Consider the forces acting on the intermediate shaft of a gearbox illustrated in Example 9.4. The maximum permissible radial deflection at any gear is limited to 1 mm. The modulus of elasticity of the shaft material is 207000 N/mm^2. Determine the shaft diameter and compare it with the solution of

Question

Consider the forces acting on the intermediate shaft of a gearbox illustrated in Example 9.4. The maximum permissible radial deflection at any gear is limited to 1 mm. The modulus of elasticity of the shaft material is 207000 N/mm². Determine the shaft diameter and compare it with the solution of Example 9.4.

Table 9.4 Bending moment and deflection of beams

	Case 1 Cantilever Beam – End load (A) Bending moments [latex]\left(M_{b} ight) ext { at } O=-P l[/latex]. (1). [latex] \left(M_{b} ight) ext { at } x=-P(l-x) [/latex]. (2). (B) Deflections [latex] \delta ext { at } x=\frac{P x^{2}}{6 E I}(x-3 l) [/latex] (3). [latex] \delta_{\max .} ext { at }(x=l)=-\frac{P l^{3}}{3 E I} [/latex] (4).
	Case 2 Cantilever Beam—Uniformly distributed load (A) Bending moments [latex]\left(M_{b} ight) ext { at } O=-\frac{w l^{2}}{2} [/latex] (5). [latex] \left(M_{b} ight) ext { at } x=-\frac{w(l-x)^{2}}{2} [/latex] (6). (B) Deflections [latex] \delta ext { at } x=\frac{w x^{2}}{24 E I}\left(4 l x-x^{2}-6 l^{2} ight) [/latex] (7). [latex] \delta_{\max .} ext { at }(x=l)=-\frac{w l^{4}}{8 E I} [/latex] (8).
	Case 3 Cantilever Beam—Moment load (A) Bending moments [latex] \left(M_{b} ight) ext { at } O=\left(M_{b} ight) ext { at } x=\left(M_{b} ight)_{B} [/latex] (9). (B) Deflections [latex] \delta ext { at } x=\frac{\left(M_{b} ight)_{B} x^{2}}{2 E I} [/latex] (10). [latex] \delta_{\max .} ext { at }(x=l)=\frac{\left(M_{b} ight)_{B} l^{2}}{2 E I} [/latex] (11).
	Case 4 Simply supported beam—Centre load (A) Bending moments [latex] \left(M_{b} ight) ext { at } x=\frac{P x}{2} [/latex] (12). [latex] \left(M_{b} ight) ext { at } x=\frac{P(l-x)}{2} \quad(x>1 / 2) [/latex] (13). [latex] \left(M_{b} ight) ext { at } B=\frac{P l}{4} [/latex] (14). (B) Deflections [latex] \delta ext { at } x=\frac{P x\left(4 x^{2}-3 l^{2} ight)}{48 E I} \quad(0 [latex] \delta_{\max .} ext { at } B=-\frac{P l^{3}}{48 E I} [/latex] (16).
	Case 5 Simply supported Beam—Uniformly distributed load (A) Bending moments [latex] \left(M_{b} ight) ext { at } x=\frac{w x(l-x)}{2} [/latex] (17). [latex] \left(M_{b} ight) ext { at } l / 2=\frac{w l^{2}}{8} [/latex] (18). (B) Deflections [latex] \delta ext { at } x=\frac{w x\left(2 l x^{2}-x^{3}-l^{3} ight)}{24 E I} [/latex] (19) [latex] \delta_{\max .} ext { at } l / 2=-\frac{5 w l^{4}}{384 E I} [/latex] (20).
	Case 6 Simply supported beam—Moment load (A) Bending moments [latex] \left(M_{b} ight) ext { at } x=\frac{\left(M_{b} ight)_{B} x}{l}(0 [latex] \left(M_{b} ight) ext { at } x=\frac{\left(M_{b} ight)_{B}(x-l)}{l}(a (B) Deflections [latex] \delta ext { at } x=\frac{\left(M_{b} ight)_{B} x\left(x^{2}+3 a^{2}-6 a l+2 l^{2} ight)}{6 E I l} \quad(0 [latex] \delta ext { at } x=\frac{\left(M_{b} ight)_{B}\left[x^{3}-3 l x^{2}+x\left(2 l^{2}+3 a^{2} ight)-3 a^{2} l ight]}{6 E I l} [/latex] (a < x < l) (24).
	Case 7 Simply supported Beam—Intermediate load (A) Bending moments [latex] \left(M_{b} ight) ext { at } x=\frac{P b x}{l} \quad(0 [latex] \left(M_{b} ight) ext { at } x=\frac{P a(l-x)}{l} \quad( a (B) Deflections [latex] \delta ext { at } x=\frac{P b x\left(x^{2}+b^{2}-l^{2} ight)}{6 E I l} \quad(0 [latex] \delta ext { at } x=\frac{P a(l-x)\left(x^{2}+a^{2}-2 l x ight)}{6 E I l}(a
	Case 8 Simply supported beam—Overhang load (A) Bending moments [latex] \left(M_{b} ight) ext { at } x=-\frac{P a x}{l} \quad(x [latex] \left(M_{b} ight) ext { at } x=P(x-l-a) \quad(x>l) [/latex] (30) (B) Deflections [latex] \delta ext { at } x=\frac{\operatorname{Pax}\left(l^{2}-x^{2} ight)}{6 E I l} \quad(x [latex] \delta ext { at } x=\frac{P(x-l)\left[(x-l)^{2}-a(3 x-l) ight]}{6 E I} \quad(x>l) [/latex] (32). [latex] \delta ext { at } C=-\frac{P a^{2}(l+a)}{3 E I} [/latex] (33).

