Chapter 4 - Bending of Straight Beams
Chapter 4 - Bending of Straight Beams
Chapter 4 - Bending of Straight Beams
Pengajian Kejuruteraan
Bending stress
Z
Fx = σx dA = 0
Z
My = zσx dA = 0
Z
Mz = −y σx dA = M
A B
L = (ρ − y )θ
K y
J 0
D E δ = L − L = (ρ − y )θ − ρθ = −y θ
A' x
B' δ −y θ y
εx = = =− varies linearly
y
L ρθ ρ
c c
εmax = or ρ =
Neutral ρ εmax
axis (NA)
c y
z εx = − εmax
O c
y
I
1
bh3 1 1
S= = 12 = bh2 = Ah
c h/2 6 6
Between two beams with the same cross sectional area, the
beam with the greater depth will be more effective in resisting
bending
Bending stress 8/40
Example 4.1
A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165 GPa and
neglecting the effects of fillets, determine the maximum tensile and compressive stresses.
90 mm
20 mm
M = 3 kNm
40 mm
30 mm
• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
The simply supported beam has the cross-sectional area as shown. Determine the ab-
solute maximum bending stress in the beam and draw the stress distribution over the
cross section at this location.
20 mm
5 kN/m B
150 mm
N.A 20 mm
a C 150 mm
20 mm
3m 250 mm D
6m
22.5
x(m)
3 6
• By symmetry, the centroid C and thus the neutral axis pass through the mid-height of the beam,
and the moment of inertia is
1 1
I = Σ(I¯ + Ad 2 ) = 2 (0.25)(0.023 ) + 0.25(0.002)(0.16)2 + (0.02)(0.33 )
12 12
= 301.3 × 10−6 m4
Mc 22.5(0.17)
σmax = = = 12.7 MPa
I 301.3 × 10−6
ΣM1 = 0; R1 = R2 = 15 kN 15
SFD
x(m)
3 6
• Apply method of section for segment chosen
-15
and generate equation of equilibrium.
5x cp M (kN.m)
BMD
22.5
M
x V x(m)
3 6
15 kN
Bending stress 15/40
Outline
Bending stress
area, defined from section where t is measured, and y’ is distance of centroid of A’,
measured from neutral axis.
The beam shown in the following figure is made from two boards. Determine the
maximum shear stress in the glue necessary to hold the boards together along the seams
where they are joined. Supports at B and C exert only vertical reactions on the beam.
150 mm
30 mm
6.5 kN/m
B D
C N.A
150 mm
Y
4m 4m
30 mm
• Internal shear: Support reactions • The centroid and therefore the neutral axis will be
and shear diagram for beam are determined from the reference axis placed at bottom
shown below. Maximum shear in of the cross sectional area.
the beam is 19.5 kN. Σȳ A
26 kN
ȳ =
ΣA
(0.075(0.15)(0.03)) + (0.165(0.03)(0.15))
=
0.15(0.03) + 0.03(0.15)
6m 2m
6.5 kN 19.5 kN = 0.12 m
V (kN)
1
SFD I = (0.03)(0.15)3 + 0.15(0.03)(0.12 − 0.075)2
6.5 12
1
5 8 x (m) + (0.15)(0.03)3 + 0.03(0.15)(0.165 − 0.12)2
4 12
= 27 × 10−6 m4
-19.5
• The top board (flange) is being held onto • Shear stress acting at top of bottom board
the bottom board (web) by the glue, which is shown in the next figure. It is the glue’s
is applied over the thickness t = 0.03m. resistance to this lateral or horizontal shear
Consequently A’ is defined as the area of stress that is necessary to hold the boards
the top board, we have; from slipping at support C.
Q = y 0 A0
= (0.18 − 0.015 − 0.12)(0.03)(0.15)
= 0.2025 × 10−3 m3
6.5(4)=26 kN M
x V
RA
4m
RA RB ΣMcp = 0, M = 6.5x
8m
ΣFy = 0, V = 6.5
ΣMB = 0, RC (8) − 26(6) = 0
• For 4 ≤ x ≤ 8 m
RC = 19.5 kN 6.5(x-4)
cp
RB = 6.5 kN
M
4m
• Apply method of section for segment chosen RA
x
V
SFD
V (kN)
6.5
8 x(m)
4
M (kN.m) 29.2
BMD
x(m)
4 8
37.5 mm
100 mm
Solution
• The moment of inertia:
Aʹ
1 3 1
I = bh = (100)(125)3 = 16.28 × 106 mm4
12 12 50 mm
12.5 mm
Q = y 0 A0 = (62.5 − 50/2)(50)(100) = 18.75 × 104 mm3 P
N.A
Aʹ
• Maximum shear stress occurs at the neutral axis, since
t is constant throughout the cross section,
62.5 mm
VQ 3(19.53 × 104 )
τmax = = = 0.36 MPa
It 16.28 × 106 (100)
Bending stress
• When a beam with a straight longitudinal axis is loaded by lateral forces, the axis is
deformed into a curve, called the deflection curve of the beam. Deflection is the
displacement in the y-direction of any point on the axis of the beam.
• The calculation of deflections is an important part of structural analysis and design.
ds = dx = Rdi B'
R i
di 1 dy
= but i = ds
dx R dx D
C
d 2y 1
• Therefore = ....(1) i
dx 2 R
A'
• simple bending theory:
A B
M 1 1 M
= → =
I R R EI x
2
d y
M = EI ....(2)
dx 2
Deflection of the beam 29/40
Macaulay’s Method
• The Macaulay’s method involves the general method of obtaining slopes and
deflections.
Characteristics
a) Term, W (x–a) is integrated with respect to (x–a) and not x
b) Mathematical operations:
(
n 0, for x < a
hx − ai = n
;n ≥ 0
(x − a) , otherwise
c) Special bracket used to represent the operation used Macaulay’s concept : i.e. hx − 2i
Consider a simple beam problem as shown and develop the expression for maximum
deflection.
w
L/2
Determine the deflection of the beam at x = 3 (middle of beam) if E = 210 kN/mm2 . The
cross-section is given below.
15 kN
5 kN/m 10 mm
40 mm
N.A 10 mm
B 40 mm
A
10 mm
1m 5m
50 mm
Determine the deflection of the beam at x = 3 (middle of beam) if E = 210 kN/mm2 . The
cross-section is given below.
15 kN 5 kN/m 10 mm
40 mm
N.A 10 mm
B 40 mm
A
10 mm
1m 5m
50 mm
15 kN 5 kN/m 10 mm
40 mm
N.A 10 mm
B 40 mm
A
10 mm
1m 4m 1m
50 mm
• Since the 5 kN/m load did not reach the end, (x-1) does not represent the actual loading.
The equivalent loading is:
Equivalent
loading cp
15 kN 5 kN/m
RA x-5
x-1
x
• Differential equations:
d 2y 5 5
EI = −15x + 30hx − 1i − hx − 1i2 + hx − 5i2
dx 2 2 2
dy 15 2 30 5 5
EI =− x + hx − 1i − hx − 1i + hx − 5i3 + A
2 3
dx 2 2 6 6
15 3 30 3 5 4 5
EIy = − x + hx − 1i − hx − 1i + hx − 5i4 + Ax + B
6 6 24 24
• Boundary conditions:
15
x = 1, y = 0 ∴ A + B = ..(1)
6
x = 6, y = 0 ∴ 6A + B = −409.99..(2)
15 3 30 5 5
• General equation: EIy = − x + hx − 1i3 − hx − 1i4 + hx − 5i4 − 82.5x + 84.99
6 6 24 24