FRICTIONAL FORCES ON SCREWS Today’s Objectives: In-Class Activities: Students will be able to: • Check Homework, if any a) Determine the forces on a squarethreaded screw. • Reading Quiz • Applications • Analysis of Impending motion • Analysis of a self locking screw • Concept Quiz • Group Problem Solving • Attention Quiz READING QUIZ 1. A screw allows a ______ moment M to lift a _________ weight W. A) (large, large) B) (small, small) C) (small, large) D) (large, small) 2. A screw is self locking if it remains in place under __________ loads. A) any axial B) small axial C) any rotational D) small rotational W APPLICATIONS Screws are sometimes used not as fasteners, but as mechanisms for transmitting power from one part of a machine to another. How can we determine the force required to turn a screw? Some screws are self locking, meaning it remains in place under any axial load. How do we determine if this is the case? APPLICATIONS (continued) The design of a turnbuckle requires knowledge of self locking properties and the minimum moment M required to turn the machine. How much friction is needed to create a self locking apparatus? ANALYSIS OF A SCREW W A screw is a simple machine in which a small Moment M is used to lift a large weight W. To determine the force required to turn the screw, it is necessary to draw an FBD of the screw thread. A square threaded screw is a cylinder with a square ridge wrapped around it. The slope of the thread is the lead angle, determined from 𝑙 𝜃 = atan 2𝜋𝑟 An FBD of the entire unraveled thread can be represented as a block. ANALYSIS OF A SCREW (continued) Four Cases can the be analyzed: 1. Upward impending motion 2. Self-Locking 3. Downward impending motion 4. Downward impending motion (not-self locking) The reaction R has both frictional and normal components. 𝐹 𝜙𝑠 = atan = atan 𝜇𝑠 𝑁 Note that this assumes impending motion ANALYSIS OF A SCREW (continued) A screw is self locking if 𝜙𝑠 = θ with no applied moment If a screw is self locking, a moment M’ must be applied to make 𝜙𝑠 > 𝜃 and lead to downward motion If a screw is not self locking, then a moment M’’ must be applied to keep the screw from falling EXAMPLE Given: The turnbuckle has a square thread with a mean radius of 5 mm and a lead of 2 mm. 𝜇𝑠 = 0.25. Find: The moment M to draw the screws closer together Plan: 1. Draw a FBD of the screw thread. 𝑙 ) 2𝜋𝑟 2. Determine the lead angle 𝜃 = atan( 3. Assume impending motion 4. Apply the E-of-E to the screw thread. EXAMPLE (continued) W M/r F Θ FX = N -N sin(Θ) + M/r – .25 N cos(Θ) = 0 FY = N cos(Θ) – .25 N sin(Θ) - W = 0 Solving the above two equations, we get M = 6.37 N * m GROUP PROBLEM SOLVING Given: The square threaded screw has a mean diameter of 0.5 in and a lead of 0.2 in. 𝜇𝑠 = 0.4. Find: The torque M that should be applied to the screw to start lifting the 6000 lb load