Mechanics of Materials- Ch. 5 (Hibbeler)

torque

moment that tends to twist a member abouts its longitudinal axis.

if the angle of twist is small...

the length of the shaft and its radius will remain unchanged.

shear strain

γ = (π/2) - θ'

γ = (ρ/c)*(γmax)

ρ= radial distance from axis of the shaft

c= radius

*shear strain at points on the cross section increases linearly with ρ

if the material is linear-elastic...

then Hooke's Law applies, τ=Gγ . And consequently a linear variation in shear strain leads to a corresponding linear variation in shear stress along any radial line on the cross section. τ = 0 at longitudinal axis to a max value at its outer surface. Due to proportionality of triangles: τ = (ρ/c)*(τmax).

polar moment of intertia

(J) of the cross sectional area

solid circular shaft: J = π/2r^4

hollow, tubular shaft: J = π/2(Ro^4-Ri^4)

maximum shear stress (torsion formula)

τmax = Tc/J

τmax= max shear stress in the shaft, which occurs at the outer surface.

T=resultant internal torque acting at the cross section. It's value is determined from the method of sections and the equation of moment equilibrium applied about the shaft's longitudinal axis

J= polar moment of inertia of the cross sectional area

c= outer radius of the shaft

shear stress (torsion formula)

shear stress at the intermediate distance of ρ:

τ=Tρ/J

complementary property of shear stress

equal shear stresses must also act on four of its adjacent faces of an isolated element. Not only does the internal torque T develop a linear distribution of shear stress along each radial line in the lane of the cross-sectional area, but also an associated shear-stress distribution is developed along an axial plane.

Power

(P) is defined as the work performed per unit of time

P = T*ω

P = 2πfT

SI system: Watts (W)

FSP system: horsepower (hp)

Watt

1 W = 1 N*m/s

Horsepower

1 hp = 550 ft*lb/s

frequency

measure of the number of revolutions or cycles the shaft makes per second, unit is hertz (Hz)

1 Hz = 1 cycle/s , 1 cycle = 2π rad > ω =2πf

torque developed in a shaft

T = P/2πf

shaft design

J/c=T/τallow

angle of twist

phi (Φ) = TL/JG

Φ= angle of twist of one end of the shaft with respect to the other end, measured in radians

T= the internal torque at the arbitrary position x, found from the method of sections and the equation of moment equilibrium applied about the shaft's axis

J= the shaft's polar moment of inertia

G= the shear modulus of elasticity for the material

shear modulus of elasticity

Usually can be found in the back of the textbook, but can also find using this formula: G = TL/JΦ

multiple torques

if subjected to several different torques or the cross-sectional area or shear modulus changes abruptly from one region of the shaft to the next, find the angle of twist for each segment of the shaft and sum them.

Φ=∑TL/JG

sign convention

we will use the right-hand rule: both the torque and angle will be positive provided the thumb is directed outward from the shaft when the fingers curl to give the tendency for rotation