Transcript Lecture 16

Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
Dr. Konstantinos A. Sierros
Problem 3.4-7
Four gears are attached to a circular shaft and transmit the
torques shown in the figure. The allowable shear stress in the
shaft is 10,000 psi.
(a) What is the required diameter d of the shaft if it has a solid
cross section?
(b) What is the required outside diameter d if the shaft is hollow
with an inside diameter of 1.0 in.?
Problem 3.5-7
The normal strain in the 45° direction on the surface of a circular
tube (see figure) is 880 x (10^-6) when the torque T = 750 lb-in.
The tube is made of copper alloy with G = 6.2 x 910^6) psi.
If the outside diameter d2 of the tube is 0.8 in., what is the inside
diameter d1?
4.4: Relationships between loads, shear forces and bending
moments
• Distributed loads and concentrated loads are positive when they act downward
on the beam and negative when they act upward
• A couple acting as a load on a beam is positive when it is counterclockwise
and negative when it is clockwise
• Shear forces V and bending moments M acting on the sides of the element are
shown in their positive directions
FIG . 4 -1 0
E le m e n t o f a b e a m
u s e d in d e r iv in g th e
r e la tio n s h ip s
b e tw e e n lo a d s , s h e a r
fo r c e s , a n d b e n d in g
m o m e n ts . (A ll lo a d s
a n d s tr e s s r e s u lta n ts a r e
s h o w n in th e ir p o s itiv e
d ir e c tio n s .)
C o p yrigh t 2 0 0 5 b y N els on , a divis ion of Th om s on C an a d a L im ite d
4.4: Distributed loads
• Consider a distributed load of intensity q and its relationship to the
shear force V
• Consider the moment equilibrium of the beam element we can relate
the shear force V with the bending moment M
Moments from left hand side
Counterclockwise +ve
Discarding products of
differentials because they are
negligible compared to the
other terms
4.4: Concentrated loads
• Consider a concentrated load P acting on the beam element
• It can be shown that the bending moment M does not change as we pass
through the point of application of a concentrated load
• At the point of application of a concentrated load P, the rate of change dM/dx
of the bending moment decreases abruptly by an amount equal to P
4.4: Loads in the form of couples
• The last case to be considered is a load in the form of a couple Mo
• From equilibrium of the element in the vertical direction we obtain V1 = 0
which shows that the shear force does not change at the point of application of
a couple
• If we take equilibrium of moments we obtain M1 = -Mo. This equation shows
that the bending moment decreases by Mo as we move from left to right through
the point of load application. Thus, the bending moment changes abruptly at the
point of application of a couple
4.5: Shear force and bending-moment diagrams
• When designing a beam, we need to know how the shear forces and bending
moments vary throughout the length of the beam. Minimum and maximum
values are of special importance
• Information of this kind is provided by graphs in which the shear force and
bending moment are plotted as ordinates (y coordinate) and the distance x along
the axis of the beam is plotted as the abscissa (x coordinate)
Shear force and
bending moment
diagrams
FIG . 4 -1 1
S h e a r -fo r c e a n d
b e n d in g -m o m e n t
d ia g r a m s fo r a s im p le
b e a m w ith a
c o n c e n tr a te d lo a d
C o p yrigh t 2 0 0 5 b y N els on , a divis ion of Th om s on C an a d a L im ite d
4.5: Shear force and bending-moment diagrams
Concentrated load
• Simply supported beam AB and concentrated load P (fig 4-11a). We can
determine the reactions of the beam
• Cut through the beam at a cross-section to
the left of the load P and at distance x from
the support at A and draw FBD (fig 4-11b)
F IG . 4 -1 1
(0 < x < α)
S h e a r-fo rc e a n d
b e n d in g -m o m e n t
d ia g ra m s fo r a s im p le
b e a m w ith a
c o n c e n tra te d lo a d
C o p yrigh t 2 0 0 5 b y N els on , a division of Th om s on C an a d a L im ite d
4.5: Shear force and bending-moment diagrams
Concentrated load
• Next cut through the beam to the right of the load P (α < x < L) and draw a
FBD (fig 4-11c)
F IG . 4 -1 1
(α < x < L)
…and
S h e a r-fo rc e a n d
b e n d in g -m o m e n t
d ia g ra m s fo r a s im p le
b e a m w ith a
c o n c e n tra te d lo a d
C o p yrigh t 2 0 0 5 b y N els on , a division of Th om s on C an a d a L im ite d