# Fig Example 8-4. Rotor shaft of a helicopter (combined torsion and axial force).

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1 PEF 309 Fundamentos de Mecânica das Estruturas Eercícios resolvidos etraídos de Mechanics of Materials, th edition, GERE, J.M. e Timoshenko, S.P.,PWS Publishing Company, 997, Boston, USA, p.08-0, Eample 8- The rotor shaft of a helicopter drives the rotor blades that provide the lifting force to support the helicopter in the air (Fig. 8-a). As a consequence, the shaft is subjected to a combination of torsion and aial loading (Fig. 8-b). For a 50-mm diameter shaft transmitting a torque T, kn.m and a tensile force P 5kN, determine the maimum shear stress in the shaft. Fig. 8- Eample 8-. Rotor shaft of a helicopter (combined torsion and aial force). Solution The stress in the rotor shaft are produced by the combined action of the aial force P and the torque T (Fig. 8-b). Therefore, the stress at any point on the surface of the shaft consist of a tensile stress 0 and shear stresses 0 as shown on the stress element of Fig. 8-c. Note that the y ais is parallel to the longitudinal ais of the shaft. The tensile stress 0 equals the aial force divided by the cross-sectional area: P P A πd (5kN) π (50mm) MPa 3.3): The shear stress 0 is obtained from the torsion formula (Eq. 3- of Section Tr I 6T 3 πd 6(.kN. m) 3 π (50mm) 0 p 97.78MPa The stress 0 and 0 act directly on cross sections of the shaft.

2 Knowing the stress 0 and 0, we can now obtain the principal stresses and maimum shear stresses by the methods described in section 7.3. The principal stresses are obtained from Eq. (7-7): + y, ± + y ( d) Substituting 0, y MPa, and y MPa, we get or 3 MPa 03MPa, ± 35 MPa 7MPa These are the maimum tensile and compressive stress in the rotor shaft. The maimum in-plane shear stresses (Eq. 7-5) are ma + y ( e) This term was evaluated above, so we see immediately that ma 03MPa Because the principal stresses and have opposite signs, the maimum inplane shear stresses are larger than the maimum out-of-plane shear stresses (see Eqs. 7-8a, b, and c and the accompanying discussion). Therefore, the maimum shear stress in the shaft is 03 MPa. Eample 8-6 A sign of dimensions.0 m. m is supported by a hollow circular pole having outer diameter 0 mm and inner diameter 80 mm (Fig. 8-0). The sign is offset 0.5 m from the centerline of the pole and its lower edge is 6.0 m above the ground. Determine the principal stresses and maimum shear stresses at points A and B at the bases of the pole due to a wind pressure of.0 kpa against the sign. Fig. 8-6 Eample 8-6. Wind pressure against a sign (combined bending, torsion, and shear of the pole).

3 Solution Stress resultants. The wind pressure against the sign produces a resultant force W that acts at the midpoint of yhe sign (Fig. 8-7a) and is equal to the pressure p times the area A over which it acts: W pa (.0kPa)(.0m.m). 8kN The line of action of this force is at height h 6.6 m above the ground and at distance b.5 m from the centerline of the pole. The wind force acting on the sign is statically equivalent to a lateral force W and a torque T acting on the pole (Fig. 8-7b). The torque is equal to the force W times the distance b: T Wb (.8kN)(.5 m) 7. kn m The stress reultants at the base of the pole (Fig. 8-7c) consist of a bending moment M, a torque T, and a shear force V. Their magnetudesn are M Wh (.8kN)(6.6m) 3.68kN m T 7.kN m V W.8kN Eamination of these stress resultants shows that maimum bending stresses occur at point A and maimum shear stresses at point B. Therefore, A and B are critical points where the stresses should be determined. (Another critical points is diametrically opposite point A, as eplained im the Note below). Stresses at points A and B. The bending moment M produces a tensile stress A at point A (Fig. 8-7d) but no stress at point B (which is located on the neutral ais). The stress A is obtained from the formula: M ( d / ) A I in which d is the outer diameter (0 mm) and I is moment of inertia of the cross section. The moment of inertia is I π 6 π 6 ( d d ) [(0mm) (80mm) ] 63.6 m 0 6 in which d is the inner diameter. Therefore, Md (3.68kN m)(0mm) A 5. 9MPa 6 I ( m )

4 Fig. 8-7 Solution to Eample 8-6. The torque T produces shear stresses at points A and B (Fig. 8-7d). We can calculate these stresses from the torsion formula: T / ) ( d I p in which I p is the polar moment of inertia: I p π 3 6 ( d d ) I 6.9 m 0 Thus,

5 Td I (7.kN m)(0mm) 6 (6.9 0 m ) p 6.MPa Finally, we calculate the shear stress at point A and B due to the shear force V. The shear stress at point A is zero, and the shear stress at point B (denoted in Fig. 8-7d) is obtained from the shear formula for a circular tube (Eq. 5- of Section 5.9): V r + r r + r ( j) 3A r r + in which r and r are the outer and innewhich r and r are the outer and inne radii, respectively, and A is the cross-sectional area: d d r 0mm r A π ( r r ),570mm 90mm Substituting numerical values into Eq. (j), we obtain 0.76 MPa All stresses acting at points A and B have now been calculed. Stress elements. The net step is to show these stresses on stress elements (Figs. 8-7e and f). For both elements, the y ais is parallel to the longitudinal ais of the pole and the ais is horizontal. At point A the stresses acting on the element are 0 y A 5.9MPa y 6. MPa At point B the stresses are y 0 y + 6.MPa MPa 7. 00MPa Since there are no normal stresses acting on the element, point B is in pure shear. Now that all stresses acting on the stress elements (Figs. 8-7e and f) are known, we can use the equations given in Section 7.3 to determine the principal stresses and maimum shear stresses. Principal stresses and maimum shear stresses at point A. The principal stresses are obtained from Eq. (7-7), which is repeated here: + y, ± + y ( k) Substituting 0, y 5.9 MPa, and y 6. MPa, we get 7.5MPa 8. MPa, ±

6 or 55.7 MPa 0. 7 MPa The maimum in-plane shear stresses may be obtained from Eq. (7-5) ma + y This term was evaluated above, so we see immediately that ma 8. MPa

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