1) The document discusses the calculation of bending and shear stresses in a beam with a rectangular cross-section. 2) Shear stress is derived to be proportional to the first moment (Q) of the cross-sectional area above the level of interest. 3) For a rectangular cross-section, the maximum shear stress occurs at the neutral axis and is 50% larger than the average shear stress.

1. Lecturer;
Dr. Dawood S. Atrushi
Dec. 2014 – Jan. 2015
Bending- and Shearing
Stresses
- y1
2
)
of the beam

2. Shear Stress in Beam of
Rectangular Cross Section
¢ For a beam subjected to M and V with
rectangular cross section having width
b and height h, the shear stress τ
acts parallel to shear force V.
January, 2015Bending Stresses - DAT2
ular Cross Section
d V with rectangular cross section
e shear stress acts parallel to the
the width of the beam
the a
small
act
lly as
subjected the a
solate a small
stresses act
d horizontally as
urfaces are free,
be vanish, i.e.
2
tion subjected the a
e isolate a small
ear stresses act
nied horizontally as
om surfaces are free,
must be vanish, i.e.
h/2

3. The top and bottom surfaces are free,
then the shear stress τmust be vanish,
i.e.;
January, 2015Bending Stresses - DAT3
shown
the top and bottom surfaces are fre
then the shear stress must be vanish, i.
= 0 at y = h/2
for two equal rectangular beams o
height h subjected to a concentrate
load P, if no friction between the beam
each beam will be in compression above i
N.A., the lower longitudinal line of th
upper beam will slide w.r.t. the upper lin
mall
act
as
ee,
i.e.
of
ted
ms,
its
the
and bottom surfaces are free,
stress must be vanish, i.e.
y = h/2
qual rectangular beams of
ubjected to a concentrated
o friction between the beams,
be in compression above its
er longitudinal line of the
ll slide w.r.t. the upper line
am
beam of height 2h, shear stress must exist along N.A. to

4. ¢ Consider a small
section of the beam
subjected M and V
in left face and M
+dM and V+dV in
right face.
¢ For the element
mm1p1p, τ acts on
p1p and no stress on
mm1.
January, 2015Bending Stresses - DAT4
onsider a small section of the beam
ected M and V in left face and
dM and V + dV in right face
or the element mm1p1p, acts on
and no stress on mm1
the beam is subjected to pure bending
= constant), x acting on mp and
must be equal, then = 0 on
or nonuniform bending, M acts on mn and M + dM acts on
, consider dA at the distance y form N.A., then on mn
small section of the beam
and V in left face and
V + dV in right face
ment mm1p1p, acts on
ess on mm1
s subjected to pure bending
x acting on mp and
equal, then = 0 on

5. ¢ For non-uniform
bending, M acts on
mn and M + dM acts
on m1n1, consider dA
at the distance y
form N.A., then on
mn
January, 2015Bending Stresses - DAT5
orm bending, M acts on mn and M + dM acts
dA at the distance y form N.A., then on mn
M y
= CC dA
I
tal horizontal force on mp is
M y
CC dA
I
(M + dM) y
p1p and no stress on mm1
if the beam is subjected to pure bending
(M = constant), x acting on mp and
m1p1 must be equal, then = 0 on
pp1
for nonuniform bending, M acts on mn and M +
m1n1, consider dA at the distance y form N.A., then o
M y
x dA = CC dA
I
hence the total horizontal force on mp is
nonuniform bending, M acts on mn and M + dM acts o
consider dA at the distance y form N.A., then on mn
M y
x dA = CC dA
I
nce the total horizontal force on mp is
M y
F1 = CC dA
I
milarly
(M + dM) y
F2 = CCCCC dA
I

