A simply supported beam carries inclined loads 100 N, 200 N, and 300 N inclined at 30 degree, 45 degree and 60 degree to the vertical as shown in fig. These loads act at 1m, 2m and 3m from the left support respectively. If the span is 4m. Draw the Shear Force, Bending Moment and Thrust (Axial Force) Diagrams.

The inclined forces are resolved into vertical and horizontal components in fig.(a)

Now, taking moment about hinged end A

$V_b*4=(86.6*1) + (141.4*2) + (150*3)$

$V_b= 204.85 N and V_a=(86.6+141.4+150)-204.85= 173.15 N$

Resolving Horizontally,

$Ha=50+141.4 + 259.8$

$Ha= 451.20 N$

SF analysis:

SF at any section in $AC = 173.15 N $

SF at any section in $CD= 173.15-86.6=86.55N $

SF at any section in $DE = 150-204.85=-54.85 N $

SF at any section in $EB= - 204.85N. $

BM analysis :

$BM_A = 0$

$BM_B=0$

$BM_C=173.15*1=173.15 Nm $

$BM_D= (173.15*2) - (86.6*1) = 259.7 Nm $

$BM_E= 204.85*1=204.85 Nm$

Thrust analysis: OR Axial face analysis:

Thrust at any section in $AC=451.2N $

Thrust at any Section in $CD=451.2-50=401.2N $

Thrust at any Section in $DE=259.8 N$