Chapter 5: Stresses in Beams
Example (4-point loading):
Pure bending: beam deflects under a
constant bending moment (M).
Recall V = dM/dx
 when M = constant, V = 0.
Non-uniform bending:
V  0  M(x) varies along the beam
Goal: given M, obtain the stresses
in the beam.
Geometric assumptions (Kirchoff):
M
M
1. beam undergoes small
deformations
2. there is a neutral surface (n.s.)
whose fibers bend without
extending/ contracting
3. cross sections remain plane and
normal to n.s.
Hence, beam deflection 
rotation of cross sectional planes
Euler-Bernoulli beam theory for long,
slender beams (for short, deep beams:
Timoshenko beam theory).
Curvature and longitudinal strains
Geometry of bending in xy plane:
z  neutral
axis (n.a.)
(intersection
of n.s. with
y-z plane)
ef / eO’ = 
 ef = eO’ 
y
Cross section of
prismatic bar
(symmetric about y)
Compression above
Neutral surface (n.s.)
Tension below
For a short fiber e-f at distance y from
n.s., its elongation is
(x,y) = ( – y)y
Hence, the strain in fiber e-f,
x =  / L0 = y
x = yy
(5-4)
where
   (5-1)
* Axial strain varies linearly with y,
regardless of material properties!
(you may ignore (5-2,5-3) for now)
 constant for pure bending, but
x) in general (“radius of
curvature”);  “curvature”
 y is measured upwards from the
n.a. (down = -ve values); y has same
value before/after bending
Example 5-1
Given:
L = 16’;
h = 12”;
bottom fiber has x
= 0.125% strain;
distance from n.a. to
bottom fiber is 6”
(i.e. y = -6”).
Determine ,  and
the deflection  of
the beam at midspan.
Answer:
Homework: Prob. 5.4-6
Normal stresses in beams
For linear elastic material that is
homogeneous & isotropic,  = E
Together with x result from (5.4)
 x = Ey= Ey  
with dFx
= xdA
n.s.
n.a.
+ve
 But where exactly is the n.a.
located (needed to evaluate y in (57))?
Apply force equilibrium: Fx = 0 
 dFx= 0
A
 A xdA =  A EydA = 0
Taking out constants E and ,
 A ydA = 0
(5-8)
Recall: y-distance from origin to
centroid, y  (A ydA) / A
 y = 0, i.e.
The neutral (z) axis goes through the
centroid of the cross-section.
This defines the origin (z0, y0).
(Appendix D gives centroid results in
terms of distance to corners)
 But how can we find  when given
M (our ultimate goal)?
Apply moment equilibrium about z:
MZ = 0
 –M + A dM = 0
M
but dM = –(–Ey)dAy
 M = A Ey2dA
 M = EA y2dA
But A y2dA  I (mmt. of inertia / 2nd
mmt. of area w.r.t. the z-axis)
 EI
(5-12)
(Moment-curvature equation;
EI  flexural rigidity)
But (5-7) gives x / Ey, hence
x y/  
(flexure formula for bending stress)
With the most remote fibers at c1 & c2
from n.a., respectively, the maximum
normal stresses are:
1 = -Mc1 / I  -M/S1 (5-15a)
2 = Mc2 / I  M/S2 (5-15b)
S1,2 are called the section moduli. 

1 = -M/(I/c1)
 -M/S1
2 = M/(I/c2)
 M/S2
If the cross section is symmetric not
only about y, but also about z 
doubly symmetric cross section
 c1 = c2 = c; S = I / c

Rectangular
cross section:
I = bh3 / 12
 Srec = I / (h/2)
= bh2 / 6
Circular cross
Circular
section: cross
section:
I = d4 / 64
I
=S
dcir4 =/ 64
I / (d/2)
 Scir==d
I /3(d/2)
/ 32
= d3 / 32
Circular cross
section:
For slender beams, presence of shear
(V  0  dM/dx  0) does not
significantly
affect
the
normal
(bending) stresses.
 For most practical purposes,
formula (5.13) can be used even when
M  constant, i.e. for non-uniform
bending (see text p. 325).
Example 5-3:
Yet another way to find the reaction
forces: use Excel Solver:
Result from previous analysis: Mmax
= 151.568 k-ft when x = 9 ft, where
the max. bending stresses occur.
S = bh2 / 6
= (1/6)(8.75”)(27”)2 = 1063 in3;
 Maximum tensile & compressive
stresses, t and c are:
t = Mmax / S
= (151.6 k-ft)(12”/ft) / (1063 in3) =
1710 psi (at bottom of beam)
c = –Mmax / S
= –1710 psi (at top of beam)
Example 5-4:
Max. values of M were found as:
M+ = M|x=1.125 = 2.025 kNm;
M- = M|x=3 = -3.6 kNm
Centroid & IZ calculations: easy, e.g.
can use AutoCAD 
c1 ~ 18.48 mm; IZ ~ 2469000 mm4.
(c2 = 80 mm – c1)
For top & bottom fiber of beam, the
respective section moduli are:
S1 = IZ / c1 = 133600 mm3;
S2 = IZ / c2 = 40100 mm3;
Hence,
at x=1.125 where M+= 2.025 kNm,
comprs. = -M+ / S1 = -15.2 MPa
tensile = M+ / S2 = 50.5 MPa
Furthermore,
at x = 3 where M- = -3.6 kNm:
tensile = -M- / S1 = +26.9 MPa
comprs. = M- / S2 = -89.8 MPa
Largest bending stresses are:
tensile, max = 50.5 MPa at x=1.125m;
comprs., max = -89.8 MPa at x = 3m
Design of Beams
(for bending stress)
1.For given material, determine
allowable bending stress, e.g.
allow = Y/ safety factor
2.From given loading, determine
M(x) and hence Mmax (self-weight
neglected at this stage)
3.Find minimum section modulus
required,
Smin = Mmax / allow
(assumed: doubly symetric; same allow
for tension/compression; if not, repeat
calculations for S2)
4.Select the minimum cross-section
(min. weight) that has the
required modulus, S  Smin
5.With selected size & shape of
cross-section, calculate beam’s
self-weight (per length), add it to
w(x) and re-calculate V, Mmax and
hence Smin. If the currently
selected S is inadequate, select
the next (larger) cross section, and
repeat until S  Mmax/allow is
satisfied.
 S can increase faster than weight
(~AL): e.g. rectangular cross section:
S = bh2 / 6 = (1/6)(bh)(h) ~ 0.17 Ah,
one may increase h and decrease b (in
this case, A can be kept constant) to
increase S.
 Theoretically “ideal” shape (of height h):
place half of the entire area A at h/2
above n.a., and the other half at h/2
below (Fig. c). This gives
S = 0.5 Ah > Ssquare.
In reality, one can achieve around
S ~ 0.35 Ah
using wide-flange beams (Fig. d)
(web has to contain some of the
material and cannot be too thin, to
prevent shear failure)
Circular beams are inefficient: with
equal areas (h = b =  r),
Ssquare / Scircle
= (bh2 / 6) / r3 / 4
~ 1.18, i.e.
Ssquare = 118% Scircle
Beams with square cross-section
outperform circular ones in
resisting bending. (Figs. a, b)
Standard beam sizes & shapes:
See Appendix E (structural steel) & F
(lumber). Example:
W
30
211

