Question

Use 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.

9 kips/ft У 6 kips/ft С В A 8 ft 22 ft

***geometrical data about cross-section of beam

Answer 1

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 kips o EMATO .. RB X 22 - 6x22x11 - 1 9 x 8X (22 + 16) 20 ... RB = 110.72 kips and RA= 57.27 kipo. AD=9.595 ft Shear diagram.

273.32- 10 192kips-ft Moment diagram Mp=57.27x9.545 - 6x9.595x9.545 - 2 X = 273-321 kips-fite Area moment of inertia of beam section is I - 4603x (1263) 0.26 X (11.45) 12 s 559.89 2nd Now manimum bending stress will be Mman Teman & max 2=section 12 z modules ...16). man 0 273.321X10X 12 X 6.15 559.89 = 360 26.879 psi The magnitude of manimum bending stress will be same in tension and compression. only sign will change.
The maximum bending stress will occure at point D which is located at distance of 9.545 ft from point A.

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