6. Hence the total horizontal force on mp
is;
January, 2015Bending Stresses - DAT6
x dA = CC dA
I
hence the total horizontal force on mp is
M y
F1 = CC dA
I
similarly
(M + dM) y
F2 = CCCCC dA
I
and the horizontal force on pp1 is
F3 = b dx
equation of equilibrium
x dA = CC dA
I
hence the total horizontal force on mp is
M y
F1 = CC dA
I
similarly
(M + dM) y
F2 = CCCCC dA
I
and the horizontal force on pp1 is
F3 = b dx
equation of equilibrium
21
M y
F1 = CC dA
I
similarly
(M + dM) y
F2 = CCCCC dA
I
and the horizontal force on pp1 is
F3 = b dx
equation of equilibrium
F3 = F2 - F1
Similarly
hence the total horizontal force on mp is
M y
F1 = CC dA
I
similarly
(M + dM) y
F2 = CCCCC dA
I
and the horizontal force on pp1 is
F3 = b dx
equation of equilibrium
And the horizontal force on pp1 is;

7. Equation of equilibrium;
January, 2015Bending Stresses - DAT7
21
F3 = b dx
equation of equilibrium
F3 = F2 - F1
(M + dM) y M y
b dx = CCCCC dA - CC dA
I I
dM 1 V
= CC C y dA = CC y dA
dx Ib I b
denote Q = y dA is the first moment of the cross s
above the level y (area mm1p1p) at which the shear stress act
V Q
= CC shear stress formula
I b
(M + dM) y M y
b dx = CCCCC dA - CC dA
I I
dM 1 V
= CC C y dA = CC y dA
dx Ib I b
denote Q = y dA is the first moment of the cro
above the level y (area mm1p1p) at which the shear stress
V Q
= CC shear stress formula
I b
Denote Q = ∫y dA is the first moment of
the cross section area above the level y
(area mm1p1p) at which the shear stress
τ, then;

8. Then, the shear stress formula will be;
January, 2015Bending Stresses - DAT8
denote Q = y dA is the first moment of
above the level y (area mm1p1p) at which the shear stre
V Q
= CC shear stress formula
I b
for V, I, b are constants, ~ Q
for a rectangular cross section
h h/2 - y1 b h2
Q = b (C - y1) (y1 + CCC) = C (C
2 2 2 4
V h2
then = CC (C - y1
2
)
2 I 4
For V, I, b are constants,
τ ~ Q for a rectangular
cross section
dx Ib I b
enote Q = y dA is the first moment of the cross section
e the level y (area mm1p1p) at which the shear stress acts, then
V Q
= CC shear stress formula
I b
or V, I, b are constants, ~ Q
or a rectangular cross section
h h/2 - y1 b h2
Q = b (C - y1) (y1 + CCC) = C (C - y1
2
)
2 2 2 4
V h2
2
dM 1 V
= CC C y dA = CC y dA
dx Ib I b
denote Q = y dA is the first moment of the cross section
above the level y (area mm1p1p) at which the shear stress acts, then
V Q
= CC shear stress formula
I b
for V, I, b are constants, ~ Q
for a rectangular cross section
h h/2 - y1 b h2
Q = b (C - y1) (y1 + CCC) = C (C - y1
2
)
2 2 2 4
V h2
then = CC (C - y1
2
)

