Shear stresses in rectangular beams
Thought experiment: consider one beam
as “two” pieces and load it:
yx
 In reality, the beam only deflects as a
whole, without slippage.
 Horizontal shear stresses must
develop along “glued” contact surface.
1
For a rectangular beam subject to shear
force V(x): examine cross section at x:
c
xy
y
2
y
Note:
yx
(horizontal)
= xy
(vertical)

b(y)

M+M)y/I
R

My/I
L
y

c1
n.a.
y
y
n.s.
M(x)+M
M(x)
x
F
x
x
 0    area of bottom face 

left face
L
dA 

R
dA
right face
1
M ( x) y
( M ( x)  M ) y
xb( y)  
b( y )dy  
b( y )dy
I
I
y
y
c1
c
M
b( y) 
 x 
Ix
c1
 yb( y)dy
y
 Shear stress on all fibers at distance y
from neutral axis (n.a.):
VQ
 ( y) 
Ib
3
(5-38)
where

V = V(x) = shear force at x,

I = moment of inertia of the whole
cross-sectional area,

b = b(y) = width of beam at height
y from n.a., and
c1

Q  Q( y)   ydA
y
Q = first moment of the partial crosssectional area above (or below if y < 0)
the level y where  is sought.
Caution:
4
 (y) should be considered as the
average shear stress over the width
 Shear stress concentrations exist near
sharp corners (e.g. junction between
web and flange of W-shape)
Recall (if y is measured from n.a.):
 ydA  0   ydA   ydA  0
all
below y
above y
Hence, whether you integrate to the top
or bottom fiber, the results only differ by
sign (unimportant for shear stress).
5
For upper cross section:
Q
h/2
h/2
y
y
 ydA   ybdy
h/2
 by 
b  h2
2

    y 
2 4
 2 y

2
V  h2
2
     y 
2I  4

  max
Vh2
V V

 1.5 
8I
A A
6
Note: accuracy of shear formula
depends on the ratio h/b:
 Reasonably accurate for h > b;
 Underestimates max (~ 13%) if b = h
Shear in webs of wide-flange beams
A1 = b(h/2 – h1/2)
A2 = t(h1/2 – y)
7
Assume: vertical shear stresses (small in
flange; ignored) are parallel to y-axis;
and uniform over web thickness t.
Hence, (5.38) still applies.
For given y, we can calculate Q(y)
individually over A1 and A2; each being
(Ai×distance from Ai’s centroid to n.a.).
Adding them and simplifying 
8

 
1
Q    b h 2  h12  t h12  4 y 2
8
 (5.45)
which, together with
bh 3  bh13  th13
I
12
(5.47)
can be used to find shear stress in the
web at level y from n.a., by  = VQ / (It)
9
In particular,
max = [V/(8It)][bh2 - h12(b - t)) at y = 0;
min = [Vb/(8It)](h2 - h12) at y = h1/2
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.
Note:  = VQ/(It) only works for vertical
shear stress in the web; doesn’t apply to
(vertical) shear stress in the flanges.
Read: Example 5-14
Homework: 5.10-3, 8, 9
10
Built-up Beams:
Made from several pieces of material:
Box beam
(wood)
©
20
01
Br
oo
ks
/C
ol
e,
a
di
vi
si
on
of
Th
o
m
so
n
Le
ar
ni
ng
,
In
c.
Th
o
m
so
n
Le
ar
ni
ng
Glued
laminated
(glulam
beam)
™
is
a
tra
de
m
ar
k
us
ed
he
rei
n
un
de
r
lic
en
se
.
11
Plate girder

Must ensure shear forces at connections
are acceptable

Calculations involve shear flow.
Concept
of shear flow: useful for
analyzing built-up beams

Connectors (e.g. nails, screws) placed
at fixed discrete intervals, or specified
by strength per meter (e.g. welding)
12
b(y)

M+M)y/I
R

My/I
L
y
c1

y
y
n.a.
n.s.
M(x)+M
M(x)
x
x
Shear flow,
Fshear
f  lim
x  0
x
bx
VQbx
 lim
 lim
 x  0 x
x  0 Ibx
VQ
 f 
I
(5-52)
Note: f is the shear force per unit
distance on the small horizontal area
(bx) along x direction; Q is for the
(blue) partial cross-sectional area
13
Example: Problem 5.11-4
Nails inserted at s = 100mm apart; each nail
can take F = 750N. Determine Vmax.
flange
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
Consider the two longitudinal contact
surfaces (in and out of the page, in red):
I = Ibox = Iout – Iin = 341.1106 mm4
Q = Qflange = Aflange y flange
= (25025)(125+25/2) = 859.4103 mm3
14
Allowable shear flow (force per distance) is
2 F
fallow = s
VQ 2  F
but f < fallow  I  s
2  FI (2)(0.75kN )(341.1 106 mm4 )
V 

 Vmax
3
3
Qs
(859.5  10 mm )(100mm)
 Vmax  5.95 kN
Example 5-16:
800 N per screw allowed; V = 10.5 kN; determine maximum s
15
Does (5-52) still work for a vertical face (and
how to use it)?
Yes, and here’s a separate derivation (not in
the book) to help you see why:
L
x
j


g
R
h
b
k
f
i
y
n.a. (z-axis)
x
F
x
x+x
 0    L dA  2bx    R dA
fghi
2bx 
fghi
M ( x  x) y
M ( x) y
dA

dA
fghi

I
I
fghi
16
1 M ( x  x)  M ( x)
2b 
ydA

I
x
fghi
In the limit x 0
 2 
VQ fghi
Ib , but
VQ fghibx
2bx
f  lim
 lim
x 0 x
x 0
Ibx
 f 
VQ fghi
I
again, but note that

Q = Qfghi

f = shear flow on the vertical faces
Now we can do Example 5-16:
Qfghi = Afghi y fghi
= (18040)(140 – 20)mm3 = 864000 mm3
I = 264.2106 mm4
17
f = VQ/I
= (10500N)(864000mm3)/(264.2106 mm4)
= 34.3 N/mm
As each nail can take F = 800N, 2 nails can
withstand 2F over a distance of s (mm)
Hence,
2F
N
2800 N 
 34.3
s
 smin
s
mm
34.3( N / mm)
 smin  46.6mm
 use a spacing of 45mm (say) for the
convenience of workers and more safety.
Homework (built-up beams and shear flow):
Do 5.11-2, 5.11-6, 5.11-10
Tutorial will cover 5.11-3, 5.11-9
18
Beams with Axial Loads
Inclined force P = Sxˆ  Qyˆ applied at end of
beam, through the centroid of cross section:
M(x) = Q(L – x) 
 = -My/I
In addition,
N = S   = N/A
 combined (normal) stress,

N My

A
I
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.

 combined 
19
(5-53)
Example 5.12-2
Determine max.
tensile &
compressive
stresses, t and c
at base of
aluminum pole
At base of pole:
P = W1 + W2 = (2300 + 330) = 2630N,
M = W2(1.2m) = 396 Nm
Simple geometry  A = 11706 mm2,
S = I/c = 63.17106mm4/(225mm/2)5.62105mm3
t  
P M

A S
20
= -2630N/11706mm2 + 396103Nmm/5.62105mm3
 0.48 MPa
c  
P M

A S
= -2630N/11706mm2 - 396103Nmm/5.62105mm3
 -0.93 MPa
Homework (beam with axial loads):
Do: 5.12-5, 5.12-9
Read: Example 5-17
Tutorial will cover: 5.12-6, 5.12-11
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