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15 kN 30 kN/m Determine the maximum deflection of the cantilevered beam. The beam is made of material having an E = 200 GPa and I = 65 • 106 mmª. A -1.5 m- 1.5 m Incorrect Umar Number Correct Answer: 12.916154 ± 2.0%

15 kN 30 kN/m Determine the maximum deflection of the cantilevered beam. The beam is made of material having an E = 200 GPa and I = 65 • 106 mmª. A -1.5 m- 1.5 m Incorrect Umar Number Correct Answer: 12.916154 ± 2.0%

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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15 kN 30 kN/m Determine the maximum deflection of the cantilevered beam. The beam is made of material having an E = 200 GPa and I = 65 * 106 mm4. A -1.5 m- - 1.5 m Incorrect Umar Number Correct Answer: 12.916154 + 2.0%
Transcribed Image Text:15 kN 30 kN/m Determine the maximum deflection of the cantilevered beam. The beam is made of material having an E = 200 GPa and I = 65 * 106 mm4. A -1.5 m- - 1.5 m Incorrect Umar Number Correct Answer: 12.916154 + 2.0%
Deflection Elastic Curve Beam Slope P - PL -Px Vmax -PL (3L – x) v = Omax 2EI Umax 3EI 6EI O max -Px? (3L – 2x) 0 < x< L/2 12EI V = Vmax -PL? -5PL3 Omax 8EI Vmax 48EI -PL? (6x – L) L/2 << L 48EI v = emax -wx? (x² – 4Lx + 6L) 24EI -wL? -wL Umax v = Omax Umax 8EI 6EI max .Omax MoL MəL? Mox? V = Omax Umax 2EI 2EI Mo Vmax EI -wx? (x² – 2Lx + L²) V = 24EI Vmax 0 <x< L/2 -wL -7wL Omax Umax 384EI -wL 48EI v = (8x – L) max 384EI L/2 < x < L Wo -w,L -wo.x? -woL Omax Umax (10L – 101?x + 5Lx? – x³) v = Umax 30EI 120EIL 24EI 0 max ||
Transcribed Image Text:Deflection Elastic Curve Beam Slope P - PL -Px Vmax -PL (3L – x) v = Omax 2EI Umax 3EI 6EI O max -Px? (3L – 2x) 0 < x< L/2 12EI V = Vmax -PL? -5PL3 Omax 8EI Vmax 48EI -PL? (6x – L) L/2 << L 48EI v = emax -wx? (x² – 4Lx + 6L) 24EI -wL? -wL Umax v = Omax Umax 8EI 6EI max .Omax MoL MəL? Mox? V = Omax Umax 2EI 2EI Mo Vmax EI -wx? (x² – 2Lx + L²) V = 24EI Vmax 0 <x< L/2 -wL -7wL Omax Umax 384EI -wL 48EI v = (8x – L) max 384EI L/2 < x < L Wo -w,L -wo.x? -woL Omax Umax (10L – 101?x + 5Lx? – x³) v = Umax 30EI 120EIL 24EI 0 max ||
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