The maximum cross-sectional normal stress occurs at point B with a value of - 2.72 MPa ( compression ) while the shear stress occurs at point c with a value of 0.5 Mpa
sum of moments is equal to zero
tan inverse ( 0.75 / 1.8 ) = 22.62 deg
summation of moments
190 * 0.16 - 10.9 * sin ( 22.62 ) * 1.5 - M = 0
M = 2.9KNM
sum of upward forces = sum of downward forces
Normal force, N
N - 10.9 * cox ( 22.62 ) = 0
N = 9.6KN
Shear force, V
V - 14 + 10.4*sin ( 22.62 ) = 0
I = ( b h^3 ) / 12
I = ( 150 * 200^3 ) / 12
I = 100,000,000 mm^4
Stress at A, Sa
Sa = - N / A + Ma / Im
Sa = - ( 9600 / ( 150 * 200 ) ) + ( 2,400,000 * 100 ) / 100,000,000
Sa = 2.08 MPa
Stress at C, Sc
Sc = - N / A + Mc / Im
Sc = - ( 9600 / ( 150 * 200 ) ) - 0
Sc = - 0.32 MPa
Stress at B, Sb
Sb = - N / A + Mb / Im
Sb = - ( 9600 / ( 150 * 200 ) ) - ( 2,400,000 * 100 ) / 100,000,000
Sb = - 2.72 MPa
comparing the values of the stresses shows that the value of the maximum stress is at point B which has the maximum value. The negative sign represent the direction and signifies compression
maximum shear stress. Ss
shear stress at the middle
Ss = 9 3 * 10,000 ) / ( 2 * 150 * 200 )
Ss = 0.5 Mpa
maximum shear stress occurs at point C
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