A transmission shaft supported between two bearings and carrying two concentrated masses is shown in Fig. 9.54. It is made of steel (E = 207 000 N/mm^2). Assuming that the shaft has negligible mass, calculate the critical speed.

Question

A transmission shaft supported between two bearings and carrying two concentrated masses is shown in Fig. 9.54. It is made of steel (E = 207 000 N/mm²). Assuming that the shaft has negligible mass, calculate the critical speed.

Accepted Answer

[latex] ext { Given } E=207000 N / mm ^{2} d=50 mm [/latex].

[latex] m_{A}=35 kg \quad m_{B}=55 kg [/latex].

Step I Deflections at mass-A and mass-B
Suppose the masses of 35 and 55 kg are attached to the shaft at points A and B respectively.

[latex] W_{A}=35 g =35(9.81)=343.35 N [/latex].

[latex] W_{B}=55 g =55(9.81)=539.55 N [/latex].

The deflections at these masses are calculated by the method of superimposition. The configuration is similar to Case 7 of Table 9.4 (simply supported beam with intermediate load). Rewriting Eqs (27) and (28) with reference to Fig. 9.55(a),

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 moment [latex] \left(M_{b} ight) ext { at } x=\frac{P b x}{l} \quad(0 (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).

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

[latex] \sigma_{c}=\frac{4 M_{t}}{d h l} [/latex] (9.28)

[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

Deflections due to force 343.35 N [Fig. 9.55 (b)]
a = 750 mm b = 1500 mm l = 2250 mm
The deflection [latex] \left(\delta_{A} ight)_{A} [/latex] at the point A due to the force at A is given by Eq. (a).

[latex] x=750 mm (0

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

[latex] \left(\delta_{A} ight)_{A}=\frac{(343.55)(1500)(750)\left(750^{2}+1500^{2}-2250^{2} ight)}{6 E I(2250)} [/latex].

[latex] =\frac{-6.44(10)^{10}}{E I} [/latex] (i).

The deflection [latex] \left(\delta_{B} ight)_{A} [/latex] at the point B due to the
force at A is given by Eq. (b).

x = 1750 mm (x > a).

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

[latex] \left(\delta_{B} ight)_{A}=\frac{(343.35)(750)(2250-1750)\left[1750^{2}+750^{2}-2(2250)(1750) ight]}{6 E I(2250)} [/latex].

[latex] =\frac{-4.05(10)^{10}}{E I} [/latex] (ii).

Deflections due to force 539.55 N [Fig. 9.55(c)]
a = 1750 mm b = 500 mm l = 2250 mm.

The deflection [latex] \left(\delta_{A} ight)_{B} [/latex] at the point A due to the
force at B is given by Eq. (a).
x = 750 mm (0 < x < a).

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

[latex] \left(\delta_{A} ight)_{B}=\frac{(539.55)(500)(750)\left(750^{2}+500^{2}-2250^{2} ight)}{6 E I(2250)} [/latex].

[latex] =\frac{-6.37(10)^{10}}{E I} [/latex] (iii).

The deflection [latex] \left(\delta_{B} ight)_{B} [/latex] at the point B due to the
force at B is given by Eq. (a).

x =1 750 mm (0 < x < a).

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

[latex] \left(\delta_{B} ight)_{B}=\frac{(539.55)(500)(1750)\left(1750^{2}+500^{2}-2250^{2} ight)}{6 E I(2250)} [/latex].

[latex] =\frac{-6.12(10)^{10}}{E I} [/latex] (iv).

Superimposing the deflections,

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

[latex] =\frac{-6.44(10)^{10}}{E I}+\frac{-6.37(10)^{10}}{E I}=\frac{-12.81(10)^{10}}{E I} [/latex].

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

[latex] =\frac{-4.05(10)^{10}}{E I}+\frac{-6.12(10)^{10}}{E I}=\frac{-10.17(10)^{10}}{E I} [/latex].

The negative sign indicates downward deflection and it is neglected.

[latex] I=\frac{\pi d^{4}}{64}=\frac{\pi(50)^{4}}{64} mm ^{4} [/latex].

Substituting values of I and E,

[latex] \delta_{A}=\frac{-12.81(10)^{10}}{E I}=\frac{-12.81(10)^{10}}{(207000)\left(\frac{\pi(50)^{4}}{64} ight)} [/latex].

[latex] =2.017 mm =2.017\left(10^{-3} ight) m [/latex].

[latex] \delta_{B}=\frac{-10.17(10)^{10}}{E I}=\frac{-10.17(10)^{10}}{(207000)\left(\frac{\pi(50)^{4}}{64} ight)} [/latex].

[latex] =1.601 mm =1.601\left(10^{-3} ight) m [/latex].

Step II Critical speed of shaft
From Eq. (9.57),

[latex] \omega_{n}=\sqrt{\frac{g\left(W_{1} \delta_{1}+W_{2} \delta_{2}+\ldots \ldots . . ight)}{\left(W_{1} \delta_{1}^{2}+W_{2} \delta_{2}^{2}+\ldots \ldots . . ight)}} [/latex] (9.57).

[latex] \omega_{n}=\sqrt{\frac{g\left(W_{1} \delta_{1}+W_{2} \delta_{2} ight)}{\left(W_{1} \delta_{1}^{2}+W_{2} \delta_{2}^{2} ight)}} [/latex].

[latex] =\sqrt{\frac{(9.81)\left[(343.35)(2.017)\left(10^{-3} ight)+(539.55)(1.601)\left(10^{-3} ight) ight]}{\left[(343.35)(2.017)^{2}\left(10^{-3} ight)^{2}+(539.55)(1.601)^{2}\left(10^{-3} ight)^{2} ight]}} [/latex].

[latex] =\sqrt{\frac{9.81(1556.36)(10)^{-3}}{2779.82(10)^{-6}}} [/latex].

= 74.11 rad/s
(1 revolution = 2p radians).

[latex] =\frac{74.11(60)}{2 \pi}=707.7 rpm [/latex].

	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 9.30: A transmission shaft supported between two bearings and carr...

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