Question

Problem #1 For the simply supported beam given below. E-200 GPa and u-o3. 18 kN 24 kN 2.4 m 4.0 m 1.8 m 150 mm 100 mm E.N.A 250 mm 200 mm Draw the shear diagram and the moment diagram For the location halfway between the right support and the 24 kN concentrated load, draw Mohr's circle for the stress states at points "a" "b", and "c" on the same axes a. b. c. For the location at midspan, draw Mohr's circle for the stress states at points "a", "b", and "c d. For the same location as part c, point "b' on the same axes a. Determine the principal stresses and maximum shear stress. b. Determine the normal and shear strains in the principal orientation. c. Determine the normal and shear strains in the maximum shear orientation. d. Determine the change in volume, assuming Vo 1

Answer 1

NOTE WE HAVE DRAWN THE SFD AND BMD USING GRAPHICAL APPROACH.

ALSO,

THE STRESS STATES HAVE BEEN SHOWN IN 3 DIFFERENT DIAGRAMS. IF YOU WANT YOU CAN COMBINE THEM. I HAVE DRAWN THEM JUST FOR THE SAKE OF CONVENIENCE AND CLARITY OF UNDERSTANDING.

NOTE WE HAVE DRAWN THE SFD AND BMD USING GRAPHICAL APPROACH. ALSO, THE STRESS STATES HAVE BEEN SHOWN IN 3 DIFFERENT DIAGRAMS. IF YOU WANT YOU CAN COMBINE THEM. I HAVE DRAWN THEM JUST FOR THE SAKE OF CONVENIENCE AND CLARITY OF UNDERSTANDING. BENDINGSTRESS = My/I where y - distance from NA I = MOI about NA SHEARSTRESS = V Ay/Ib CALCULATIONS ARE SHOWN BELOW\.\n\null finding Reactions sigma Fy = 0 R_A + R_C = 42 KN sigma M_A 2 = 0 18 times 2.4 + 24 times 6.4 - R_C times 8.2 = 0 therefore RC = 24 KN R_A = 18 KN SFD /At A there is upward point road of 18 KN. SFD jumps up by 18 KN AT B Downward points load of 18 kN. SFD jumps down by 18 KN. At D Downward point load of 24 KN SFD jumps down by 24 KN At C upward point load as 24 kN

where y- distance from NA

I- MOI about NA

CALCULATIONS ARE SHOWN BELOW

NOTE THAT THE STRESS STATE AT C HERE IS REPRESENTED BY A POINT AT THE INTERSECTION OF THE SHEAR AND NORMAL STRESS AXIS.