9. Then;
January, 2015Bending Stresses - DAT9
1
Q = b (C - y1) (y1 + CCC) = C (C - y1
2
)
2 2 2 4
V h2
then = CC (C - y1
2
)
2 I 4
= 0 at y1 = h/2, max occurs
at y1 = 0 (N.A.)
V h2
V h2
3 V 3
max = CC = CCCC = CC = C ave
8 I 8 b h3
/12 2 A 2
max is 50% larger than ave
V = resultant of shear stress, V and in th
direction
Q = b (C - y1) (y1 + CCC) = C (C
2 2 2 4
V h2
then = CC (C - y1
2
)
2 I 4
= 0 at y1 = h/2, max occurs
at y1 = 0 (N.A.)
V h2
V h2
3 V
max = CC = CCCC = CC =
8 I 8 b h3
/12 2 A
max is 50% larger than ave
V = resultant of shear stress, V
direction
h/2 - y1 b h2
y1) (y1 + CCC) = C (C - y1
2
)
2 2 4
h2
C (C - y1
2
)
I 4
= h/2, max occurs
A.)
V h2
3 V 3
= CCCC = CC = C ave
8 b h3
/12 2 A 2
ger than ave
nt of shear stress, V and in the same
h h/2 - y1 b h2
Q = b (C - y1) (y1 + CCC) = C (C - y1
2
)
2 2 2 4
V h2
then = CC (C - y1
2
)
2 I 4
= 0 at y1 = h/2, max occurs
at y1 = 0 (N.A.)
V h2
V h2
3 V 3
max = CC = CCCC = CC = C
8 I 8 b h3
/12 2 A 2
max is 50% larger than ave
V = resultant of shear stress, V and
h h/2 - y1 b h2
Q = b (C - y1) (y1 + CCC) = C (C - y1
2
)
2 2 2 4
V h2
= CC (C - y1
2
)
2 I 4
0 at y1 = h/2, max occurs
= 0 (N.A.)
V h2
V h2
3 V 3
max = CC = CCCC = CC = C ave
8 I 8 b h3
/12 2 A 2
is 50% larger than ave
V = resultant of shear stress, V and in the sam
for a rectangular cross section
h h/2 - y1 b h2
Q = b (C - y1) (y1 + CCC) = C (C - y1
2
)
2 2 2 4
V h2
then = CC (C - y1
2
)
2 I 4
= 0 at y1 = h/2, max occurs
at y1 = 0 (N.A.)
V h2
V h2
3 V 3
max = CC = CCCC = CC = C ave
8 I 8 b h3
/12 2 A 2
max is 50% larger than ave
V = resultant of shear stress, V and in the
h h/2 - y1 b h
Q = b (C - y1) (y1 + CCC) = C (C - y1
2
)
2 2 2 4
V h2
then = CC (C - y1
2
)
2 I 4
= 0 at y1 = h/2, max occurs
at y1 = 0 (N.A.)
V h2
V h2
3 V 3
max = CC = CCCC = CC = C
8 I 8 b h3
/12 2 A 2
max is 50% larger than ave
V = resultant of shear stress, V and

10. Example 1
A metal beam with span L = 1m,
q=28kN/m, b=25mm and h=100mm.
Determine bending and shear stresses
at point C.
January, 2015Bending Stresses - DAT10
23
ear force varies continuously along the beam, the warping of cross
due to shear strains does not substantially affect the longitudinal
y more experimental investigation
it is quite justifiable to use the flexure formula in the case of
rm bending, except the region near the concentrate load acts of
ly change of the cross section (stress concentration)
5-11
al beam with span L = 1 m
8 kN/m b = 25 mm h = 100 mm

11. Example 1
Solution
¢ The shear force VC
and bending
moment MC at the
section through C
are found;
MC = 2.24 kNm
VC = - 8.4 kN
January, 2015Bending Stresses - DAT11
C and C at point C
force VC and bending
C at the section through C
= 2.24 kN-m
= - 8.4 kN
nt of inertia of the section is

12. The moment of inertia of the section is;
January, 2015Bending Stresses - DAT12
VC = - 8.4 kN
the moment of inertia of the section is
b h3
1
I = CC = C x 25 x 1003
= 2,083 x 10
12 12
normal stress at C is
M y 2.24 x 106 N-mm x 25 mm
C = - CC = - CCCCCCCCCCC =
I 2,083 x 103
mm4
shear stress at C, calculate QC first
AC = 25 x 25 = 625 mm2
yC = 37.5 mm
QC = AC yC = 23,400 mm3
kN-m
kN
ia of the section is
1
= C x 25 x 1003
= 2,083 x 103
mm4
12
is
2.24 x 106 N-mm x 25 mm
= - CCCCCCCCCCC = - 26.9 MPa
2,083 x 103
mm4
calculate QC first
The normal stress at C is;
VC = - 8.4 kN
the moment of inertia of the section is
b h3
1
I = CC = C x 25 x 1003
= 2,083 x 103
m
12 12
normal stress at C is
M y 2.24 x 106 N-mm x 25 mm
C = - CC = - CCCCCCCCCCC = - 2
I 2,083 x 103
mm4
shear stress at C, calculate QC first
AC = 25 x 25 = 625 mm2
yC = 37.5 mm
MC = 2.24 kN-m
VC = - 8.4 kN
e moment of inertia of the section is
b h3
1
I = CC = C x 25 x 1003
= 2,083 x 103
mm4
12 12
ormal stress at C is
M y 2.24 x 106 N-mm x 25 mm
C = - CC = - CCCCCCCCCCC = - 26.9 MPa
I 2,083 x 103
mm4
hear stress at C, calculate QC first

