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- 1. TORSION FORCE Presented by TASFIA AHMED 10.01.03.069
- 2. INTRODUCTION • Torsion force is a twisting force that is applied on an object by twisting one end when the other is held in position or twisted in the opposite direction. Different materials have a different way of responding to torsion. Some will deform, crack or even break depending on the type of material.
- 3. EXAMPLES • If a plastic ruler is twisted between both hands. The ruler is said to be in a state of torsion. • Whenever we turn a key in a lock the handle /shank of the key is in torsion. • Propeller shaft on a ship. The engine turns the shaft, but the water resists the movement of the propeller. That induces torsion in the propeller shaft.
- 4. IMAGES
- 5. TORQUE • In solid mechanics torsion is the twisting of an object due to applied torque. • Twisting moments or torques are forces acting through distance so as to promote rotation. Simple torque : T = F * l
- 6. TORSIONAL DEFORMATION OF CIRCULAR SHAFT • • • • Consider a cylindrical bar of circular cross section twisted by the torques T at both the ends. Since every cross section of the bar is symmetrical and is applied to the same internal torque T we can say that the bar is in pure torsion. Under action of torque T the right end of the bar will rotate through small angle known as angle of twist. The angle of twist varies along the axis of the bar at intermediate cross section denoted byφ ( x ) φ ( x)
- 7. TORSIONAL FORCE FOR CIRCULAR BAR • Rate of twist θ= dφ dx • Shear Strain at the outer surface of the bar γ max = bb | rdφ = = rθ ab dx • For pure torsion the rate of twist is constant and equal to the total angle of twist divided by the length L of the bar rφ γ max = rθ = L
- 8. BEAMS ARE SUBJECTED TO TORSION A member subjected to torsion moments would twist about a longitudinal axis through the shear centre of the cross section. It was also pointed out that when the resultant of applied forces passed through the longitudinal shear centre axis no torsion would occur. In general, torsion moments would cause twisting and warping.
- 9. BAR SUBJECTED TO TORSION Let us now consider a straight bar supported at one end and acted upon by two pairs of equal and opposite forces. Then each pair of forces P and P2 1 form a couple that tend to twist the bar about its longitudinal axis, thus producing surface tractions and moments. Then we can write the moments as T = 1 d1 P 1 T2 = 2 d 2 P
- 10. TORSIONAL FORMULA • Since the stresses act continuously they have a resultant in the form of moment. The Moment of a small element dA located at radial distance ρ and is given by dM = τ max 2 ρ dA r The resultant moment ( torque T ) is the summation over the entire cross sectional area of all such elemental moments. T = ∫ dM = A τ max r ∫ρ 2 dA = A τ max Ip r Polar moment of inertia of circle with radius r and diameter d Ip = π r4 π d 4 = 2 32 Maximum Shear Stress τ max = Tr 16T = Ip πd3
- 11. EXAMPLE • A solid steel bar of circular cross section has diameter d=1.5in, l=54in,G=11.5*10^6psi.The bar is subjected to torques T acting at the ends if the torques have magnitude T=250 lbft.a) what is the maximum shear stress in the bar b) what is the angle of twist between the ends? • From torsion formula τ max = 16T 16* 250*12 = = 4530 psi πd3 π *1.53 Angle of twist Ip = π d 4 π *1.54 = = .4970in 4 32 32 φ= TL 250*12*54 = GI p 11.5*106
- 12. TORSIONAL FAILURE • Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear. • When subjected to torsion a ductile specimen breaks along the plane of maximum shear, i.e. a plane perpendicular to the shaft axis. • When subjected to torsion a brittle material breaks along planes perpendicular to the direction in which tension is maximum, i.e. along surfaces at 45 to the shaft axis.
- 13. CONCLUSION • Because of torsion force an object can be turned and twisted. • Because if torsion force we can tighten a nut in a bolt by using a wrench.
- 14. THANK YOU

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