wide-flange shape nominal depth
(inches)
weight per
length (lb/ft)
Table E-1: other details such as actual
depth, flange width, moment of
inertia, section modulus, etc.
Other standard shapes:
S shapes: C shapes:
I-beams
channels
L shapes:
angles
Example 5-5:
Span length, L = 12 ft
Loading: UDL, q = 420 (lb/ft)
Neglecting self-weight for now, use
result in (4-15) for the max. bending
moment (for uniform q):
Mmax = q L2 / 8 (at mid-span)
= (420/12 lb/in) (12*12 in)2 / 8
= 90720 lb-in
 Allowable stress,
allow = 1800 psi, while
max = Mmax / S < allow
 S > (90720/1800)in3
= 50.40 in3
Appendix F gives S = 52.72 in3 for a
312 in (nominal) beam, whose
weight per unit length is
wself = 2.5” 11.25”  (35 lb/(12”)3)
= 0.5697 lb/in (or 6.8359 lb/ft)
(Note: wself < 2% of q)
Hence, new total uniform load is
q + wself = (35 + 0.5697)
= 35.5697 lb/in,
Recalculate:
Mmax = (q + wself) L2 / 8
= 92196.56 lb-in
Sreq = (92196.56/1800)in3
= 51.22in3
Current selection:
S = 52.72 in3 > Sreq, hence the 312
in. beam suffices.
Example 5-7:
The calculations are routine yet a bit
tedious, as we will need to repeat them
(with self-weight included) after
selecting the cross section. We can use a
computer algebra system (CAS) to
expedite the work (e.g. TI-92,
Mathematica, Maple, etc.):
Homework:
Read:
Example 5-6 (you may ignore
Example 5-8).
Do:
Problem 5-6-3, 5-6-5, 5-6-12, 5-6-14.
(Use of CAS is optional)
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.
 No slippage  horizontal shear
stresses must develop along “glued”
contact surface.
For a rectangular beam subject to shear
force V(x): examine cross section at x:
c
xy
y
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
x
Note:  is interpreted as the average shear stress on area bx, as it may vary along the width b
F
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
 x  b( y) 
Ix
As
c1
 yb( y)dy
y
M
V 
x 0 x
lim
formula for average shear
stress on fibers at distance y from n.a.:
VQ
 ( y) 
Ib
(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, i.e. from y
to extreme fibre.
Caution:
 (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).
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
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)
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 

 
1
Q    b h 2  h12  t h12  4 y 2
8
which, together with
 (5.45)
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)
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; it does not apply
to (vertical) shear stress in the flanges.
Read: Example 5-14
Homework: 5.10-3, 8, 9
Built-up Beams & Shear Flow:
Built-up beams: made from several pieces:
Box beam
(wood)
Glued
laminated
(glulam
beam)
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Plate girder

Must ensure shear forces at connections
are acceptable

Calculations involve shear flow – a
useful concept for analyzing the
connection areas in built-up beams

Connectors (e.g. nails, screws) usually
placed at fixed discrete intervals, or
specified by strength per meter (e.g.
welding)

Require: strength of connection (force
that can be resisted per meter along the
beam)  loading (shear force per meter)
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
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
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
Does (5-52) still work for a vertical face?
Yes! Derivation: let   average shear stress
over each side area of bx (e.g. jghk):
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
fghi
M ( x  x) y
M ( x) y
 2bx  
dA  
dA
I
I
fghi
fghi
1 M ( x  x)  M ( x)
 x  2b 
ydA

I
x
fghi
In the limit x 0
 2 
VQ fghi
Ib , hence
VQ fghibx
2bx
f  lim
 lim
x 0 x
x 0
Ibx
 f 
VQ fghi
I
again, but note that
 f = shear flow on the vertical faces
 Q = Qfghi
  = average shear stress along g-h (or f-i)
Now we can do Example 5-16:
Qfghi = Afghi y fghi
= (18040)(140 – 20)mm3 = 864000 mm3
I = 264.2106 mm4
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