13. Shear stress at C, calculate Qc first;
January, 2015Bending Stresses - DAT13
I 2,083 x 103
mm4
shear stress at C, calculate QC first
AC = 25 x 25 = 625 mm2
yC = 37.5 mm
QC = AC yC = 23,400 mm3
VC QC 8,400 x 23,400
C = CCC = CCCCCCC = 3.8 MPa
I b 2,083 x 103
x 25
the stress element at point C is shown
Example 5-12
a wood beam AB supporting two
concentrated loads P
b = 100 mm h = 150 mm
M y 2.24 x 106 N-mm x 25 mm
- CC = - CCCCCCCCCCC = - 26.9 MPa
I 2,083 x 103
mm4
s at C, calculate QC first
25 x 25 = 625 mm2
yC = 37.5 mm
= AC yC = 23,400 mm3
VC QC 8,400 x 23,400
= CCC = CCCCCCC = 3.8 MPa
I b 2,083 x 103
x 25
element at point C is shown
2
beam AB supporting two
loads P
0 mm h = 150 mm
M y 2.24 x 106 N-mm x 25 mm
C = - CC = - CCCCCCCCCCC = - 26.9 MP
I 2,083 x 103
mm4
shear stress at C, calculate QC first
AC = 25 x 25 = 625 mm2
yC = 37.5 mm
QC = AC yC = 23,400 mm3
VC QC 8,400 x 23,400
C = CCC = CCCCCCC = 3.8 MPa
I b 2,083 x 103
x 25
the stress element at point C is shown
Example 5-12
a wood beam AB supporting two
concentrated loads P
b = 100 mm h = 150 mm
M y 2.24 x 106 N-mm x 25 mm
C = - CC = - CCCCCCCCCCC = - 26.9 MP
I 2,083 x 103
mm4
shear stress at C, calculate QC first
AC = 25 x 25 = 625 mm2
yC = 37.5 mm
QC = AC yC = 23,400 mm3
VC QC 8,400 x 23,400
C = CCC = CCCCCCC = 3.8 MPa
I b 2,083 x 103
x 25
the stress element at point C is shown
Example 5-12
a wood beam AB supporting two
concentrated loads P
b = 100 mm h = 150 mm
h3
1
C = C x 25 x 1003
= 2,083 x 103
mm4
2 12
t C is
y 2.24 x 106 N-mm x 25 mm
C = - CCCCCCCCCCC = - 26.9 MPa
2,083 x 103
mm4
C, calculate QC first
25 = 625 mm2
yC = 37.5 mm
AC yC = 23,400 mm3
C QC 8,400 x 23,400
The stress
element at point
C is shown
3.8 MPA
3.8 MPA
26.9 MPA26.9 MPA

14. Example 2
A wood beam AB supporting two
concentrated loads P. Cross sectional
dimentions; b = 100mm, h = 150mm, a =
0.5 m. Given σallow = 11MPa, τallow = 1.2
Mpa, determine Pmax.
January, 2015Bending Stresses - DAT14
ate QC first
m2
yC = 37.5 mm
3,400 mm3
8,400 x 23,400
CCCCCCC = 3.8 MPa
2,083 x 103
x 25
C is shown
pporting two
50 mm
a = 0.5 m allow = 11 MPa allow = 1.2 MPa
determine Pmax
the maximum shear force and bending moment are
Vmax = P Mmax = P a
the section modulus and area are
b h2
S = CC A = b h
6

