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# 4.3-4 - Problem 4.8-4 Coleuhnc the shear l'oree V and...

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Unformatted text preview: Problem 4.8-4 Coleuhnc the shear l'oree V and bending moment M in a cross section located 0.5 m from the fixed Stlpporl ot' the cantilever beam AB shown in the ﬁgure. Solution 4.3-4 Gantiieuer beam [.5 kN/m | ‘ I L- t .0 111-310 111+L— 2.0 m J FREE-BODY DIAGRAM or SEGMENT DB Point D is; 0.5 m from support .41. #0 RN 7 i tel—ml- 11-31 .3“ 4W ,3 E SECTION 4.3 Shear Forces and Bencfing Moments 345 Eer-le : 0: V = 4.0 kN -t- (1.5 kN/mHlU m') = 4.0 kN + 3.0 kN x 7.0 kN *— EMU = 01' M = —{4.0 kN)t_0.5 m) m ([5 kN/m)(1.0 m)(2.5 m) 2 “-2.0 kN ‘ m - 7.5 kN ' m —95 RN - m *— Problem 4.3-5 Determine the sheur l'oree V zmd bending moment M at a cross 4000301 I 300 thin section located 18 ft from [he let‘t-hzmri end .4 of the. beam with an overhung Shown in the ﬁgure. Solution 4.3«5 491.] Ib/f'r I _ If”! 77'. E . .4..- i h—ID ft -—M~—£D own—«s rpm-rs {1- {RA “E565 EM“ 2 0; R.l : 2190111 20414 2 0. RH : 36”) “1 «~10 n—L—m f’t—+—ﬁ MIL)" n--‘ FREE-BODY DIAGRAM 0F SEGMENT AD Point D is: E811 from support A. EFVERT : o: v = 2190113 — (400 lb/ft_](10 n) z mtg-10:13 «'— EM‘. = 0: M = (2190 lh)(18 ft) W (400 lh/l‘IJt'lO l‘t)(.13 ft) = -12580 lh-l't <— ...
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