Accepted Answer

[latex] ext { Given } \delta_{\max .}=1 mm \quad E=207000 N / mm ^{2} [/latex].

Step I Deflection at Gear -B
The deflection at gears B or C (Fig. 9.6) is calculated by the principle of superimposition.
Consider the deflection at B.
Vertical plane
(i) Deflection due to a force of 1609 N [Fig. 9.47(a), Eq. (27) of Table 9.4]:

[latex] au=\frac{2 M_{t}}{d b l} [/latex] (9.27).

[latex] \left(\delta_{B} ight)_{1}=\frac{P b x\left(x^{2}+b^{2}-l^{2} ight)}{6 E I l} [/latex].

[latex] =\frac{(1609)(1800)(900)\left(900^{2}+1800^{2}-2700^{2} ight)}{6 E I(2700)} [/latex].

[latex] =-\frac{5.2131 imes 10^{11}}{E I} mm [/latex] (i).

(ii) Deflection due to a force of 6631.5 N [Fig. 9.47(b), Eq. (27) of Table 9.4]:

[latex] au=\frac{2 M_{t}}{d b l} [/latex] (9.27).

[latex] \left(\delta_{B} ight)_{2}=\frac{P b x\left(x^{2}+b^{2}-l^{2} ight)}{6 E I l} [/latex].

[latex] =\frac{(-6631.5)(900)(900)\left(900^{2}+900^{2}-2700^{2} ight)}{6 E I(2700)} [/latex].

[latex] =+\frac{18.8 imes 10^{11}}{E I} mm [/latex] (ii).

The vertical deflection at B denoted by [latex] \left(\delta_{B} ight)_{v} [/latex] is given by the principle of superimposition.

[latex] \left(\delta_{B} ight)_{v}=\left(\delta_{B} ight)_{1}+\left(\delta_{B} ight)_{2} [/latex].

[latex] =-\frac{5.2131 imes 10^{11}}{E I}+\frac{18.8 imes 10^{11}}{E I} [/latex].

[latex] =\frac{13.5869 imes 10^{11}}{E I} mm [/latex] (a).

Horizontal plane
(i) Deflection due to a force of 4421 N [Fig. 9.47(c), Eq. (27) of Table 9.4]:

[latex] au=\frac{2 M_{t}}{d b l} [/latex] (9.27).

[latex] \left(\delta_{B} ight)_{1}=\frac{P b x\left(x^{2}+b^{2}-l^{2} ight)}{6 E I l} [/latex].