15. Example 2
Solution
The maximum shear force and bending
moment are
Vmax = P Mmax = Pa
January, 2015Bending Stresses - DAT15
The section modulus and are are;
a = 0.5 m allow = 11 MPa al
determine Pmax
the maximum shear force and bending momen
Vmax = P Mmax = P a
the section modulus and area are
b h2
S = CC A = b h
6
maximum normal and shear stresses are

16. Maximum normal and shear stresses
are;
January, 2015Bending Stresses - DAT16
b h2
S = CC A = b h
6
maximum normal and shear stresses are
Mmax 6 P a 3 Vmax 3 P
max = CC = CC max = CCC = CCC
S b h2
2 A 2 b h
allow b h2
11 x 100 x 1502
Pbending = CCCC = CCCCCCC = 8,250 N = 8.25 kN
6 a 6 x 500
2 allow b h 2 x 1.2 x 100 x 150
Pshear = CCCC = CCCCCCCC = 12,000 N = 12 kN
3 3
Pmax = 8.25 kN
8-9 Shear Stresses in Beam of Circular Cross Section
V Q r4
b h2
S = CC A = b h
6
maximum normal and shear stresses are
Mmax 6 P a 3 Vmax 3 P
max = CC = CC max = CCC = CCC
S b h2
2 A 2 b h
allow b h2
11 x 100 x 1502
Pbending = CCCC = CCCCCCC = 8,250 N = 8.25 kN
6 a 6 x 500
2 allow b h 2 x 1.2 x 100 x 150
Pshear = CCCC = CCCCCCCC = 12,000 N = 12 kN
3 3
Pmax = 8.25 kN
8-9 Shear Stresses in Beam of Circular Cross Section
V Q r4
= CC I = CC for solid section
ce and bending moment are
Mmax = P a
d area are
A = b h
shear stresses are
6 P a 3 Vmax 3 P
= CC max = CCC = CCC
b h2
2 A 2 b h
b h2
11 x 100 x 1502
CC = CCCCCCC = 8,250 N = 8.25 kN
6 x 500
h 2 x 1.2 x 100 x 150
C = CCCCCCCC = 12,000 N = 12 kN
3
Vmax = P Mmax = P a
the section modulus and area are
b h2
S = CC A = b h
6
maximum normal and shear stresses are
Mmax 6 P a 3 Vmax 3 P
max = CC = CC max = CCC = CCC
S b h2
2 A 2 b h
allow b h2
11 x 100 x 1502
Pbending = CCCC = CCCCCCC = 8,250 N = 8.25 kN
6 a 6 x 500
2 allow b h 2 x 1.2 x 100 x 150
Pshear = CCCC = CCCCCCCC = 12,000 N = 12 kN
3 3
Pmax = 8.25 kN
Vmax = P Mmax = P a
the section modulus and area are
b h2
S = CC A = b h
6
maximum normal and shear stresses are
Mmax 6 P a 3 Vmax
max = CC = CC max = CCC =
S b h2
2 A
allow b h2
11 x 100 x 1502
Pbending = CCCC = CCCCCCC = 8,250 N
6 a 6 x 500
2 allow b h 2 x 1.2 x 100 x 150
Pshear = CCCC = CCCCCCCC = 12,000
3 3
Pmax = 8.25 kN

17. Shear Stresses in Beam of
Circular Cross Section
The shear stress at the neutral axis;
January, 2015Bending Stresses - DAT17
Pmax = 8.25 kN
8-9 Shear Stresses in Beam of Circular Cross Section
V Q r4
= CC I = CC for solid section
I b 4
the shear stress at the neutral axis
r2
4 r 2 r3
Q = A y = (CC) (CC) = CC
2 3 3
hear = CCCC = CCCCCCCC = 12,000 N = 1
3 3
max = 8.25 kN
Stresses in Beam of Circular Cross Section
V Q r4
= CC I = CC for solid section
I b 4
ar stress at the neutral axis
r2
4 r 2 r3
= A y = (CC) (CC) = CC b = 2 r
Pshear = CCCC = CCCCCCCC = 1
3 3
Pmax = 8.25 kN
8-9 Shear Stresses in Beam of Circular Cross Section
V Q r4
= CC I = CC for solid section
I b 4
the shear stress at the neutral axis
r2
4 r 2 r3
Q = A y = (CC) (CC) = CC
2 3 3
25
Q r
C I = CC for solid section
4
at the neutral axis
r2
4 r 2 r3
y = (CC) (CC) = CC b = 2 r
2 3 3

