Bending Moments, Shear Forces and Deflections with Any Combination of Loads
Caption Calculation Symbol =ValueUnitReference OR Explanation
Load No: 1
1Loading PatternL Ty=4Pt = 1; UDL = 2; TDL = 3; VDL= 4
2Distributed Loadw=10kN/m
3Start Locationa=4m
4Length of Spreads=4m
5DirectionDir=270Deg
6Increasing/DecreasingRate=21-Constant, 2-Increasing, 3-Decreasing
7Equation exponent Eq=2For example Eq=1 for y=mx+b AND Eq=3 for y=cx^3
Inputs: Beam Specifications
Moment of InertiaIz=562500000mm^4
Loads & Reactions=
Shear Force =
SF zero positionIteration SF(~0)=0.00Occurs at 6.40
Results
Support ReactionIterationR1=2.88kN2.88 kN
Support ReactionSum of All loads - R1R2=10.453kN10.453 kN; R2 = P - R1
FEM at support AIteration MethodMab=-8.427kNm
FEM at support BIteration MethodMba=-18.775kNm
Bending Moment=
Max Sagging MomentIteration MethodMx+=13.020kN-mOccurs at: 6.051
Deflection Diagram=0.404 mm occurs at 5.751