[latex] =\frac{(4421)(1800)(900)\left(900^{2}+1800^{2}-2700^{2} ight)}{6 E I(2700)} [/latex].

[latex] =-\frac{14.324 imes 10^{11}}{E I} mm [/latex] (iii).

(ii) Deflection due to a force of 2413.67 N [Fig. 9.47 (d), Eq. (27) of Table 9.4]:

[latex] au=\frac{2 M_{t}}{d b l} [/latex] (9.27).

[latex] \left(\delta_{B} ight)_{2}=\frac{P b x\left(x^{2}+b^{2}-l^{2} ight)}{6 E I l} [/latex].

[latex] =\frac{(-2413.67)(900)(900)\left(900^{2}+900^{2}-2700^{2} ight)}{6 E I(2700)} [/latex].

[latex] =+\frac{6.8427 imes 10^{11}}{E I} mm [/latex] (iv)

By the principle of superimposition,

[latex] \left(\delta_{B} ight)_{h}=\left(\delta_{B} ight)_{1}+\left(\delta_{B} ight)_{2} [/latex].

[latex] =-\frac{14.324 imes 10^{11}}{E I}+\frac{6.8427 imes 10^{11}}{E I} [/latex].

[latex] =-\frac{7.4813 imes 10^{11}}{E I} mm [/latex] (b).

The radial deflection at the gear B consists of two components—vertical deflection [latex] \left(\delta_{B} ight)_{v} [/latex] and horizontal component [latex] \left(\delta_{B} ight)_{h} [/latex]. The radial deflection [latex] \left(\delta_{B} ight) [/latex] is given by,

[latex] \delta_{B}=\sqrt{\left[\left(\delta_{B} ight)_{v} ight]^{2}+\left[\left(\delta_{B} ight)_{h} ight]^{2}} [/latex].

[latex] =\sqrt{(13.5869)^{2}+(7.4813)^{2}\left(\frac{10^{11}}{E I} ight)} [/latex].

[latex] =\frac{15.51 imes 10^{11}}{E I} mm [/latex] (c).

Step II Deflection at Gear-C
A similar procedure is used to calculate the deflection at C. The final values of vertical and horizontal deflections are as follows:

[latex] \left(\delta_{C} ight)_{v}=+\frac{16.9245 imes 10^{11}}{E I} mm [/latex].

[latex] \left(\delta_{C} ight)_{h}=-\frac{4.7128 imes 10^{11}}{E I} mm [/latex].

[latex] \delta_{C}=-\frac{17.5684 imes 10^{11}}{E I} mm [/latex] (d).

Step III Shaft diameter
From (c) and (d), the radial deflection is maximum at the gear C. Equating it with permissible deflection,

[latex] 1=\frac{17.5684 imes 10^{11}}{(207000)\left(\pi d^{4} / 64 ight)} [/latex].

d = 114.67 mm
In Example 9.4, the shaft diameter was calculated as 68.59 mm on the basis of strength, while in this problem the same has been calculated as 114.67 mm on the basis of rigidity. Therefore, rigidity becomes the criterion of design in this case.