18. January, 2015Bending Stresses - DAT18
V (2 r3
/ 3) 4 V 4 V
max = CCCCCC = CCC = CC
( r4
/ 4) (2 r) 3 r2
3 A
for a hollow circular cross section
2
I = C (r2
4
- r1
4
) Q = C (r2
3
- r1
3
)
4 3
b = 2 (r2 - r1)
then the maximum shear stress at N.A. is
V Q 4 V r2
2
+ r2r1 + r1
2
max = CC = CC (CCCCCC)
I b 3 A r2
2
+ r1
2
V (2 r3
/ 3) 4 V 4 V 4
CCCCCC = CCC = CC = C ave
( r4
/ 4) (2 r) 3 r2
3 A 3
rcular cross section
4 V 4 V 4
C = CCC = CC = C ave
r) 3 r2
3 A 3
ction
2
Q = C (r2
3
- r1
3
)
3
ss at N.A. is
2 2

19. For a hollow circular cross section;
January, 2015Bending Stresses - DAT19
for a hollow circular cross section
2
I = C (r2
4
- r1
4
) Q = C (r2
3
- r1
3
)
4 3
b = 2 (r2 - r1)
then the maximum shear stress at N.A. is
V Q 4 V r2
2
+ r2r1 + r1
2
max = CC = CC (CCCCCC)
I b 3 A r2
2
+ r1
2
where A = (r2
2
- r1
2
)
Example 5-13
circular cross section
2
C (r2
4
- r1
4
) Q = C (r2
3
- r1
3
)
4 3
2 (r2 - r1)
mum shear stress at N.A. is
V Q 4 V r2
2
+ r2r1 + r1
2
CC = CC (CCCCCC)
I b 3 A r2
2
+ r1
2
A = (r2
2
- r1
2
)
Then the maximum shear
stress at N.A. is;
then the maximum shear stress at N.A.
V Q 4 V r2
2
+
max = CC = CC (CCC
I b 3 A r2
2
where A = (r2
2
- r1
2
)
Example 5-13
a vertical pole of a circular tube
d2 = 100 mm d1 = 80 mm P = 6,675
(a) determine the max in the pole
(b) for same P and same max, calcula
( r / 4) (2 r) 3 r 3 A
for a hollow circular cross section
2
I = C (r2
4
- r1
4
) Q = C (r2
3
- r1
3
)
4 3
b = 2 (r2 - r1)
then the maximum shear stress at N.A. is
V Q 4 V r2
2
+ r2r1 + r1
2
max = CC = CC (CCCCCC)
I b 3 A r2
2
+ r1
2
where A = (r2
2
- r1
2
)
Example 5-13
V (2 r3
/ 3) 4 V 4 V
max = CCCCCC = CCC = CC
( r4
/ 4) (2 r) 3 r2
3 A
for a hollow circular cross section
2
I = C (r2
4
- r1
4
) Q = C (r2
3
- r1
3
)
4 3
b = 2 (r2 - r1)
then the maximum shear stress at N.A. is
V Q 4 V r2
2
+ r2r1 + r1
2
max = CC = CC (CCCCCC)
I b 3 A r2
2
+ r1
2
where A = (r 2
- r 2
)

