1 answer

Use this cross section and the associated M(x) and V(x) diagrams to determine the maximum bending...

Question:

9 kips/ft У 6 kips/ft С В A 8 ft 22 ftUse this cross section and the associated M(x) and V(x) diagrams to determine the maximum bending stress (in Tension and in Compression) and the location along the beam where this occurs.

***geometrical data about cross-section of beam

X Section Properties Module Back File Simple Flanged Double Standard Help Axes AISC Shape W12x22 с у-z 4.0300 CX-y У .4250 to

9 kips/ft У 6 kips/ft С В A 8 ft 22 ft
X Section Properties Module Back File Simple Flanged Double Standard Help Axes AISC Shape W12x22 с у-z 4.0300 CX-y У .4250 to mm Rotate 0 .2600 90 11.4500 180 270 Rotate Mohr's Circle Compute To scale Click Compute to update section properties W12x35 W12x30 W12x26 W12x22 W12x19 W12x16 W12x14 W10x112 W10x100 W10x88 А36 Yield Stress 36 ksi Ultimate Stress = 58 ksi

Answers

9 kips/ft 6 kips/ft А ТА — B 777777 PRA 1 Consider equilibrium of beam Efy =o .. RA TRB = 6x2 2 + 1 x 9x8 .:RA + R2 = 168 kip273.32- 10 192kips-ft Moment diagram Mp=57.27x9.545 - 6x9.595x9.545 - 2 X = 273-321 kips-fite Area moment of inertia of beamThe maximum bending stress will occure at point D which is located at distance of 9.545 ft from point A.

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