	Case 1 Cantilever Beam – End load (A) Bending moments $\left(M_{b}\right) \text { at } O=-P l$ . (1). $\left(M_{b}\right) \text { at } x=-P(l-x)$ . (2). (B) Deflections $\delta \text { at } x=\frac{P x^{2}}{6 E I}(x-3 l)$ (3). $\delta_{\max .} \text { at }(x=l)=-\frac{P l^{3}}{3 E I}$ (4).
	Case 2 Cantilever Beam—Uniformly distributed load (A) Bending moments $\left(M_{b}\right) \text { at } O=-\frac{w l^{2}}{2}$ (5). $\left(M_{b}\right) \text { at } x=-\frac{w(l-x)^{2}}{2}$ (6). (B) Deflections $\delta \text { at } x=\frac{w x^{2}}{24 E I}\left(4 l x-x^{2}-6 l^{2}\right)$ (7). $\delta_{\max .} \text { at }(x=l)=-\frac{w l^{4}}{8 E I}$ (8).
	Case 3 Cantilever Beam—Moment load (A) Bending moments $\left(M_{b}\right) \text { at } O=\left(M_{b}\right) \text { at } x=\left(M_{b}\right)_{B}$ (9). (B) Deflections $\delta \text { at } x=\frac{\left(M_{b}\right)_{B} x^{2}}{2 E I}$ (10). $\delta_{\max .} \text { at }(x=l)=\frac{\left(M_{b}\right)_{B} l^{2}}{2 E I}$ (11).
	Case 4 Simply supported beam—Centre load (A) Bending moments $\left(M_{b}\right) \text { at } x=\frac{P x}{2}$ (12). $\left(M_{b}\right) \text { at } x=\frac{P(l-x)}{2} \quad(x>1 / 2)$ (13). $\left(M_{b}\right) \text { at } B=\frac{P l}{4}$ (14). (B) Deflections $\delta \text { at } x=\frac{P x\left(4 x^{2}-3 l^{2}\right)}{48 E I} \quad(0<x<1 / 2)$ (15). $\delta_{\max .} \text { at } B=-\frac{P l^{3}}{48 E I}$ (16).
	Case 5 Simply supported Beam—Uniformly distributed load (A) Bending moments $\left(M_{b}\right) \text { at } x=\frac{w x(l-x)}{2}$ (17). $\left(M_{b}\right) \text { at } l / 2=\frac{w l^{2}}{8}$ (18). (B) Deflections $\delta \text { at } x=\frac{w x\left(2 l x^{2}-x^{3}-l^{3}\right)}{24 E I}$ (19) $\delta_{\max .} \text { at } l / 2=-\frac{5 w l^{4}}{384 E I}$ (20).
	Case 6 Simply supported beam—Moment load (A) Bending moments $\left(M_{b}\right) \text { at } x=\frac{\left(M_{b}\right)_{B} x}{l}(0<x<a)$ (21). $\left(M_{b}\right) \text { at } x=\frac{\left(M_{b}\right)_{B}(x-l)}{l}(a<x<l)$ (22). (B) Deflections $\delta \text { at } x=\frac{\left(M_{b}\right)_{B} x\left(x^{2}+3 a^{2}-6 a l+2 l^{2}\right)}{6 E I l} \quad(0<x< a )$ (23). $\delta \text { at } x=\frac{\left(M_{b}\right)_{B}\left[x^{3}-3 l x^{2}+x\left(2 l^{2}+3 a^{2}\right)-3 a^{2} l\right]}{6 E I l}$ (a < x < l) (24).
	Case 7 Simply supported Beam—Intermediate load (A) Bending moment $\left(M_{b}\right) \text { at } x=\frac{P b x}{l} \quad(0<x<a)$ (25). $\left(M_{b}\right) \text { at } x=\frac{P a(l-x)}{l} \quad( a <x<l)$ (26). (B) Deflections $\delta \text { at } x=\frac{P b x\left(x^{2}+b^{2}-l^{2}\right)}{6 E I l} \quad(0<x<a)$ (27). $\delta \text { at } x=\frac{P a(l-x)\left(x^{2}+a^{2}-2 l x\right)}{6 E I l}(a<x<l)$ (28).
	Case 8 Simply supported beam—Overhang load (A) Bending moments $\left(M_{b}\right) \text { at } x=-\frac{P a x}{l} \quad(x<l)$ . (29) $\left(M_{b}\right) \text { at } x=P(x-l-a) \quad(x>l)$ (30) (B) Deflections $\delta \text { at } x=\frac{\operatorname{Pax}\left(l^{2}-x^{2}\right)}{6 E I l} \quad(x<l)$ (31). $\delta \text { at } x=\frac{P(x-l)\left[(x-l)^{2}-a(3 x-l)\right]}{6 E I} \quad(x>l)$ (32). $\delta \text { at } C=-\frac{P a^{2}(l+a)}{3 E I}$ (33).

Question 9.26: Consider the forces acting on the intermediate shaft of a ge...

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