20. Example 3
A vertical pole of a
circular tube d2 =100mm,
d1=80mm, P=6,675 N.
(a) determine the σmax in
the pole
(b) for same P and same
τmax, calculate d0 of a
solid circular pole
January, 2015Bending Stresses - DAT20
A = (r2 - r1 )
le of a circular tube
m d1 = 80 mm P = 6,675 N
e the max in the pole
e P and same max, calculate d0
rcular pole
imum shear stress of a circular

21. Shear Stress in the Webs
of Beams with Flanges
January, 2015Bending Stresses - DAT21
circular pole has a diameter approximately 5/8 that of
ress in the Webs of Beams with Flanges
am of wide-flange shape
shear force V, shear stress
complicated
he shear force is carried by shear
web
he shear stress at ef, the same
in the case in rectangular beam, i.e.
For a beam of wide-
flange shape subjected
to shear force V, shear
stress is much more
complicated. Most of
the shear force is
carried by shear
stresses in the web

22. mly distributed across t
valid with b = t
of
vided
the
area
n the
January, 2015Bending Stresses - DAT22
Consider the shear stress at ef, the same
assumption as in the case in rectangular
beam, i.e. τ parallel to y-axis and
uniformly distributed across t.
stresses in the web
consider the shear st
assumption as in the case
// y axis and uniform
V Q
= CC is still
I b
the first moment Q
the shaded area is di
into two parts, i.e.
upper flange and the
is much more complicated
most of the shear force is carried
stresses in the web
consider the shear stress at ef,
assumption as in the case in rectangular
// y axis and uniformly distributed
V Q
= CC is still valid with b
I b
the first moment Q of
the shaded area is divided
subjected to shear force V, shear stress
is much more complicated
most of the shear force is carried by shear
stresses in the web
consider the shear stress at ef, the same
assumption as in the case in rectangular beam, i.e.
// y axis and uniformly distributed across t
V Q
= CC is still valid with b = t
I b
the first moment Q of

23. mption as in the case in rectangular beam, i.e.
/ y axis and uniformly distributed across t
V Q
= CC is still valid with b = t
I b
he first moment Q of
shaded area is divided
two parts, i.e. the
er flange and the area
ween bc and ef in the January, 2015Bending Stresses - DAT23
Then the first moment of A1 and A2 w.r.t.
N.A. is;
A1 = b (C - C) A2 = t (C - y1)
2 2 2
then the first moment of A1 and A2 w.r.t. N.A. is
h1 h/2 - h1/2 h1/2 - y1
Q = A1 (C + CCCC) + A2 (y1 + CCCC)
2 2 2
b t
= C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)
8 8
V Q V b t
= CC = CC [C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)]
I b 8 I t 8 8
b h3
(b - t) h1
3
1
where I = CC - CCCC = C (b h3
- b h1
3
+ t h1
12 12 12
maximum shear stress in the web occurs at N.A., y1 = 0
2 2 2
then the first moment of A1 and A2 w.r.t. N.A. is
h1 h/2 - h1/2 h1/2 - y1
Q = A1 (C + CCCC) + A2 (y1 + CCCC)
2 2 2
b t
= C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)
8 8
V Q V b t
= CC = CC [C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)]
I b 8 I t 8 8
b h3
(b - t) h1
3
1
where I = CC - CCCC = C (b h3
- b h1
3
+ t h1
12 12 12
maximum shear stress in the web occurs at N.A., y1 = 0
V
then the first moment of A1 and A2 w.r.t. N.A. is
h1 h/2 - h1/2 h1/2 - y1
Q = A1 (C + CCCC) + A2 (y1 + CCCC)
2 2 2
b t
= C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)
8 8
V Q V b t
= CC = CC [C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)]
I b 8 I t 8 8
b h3
(b - t) h1
3
1
where I = CC - CCCC = C (b h3
- b h1
3
+ t h1
3
)
12 12 12
maximum shear stress in the web occurs at N.A., y1 = 0
V 2 2 2
h h1 h1
A1 = b (C - C) A2 = t (C - y1)
2 2 2
he first moment of A1 and A2 w.r.t. N.A. is
h1 h/2 - h1/2 h1/2 - y1
Q = A1 (C + CCCC) + A2 (y1 + CCCC)
2 2 2
b t
= C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)
8 8
V Q V b t
= CC = CC [C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)]
I b 8 I t 8 8
b h3
(b - t) h1
3
1
I = CC - CCCC = C (b h3
- b h1
3
+ t h1
3
)
12 12 12
h1
) A2 = t (C - y1)
2
A1 and A2 w.r.t. N.A. is
/2 - h1/2 h1/2 - y1
CCCC) + A2 (y1 + CCCC)
2 2
t
+ C (h1
2
- 4 y1
2
)
8
b t
C [C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)]
t 8 8
3

