In a centrifugal pump system used by DWASA a rotating shaft plays a critical role in power transmission and structural support. The shaft is simply supported over a span of 2 m and a concentrated transverse load of 300 N at its midpoint. It rotates at 1440 rpm while delivering 15 kW of Mechanical power. Determine (a) the torque generated by the shaft due to power transmission, (b) the maximum bending moment (c) the bending stress and torsional stresses assuming a shaft diameter of 30 mm (d) the required shaft diameter to safely carry the loads if the maximum allowable stress is 60 MPa using Maximum shear stress theory (e) Evaluate the actual factor of safety.

Created: 9 months ago | Updated: 9 months ago

Updated: 9 months ago

BUET DWASA – Assistant Engineer - 2025 টেকনিক্যাল 2025

175

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(a)

$P o w e r, P = Τ ω = 15 \times 10^{3} = T \times \frac{2 π \times 1440}{60}$

$\Rightarrow T = 99, 47 \approx 100 N m$

(b)

$M o m e n t, M = \frac{F L}{4} = 300 N \times \frac{2}{4} = 150 N m$

(c)

Equivalent Torque, T_e = $\sqrt{M^{2} + T^{2}} = \sqrt{{(150)}^{2} + {(100)}^{2}} = 180.28 N m$

Bending Moment,

$M_{e q} = \frac{1}{2} \{M + \sqrt{M^{2} + T^{2}}\} = \frac{1}{2} \{150 + \sqrt{{(150)}^{2} + {(100)}^{2}}\} = 165.14 N m$

Bending Stress,

$σ_{m a x} = \frac{4}{{πr}^{3}} \times M_{e q} = \frac{4}{{πr}^{3}} \times 165.14 = 62.3 M P a$

Torsion Shear Stress,

$τ_{m a x} = \frac{2 T_{e q}}{{πr}^{3}} = \frac{2}{{πr}^{3}} \times 180.28 = 34 . 01 M P a$

(d)

Maximum allowable stress means we can consider it as yield stress. According to MSS Theory,

$τ_{m a x} = 0.5 S_{y} = 0.5 \times 60 = 30 M P a$

$\Rightarrow \frac{16}{{πd}^{3}} \times \sqrt{M^{2} + T^{2}} = 30 \times 10^{6}$

$\Rightarrow$ d = 31.27 mm

(e)

Here, applied stresses are,

$σ_{χ} = \frac{32 M}{π d^{3}} = 56.6 M P a & T_{x y} = \frac{16 T}{π d^{3}} = 18.9 M P a$

According to MSS:

$σ_{1}, σ_{2} = \frac{σ_{χ}}{2} \pm \sqrt{{(\frac{σ_{χ}}{2})}^{2} + τ_{x y}^{2}} = 28.3 \pm 34.03 = 62.33 - 5.73$

$τ_{m a x} = \frac{σ_{1} - σ_{3}}{2} = 34.03$

$∴ n = \frac{S_{y}}{2 τ_{m a x}} = \frac{60}{2 \times 34.03} = 0.88$

According to DET:
Von Mises strees.

$σ^{'} = \sqrt{σ_{x}^{2} + {3 τ}_{x y}^{2}} = 65.38$

$n = \frac{s_{y}}{σ^{'}} = \frac{60}{65.38} = 0.917$

9 months ago

175

MCQ: 876 CQ: 1k

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