16-6.Distributed load such as double-point support beam
16-6. Distributed load such as double-point support beam
Symbol description and unit
Maximum deflection�F(mm)
ƒÂmax(x=0)=1/24*‚—*L
1
*(3*L
1
3
+6*L
1
2
*L
2
-L
2
3
)/(E*I)
ƒÂmax(x=0)=
ƒÂmax(x=L/2)=1/384*‚—*L
2
2
*(5*L
2
2
-24*L
1
2
)/(E*I)
ƒÂmax(x=L/2)=
Maximum stress�F (Kg/mm
2
)
ƒÐmax=M(x=L
1
)/Z=1/2*‚—*L
1
2
/Z
ƒÐmax=
ƒÐmax=M(x=L/2)/Z=1/2*‚—*L
2
*(L
1
/L-1/4)/Z
ƒÐmax=
E�FYoung's modulus(Kg/mm
2
)
Steel:21000
Stainless steel:19000
Aluminum:7000
cast iron:10000
I�F Moment of inertia of area(mm
4
)
Z�F Moment of inertia (mm
3
)
‚—�F Load B (Kg/mm)
L�F Beam length (mm)
L
1
�FBeam length to load position (mm)
L
2
�FBeam length to load position (mm)
Reference: "Machine Design Handbook for Engineers" (Revised version by Saburo Kano) Applied Structural Mechanics Beam Calculation Table, page 142, refer to beams and columns.