24. January, 2015Bending Stresses - DAT24
¢ Maximum shear stress in the web occurs
at N.A., y1 = 0
I b 8 I t 8 8
b h3
(b - t) h1
3
1
where I = CC - CCCC = C
12 12 12
maximum shear stress in the web occurs at N.A.,
V
max = CC (b h2
- b h1
2
+ t h1
2
)
8 I t
minimum shear stress occurs where the web mee
h1/2
V b
min = CC (h2
- h1
2
)
= CC = CC [C (h2
- h1
2
) + C (h1
2
- 4 y1
2
)]
I b 8 I t 8 8
b h3
(b - t) h1
3
1
where I = CC - CCCC = C (b h3
- b h1
3
+ t h1
3
)
12 12 12
maximum shear stress in the web occurs at N.A., y1 = 0
V
max = CC (b h2
- b h1
2
+ t h1
2
)
8 I t
minimum shear stress occurs where the web meets the flange, y1 =
h1/2
V b
min = CC (h2
- h1
2
)
8 I t
the maximum stress in the web is from 10% to 60% greater than the
I b 8 I t 8 8
b h3
(b - t) h1
3
1
where I = CC - CCCC = C (b h3
12 12 12
maximum shear stress in the web occurs at N.A., y1
V
max = CC (b h2
- b h1
2
+ t h1
2
)
8 I t
minimum shear stress occurs where the web meets th
h1/2
V b
min = CC (h2
- h1
2
)
8 I t
¢ Minimum shear stress occurs where the
web meets the flange, y = ±h1/2;
Where

25. with b = t
f
d
e
a
e
Shear stress distribution on the web;
January, 2015Bending Stresses - DAT25

26. January, 2015Bending Stresses - DAT26
¢ The shear force carried by the web
consists two parts, a rectangle of area h1
τmin and a parabolic segment of area
2/3 h1 (τmax - τmin)
28
the maximum stress in the web is from 10% to
minimum stress
the shear force carried by the web consists two pa
h1 min and a parabolic segment of area b h1 ( max
Vweb = h1 min + b h1 ( max - min)
t h1
= CC (2 max + min)
3
Vweb = 90% ~ 98% of total V
for design work, the approximation to calculate
V <= total shear force
¢ For design work, the approximation to
calculate τmax is
t h1
= CC (2 max + min)
3
Vweb = 90% ~ 98% of total V
for design work, the approximation to calculate
V <= total shear force
max = CC
t h1 <= web area

27. Example 4
January, 2015Bending Stresses - DAT27
a beam of wide-flange shape wit
and h1 = 290 mm, vertical shear forc
determine max, min and total
the moment of inertia of the cros
A beam of wide-flange
shape with b = 165 mm,
t = 7.5mm, h = 320 mm,
and h1 = 290 mm,
vertical shear force
V = 45 kN.
Determine τmax, τmin
and total shear force in
the web.

28. V
CC = 20.7 MPa
h1
a T-shaped cross section
= 24 mm h = 200 mm
(top of the web) and max
6 x 24 x 12 + 200 x 24 x 100
CCCCCCCCCCC = 75.77 mm
Example 5
A beam having a T-
shaped cross section
b=100mm, t=24mm,
h=200mm,
V = 45kN. Determine
τmin (top of the web)
and τmax.
January, 2015Bending Stresses - DAT28

29. Thank You
January, 2015Bending Stresses - DAT29