Excessively weak rigidity of the screw's peripheral structure is one of the primary causes that result in lost motion. Therefore in order to achieve excellent positioning accuracy for the precision machines such as NC working machines, etc., axial rigidity balance as well as torsional rigidity for the parts at various portions of the transmission screw have to be taken intoconsideration at time of designing.
Static Rigidity K
The axial elastic deformation and rigidity of the
transmission screw system can be determined from the formula below.
K = p/e (kgf/mm)
p: Axial load (kgf) borne by the transmission screw system
e: Axial flexural displacement (mm)
1/K=1/Ks+1/KN+1/KB+1/KH (mm/kgf)
KS: Axial rigidity of screw shaft (1)
KN: Axial rigidity of nut (2)
KB: Axial rigidity of bracing shaft (3)
KH: Axial rigidity of installation portions of nuts and bearings (4)
(1) Axial rigidity Ks And displacement  s screw shaft
KS = P/ s (kgf/mm)
P: Axial load (kgf)
KS = P/ s (kgf/mm)
For places of Fixed - Fixed installation
sf = PL/4AE (kgf/mm)
For places other than Fixed - Fixed installation
ss=PLO/4AE(mm)
ss=4 sf
sf: Directional displacement at places of fixed-fixed installation
ss: Directional displacement at places other than fixed-fixed installation
A: Cross-sectional area of the screw shaft tooth root diameter (mm2)
E: Longitudinal elastic modulus(2.1 x 104kgf/mm2)
L: Distance between installations(mm)
LO: Distance between load applying points (mm)
(2)Axial rigidity KN and displacement N of nut
KN = P/ s(kgf/mm)
(a) In Case of Single Nut
Q: Load of one steel aball¡]kgf¡^
n: Number of steel ball
K: Constant determined based on material,shape dimensions
K  5.7 x 10-4
ß: Angle of contact (45°)
P: Axial load (kgf)
d: Steel ball diameter (mm)
 : Accuracy, internal structure coefficient
m: Effective number of balls
Do: Steel ball center diameter (mm)
l: Lead (mm)
 : Lead angle
(b) In Case of Double Nuts
When an axial load P of approximately 3 times of the preload load Ppl is exerted, for the purpose of eliminating the preload Ppl on nut B, please set the preload load Ppl at no more than 1/3 of the maximal axial load (0.25Ca should be taken as the standard maximal preload load). Witn respect to the displacement value,it should be of 1/2 of the single nut displacement when axial load is 3 times of the preload.
NS: Displacement of single nut (mm)
NW: Displacement of double nuts (mm)
(Explanation of the rigidity of double nuts)
As shown in Diagram 3.27 and 3.28¡Awhen a preload Ppl is applied on the 2 nuts A, B, both nuts A & B would produce flexural deformations that will reach point X. If an external force P is exerted from here, nut A would move from point X to point X1, while nut B would move from X to X2.
Then,based on the computing formula for displacement
NS of the single nut,we can obtain:
o=aPpl2/3
while displacements of nuts A & B are: A=aPpl2/3
since displacements of nuts A & B generated due to exertion of external force P are equal,therefore
A- o= o- B
or if P is the only external force P that exerts on nuts A ,B, if PAincreases
PA-PB=P
B=0
for preventing the external force applied on nut B being absorbed by nut A thus decreasing,so
when, B=0
aPA2/3 -aPpl2/3=aPpl2/3
PA2/3=2Ppl2/3
or based on A- o= o
o= A/2
thus it can also be judged from Fig.3.29 that,when axial load is 3 times of preload load,for a single nut with 1/2 displacement,the rigidity is 2 times as high.
(3) Axial rigidity KB and displacement B of bracing shaft
KB=P/ B(kgf/mm)
The rigidity of the assembled diagonal thrust ball bearing that is used as the bracing bearing for the ball screw and is widely utilized in the field of precision machines can be found from the following formula
Q: Load of one steel ball (kgf)
ß: Angle of contact (45°)
d: Steel ball diameter (mm)
P: Axial load (kgf)
n: Number of steel ball
(4) Axial rigidity KH and displacement H of installation portions of nuts and
bearings.
In early stage of machine development,special
attentions should be paid to the requirement of high rigidity for
the installation portion.
KH=P/ H(kgf/mm) | 
