwardneed help engineering of materialsarrow_forward
- Determine the absolute maximum normal stress (in unit of MPa) in the beam with external loadings shown below. The beam has a uniform square cross-section with lateral size a = 0.2 m. Note: (1) the shear and moment diagrams can be calculated by either section method or graphical method; (2) there is a concentrated load at point A and a bending moment at point C. 20 kN w = 20 kN/m Mc = = 80 kN · m || В A¬Q C y 2 m 2 m 2 marrow_forwardF1 F2 M B E Determine the shear force and bending moment in the beam. Use F1 = 1100 N, F2 = 900 N, M=900 N.m, and a = 1.5 m. The shear force in the section AC in N is (round to the nearest integer): N The shear force in the section CD in N is (round to the nearest integer): N The shear force in the secetion DE in N is (round to the nearest integer): N The shear force in the section EB in N is (round to the nearest integer): N The bending moment at the point C in N.m is (round to the nearest integer): N.m The bending moment at the point E in N.m is (round to the nearest integer): N.marrow_forwardFind reactions at A and Ma. Give the (signed) values of the largest shear and bending moment occurring anywhere within the beam. Draw the corresponding shear and bending moment diagram.arrow_forward
- Question 3: The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take: A = C kN/m 2 B = 1.5 m C = kN /m B В m A m 1m Solution: Equation of Equilibrium: C(A+1) kN C kNlm 0.5(A+1) m Ax 4 AM Im Fec A meter (a) (b) 2MA = 0; Fec (3 /5)A – 0.5C(A +1)° = 0 F, = 0; A, + Fec (3/5)- C(A +1) = 0 kN A, = kN kN V, = M(KN.M) m X2 v3 M1 V2 kN vl V, = kN x2 x2 м, M3 M, = v2 (C) (d)arrow_forward3.0 m 7.95 KN B -9 kN T 2.0 m 1.95 KN 1. Draw the load diagram. 2. Draw the moment diagram. 6.0 m 2º curve D -10.05 KN 2.0 m 16 KN E 2.0 m 3. Determine reaction supports at point A and B. 4. Determine the location of the point of zero shear between supports from support B. Load Diagram 20 KN V Diagram The shear diagram of beam ABCDEF with overhang at both ends is shown in the figure.arrow_forwardDraw the shear force diagram and bending moment diagram using differential relationship method between loads, shear forces, and bending momentsarrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY