材料力学06梁的剪应力

• For most beams, both the bending moment and shear force will present.
y
txy txz
z
• The possible stress components at any point on the cross
section will be sx, txy and txz.
tyx= 0. It follows that txy= 0 on the upper and
lower edges of the transverse sections.
Shear Stress in a Narrow Rectangular Beam
Free Body Diagram
txy sx
pair, arrow to arrow and tail to tail, as shown. If we
can determine tyx, then from txy = tyx, we can
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obtain txy.
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Shear on the Horizontal Face of a Beam Element
A*: Area of the
*
shaded area. A: Area of the
entire cross section
Shear Stress t VQ
bI
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• Q A*y b(c y) c y b (c2 y2) 22
• I bh3 Ac2 12 3

A2* y2

bt1(c

t1 ) 2

t [(c 2

t1)2

y2]
t

QV It

V 2I
[bt1 t
(2c

t1)
(c
t1)2

y2]
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I bH 3 (b t)(H 2t1)3
12
12
t max

V Aweb
Example 2
Fz t xzdA 0
( txy 0)
txz may not be zero, but the average is zero
• When shearing stresses are
exerted on the vertical faces of an
element, equal stresses must be
I is the second moment of area of the entire section about the centroidal axis of the entire section.
Example 1
Problem
A beam is made of three planks, nailed together. Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam is V = 500 N, determine the shear force in each nail.
3704
N/m
Shear Force in Nail. From the solution strategy, we have
F qs q(0.025 m) (3704 N m)(0.025 m) F = 92.6 N◄
Shearing Stress in a Beam
dx CD
• Let’s consider the beam element C′D′C″D″. On the surface C′D′, we
• The nails will carry the force resulted from the shear flow.
Example 1: Solution
Moment of Inertia. The moment of inertia of the entire area • can be calculated as the moment of outer rectangle subtracting
C" D"
M
M+dM
V V+dV
F1
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s txy tyx
C C′
D
A*
D′
s + ds F2
dH (Resultant of tyx)
dH F2 F1

M
dM I
A*
ydA
M I
A*
ydA

dM I
A*
ydA

Vdx I
A*
ydA
dH dx

V I
A*
ydA
t ave

dH dA

VQ I
dx t dx

VQ It
t ave

VQ It
• If the width of the beam is comparable or large
dx
relative to its depth, the shearing stresses at D1′ and D2′ are significantly higher than at D′.
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Example 1: Solution Strategy
Solution Strategy
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• If the beam were one piece, there would be shear stress (shear flow) between the cap and the web.
s + ds
q VQ I
q
First Moment of Area
• Q A* ydA y*A*
= the 1st moment of area of the
y*
y1
y
A*
C*
isolated area A*.
• For built up sections
Q Qi yi*Ai*
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A* is area of the isolated area, from where the shear flow is calculated to a free surface.
y* is the distance from the centroidal axis (neutral axis, N.A.) to the centroid (C*) of the isolated area.
sx
z
tzx = 0
x
• However, for a narrow beam (width much smaller
txz
than height), we can assume that txz = 0 zero, since tzx 0 on the left and right surfaces. Therefore, this•t xy来自VQ bI
3V (1 2A
y2 c2
)
•
t max

3V 2A

1.5t
ave
parabolic
I Beam: Shear Stress txy
*
* *
Shear Stress
Q A*y b (c2 y2 ) 2
t QV V (c2 y2)
Ib 2I
Q

A1* y1

VQ I
q VQ I

q dH dx
Shear Flow
Q A* ydA 1st momentof area
Shear on the Horizontal Face: First Moment of Area
Shear Flow • Horizontal shear flow, the horizontal s shear force per unit length, in a beam is given by
that of the 2 small rectangles.
I

1 12
0.100 m0.14 m3

2[112
0.04 m0.10
m 3 ]
16.20 106 m4
* *
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First Moment of Area. For the isolated area (the flange), Q can be calculated as
topic will be focusing on txy only.
• The distribution of txy can not be determined from
statics alone. Analysis of deformation may not help
too much.
• However, we know that shearing stress comes in
If V = 0 (pure bending), txy = txz =0
Now V 0, both txy and txz may not be zero.
Distribution of shearing stresses is unknown but
sx
satisfies
x
Fy t xydA V
have the same horizontal force dH.
dH VQ dx I
• The average shearing stress on the horizontal face
dA dH
of the element is obtained by dividing the shearing force on the element by the area of the face.
Problem
A timber beam is to support the three concentrated loads shown.
Knowing that for the grade of timber used, the allowable stresses are
sall = 1800 psi and tall = 120 psi, determine the minimum required
• For the beam shown, consider an element
CDD′C′ to calculate shear stress at y1.
F1

A* s
dA
M I
A*
y
dA
dx
y A*
F2

A*
(s

ds
)
dA

M
dM I
A*
ydA
Fx 0 : dH F1 F2 0
TOPIC
06
School of Engineering Mechanical Engineering
ENGR 323
Mechanics of Deformable Bodies
Shearing Stress in Beams
© Tulong Zhu, All rights reserved.
Introduction
• For a narrow beam (width << height), we can assume
that txy is uniform through the thickness, and
t xy
t
yx

VQ It
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• On the upper and lower surfaces of the beam,
depth d of the beam based on the maximum normal stress and shear
stress.
Solution Strategy
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• Develop shear and bending moment diagrams. Identify the maximums.
Q A*y* 0.020 m 0.100 m0.060 m 120 106 m3
Shear Flow. The horizontal shear flow q between the flange
and web can be calculated as
q
VQ I

(500N)(120 106 m3) 16.20 10-6 m4
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exerted on the horizontal faces.
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Calculation of Shearing Stress: Preliminary
y
txy
• We know that txz may not be zero, but the average is
zero.
txz