Transcript Chapter 8, Sections 8.3, 8.4, & 8.5

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