- Problem 2 A P L 2 W B Z Tw Y B H L 2 Figure 2 Not True To Scale A Simply Supported Beam Of Total Length L 4 M Is Su 1 (52.23 KiB) Viewed 28 times
Q2.1: Calculate the statically equivalent replacement force for the concentrated load Pw (pick in table 2.1) Q2.2: Calculate Ax (pick in table 2.1) Q2.3: Calculate Ay (pick in table 2.1) Q2.4: Calculate By (pick in table 2.1) 0 20.0 kN 40.0 kN a. d. g. Table 2.1 Statics and internal forces b. e. h. Ax 10.0 kN 25.0 kN 50.0 kN Ay Free-body diagram C. f. i. P 15.0 kN 30.0 kN 75.0 KN W Cross-sectional constants Q2.7: Calculate the moment of inertia lange) of a single flange around it's local bending axis, z' (pick in table 2.2) in mm4 Q2.8: Calculate the moment of inertia (web) of the beam web around the global z-axis (pick in table 2.2) in mm4 Q2.9: Calculate the moment of inertia of the entire beam cross-section around the global z-axis (pick in table 2.2) in mm4 Q2.10: Calculate the max. first order moment Q for a section above the global z-axis (pick in table 2.2) in mm³ By Q2.5 Sketch the internal shear force diagram for the beam in the answer sheets. Clearly mark the values Ay. By.w and P Q2.6 Sketch the internal bending moment diagram for the beam in the answer sheets HI
a. 2.330.10³ d. 2.500 105 2.681 106 1.572 107 j. Table 2.2 a. d. j. Table 2.3 Stress calculation Q2.11: Calculate the maximum bending moment (pick the correct value in table 2.3) 0.5.107 Nmm b. 1.5.107 Nmm 3.0.107 Nmm 4.5.107 Nmm 6.0.107 Nmm a. d. 2.0.107Nmm 3.5. 107 Nmm 5.0.107Nmm a. d. Table 2.4 g. 0 20.0 KN 40.0 KN j. Q2.12: Calculate the maximum shear force (pick the correct value in table 2.4) b. 10.0 KN 25 kN 50.0 KN b. e. h. k. 5.7 N/mm² 22.7 N/mm² 39.7 N/mm² 159.0 N/mm² 254.4 N/mm² e. h. m. Table 2.5 2.858 103 5.00 105 6.521 106 6.670 107 e. h. Q2.13: Calculate the maximum normal stress due to bending (pick the correct value in table 2.5) Q2.14: Calculate the maximum shear stress (pick the correct value in table 2.5) b. 1.0.107 Nmm 2.5 107 Nmm 4.0 107 Nmm 5.5.107 Nmm e. h. k. n. C. f. i. 1. 10.7 N/mm² 28.4 N/mm² 63.6 N/mm² C. f. i. 1. 190.8 N/mm² 318.0 N/mm² 8.917.104 1.671 106 9.203.106 3.145 108 C. f. i. C. f. i. 1. 0. 15.0 KN 30.0 KN 75.0 kN 21.5 N/mm² 34.0 N/mm² 127.2 N/mm² 222.6 N/mm² 512.5 N/mm²
Q2.15: Where in the cross-section would the maximum tensile normal stress occur (pick the correct value in table 2.6) Q2.16: Where in the cross-section would the max. shear stress occur (pick the correct value in table 2.6) a. d. Along P1-P2 Along P9-P10 Along P5-P6 Table 2.6 b. e. h. Along P3-P4 Along P1-P9 Along P2-P9 C. f. i. P1 P3 P7 P9 P5 P6. Along P7-P8 Along P2-P10 Along P1-P10 P2 P4 P8 P10
The maximum deflection of the considered beam can be obtained by superposition of standard beam problems. Q2.17: Identify the two beam cases in table 2.7 which are required to calculate the max. beam Deflection (Select two cases in table 2.7) Q2.18: Calculate the maximum deflection of the beam in mm (write the value) (Hint: Do not use direct integration, use the deflection solutions in table 2.7) Load case and BCDs Max. Deflection End slope P PL³ PL² 3EI 2EI a. b. C. d. e. f. 60 58 Table 2.7 A 27 L W W L L L M M - WLA 8EI ML² 2EI 5wL4 384EI PL³ 48EI ML² 9√3EI - + + OB WL³ 6EI ML ΕΙ WL³ 24EI PL² 16EI 0A = +₁ ML = - 6EI ML 3EI Equation of elastic curve P u= (x³ - 3Lx²) 6EI U= U= W 24EI W 24EI For x ≤ باب u= U= (x4 - 4Lx³ +6L²x²) U= M 2EI (x4-2Lx³ + L³x) P 48EI (4x³ - 3L²x) M 6EIL -(x³ - L²x)
The transverse loads are removed from the beam in figure 2 and replaced by a centric compressive load N. The boundary conditions shown in the figure to the right are the same in the xy- and xz-planes. a. L Q2.19: Determine the Euler length of the beam column (select the correct value in table 2.8) L LE == b. LE = 0.699L C. LE = L LE = 2L f. i. d. LE = 11L Table 2.8 a. d. 4.82 10¹N 5.48 105N 2.04.106N 1.93.108 N e. h. j. Table 2.9 A LE = πL LE = 23 Q2.20: Calculate the lowest critical load (Euler load) for which the beam will buckle sideways (select the correct value in table 2.9) b. e. h. k. 1.51 105N 6.06 105 N 4.17. 106 N 6.06.108 N LE=L 2 LE =8 C. f. i. 1. B 3.10.105N 6.48 105N 8.15 106N 8.15.10°N N