Chapter 4: Pure Bending of Beams
There are two ways to bend a beam:
1. With a pure bending moment (internal moment is constant through the length of
the beam - Chapter 4)
2. With transverse loads (forces are applied perpendicular to the beam - Chapter 5)
Pure bending:
 Grip the two ends and apply a moment
Sign convention: M(+) for concave up
 Two point support
The portion of the bar between
the hands is under a pure
bending moment.
Ch4 Page 1
What is the main type of stress that results from bending a beam?
Imagine drawing a grid of small squares on the side of the beam. Look at how the
squares deform.
Towards the top of the beam, the squares are increasingly compressed in the horizontal
direction. Towards the bottom of the beam, the squares are increasingly stretched in the
horizontal direction.
Because the squares are so small, changes in angle are negligible - there is virtually no
shear!
A simpler way is to look only at the horizontal lines that deform into circular arcs. The arc
lengths get increasingly shorter towards the top of the beam and increasingly longer
towards the bottom of the beam, indicating tension and compression, respectively.
The arc corresponding to the centroid of the cross section does not change length,
indicating that stress and strain are zero along this arc. This is known as the neutral axis
(NA).
In the front view, the neutral axis appears to be a line, hence the name "axis". In
actuality, it's a plane:
Ch4 Page 2
Important bending equations
(See Chapter 4 in textbook for derivations.)
Chapters 1 and 2
(Axial loading of rods)
𝑃
Stress
𝜎 = ⎯⎯
𝐴
𝛿
Strain
𝜀 = ⎯⎯
𝐿
Hooke’s law
𝜎=𝐸𝜀
𝑃𝐿
Deflection
𝛿 = ⎯⎯⎯
𝐴𝐸
Chapter 3
(Torsion of shafts)
𝑇𝑟
𝑇𝑐
𝜏 = ⎯⎯⎯ , 𝜏
= ⎯⎯⎯
𝐽
𝐽
𝑟𝜙
𝑐𝜙
𝛾 = ⎯⎯⎯ , 𝛾
= ⎯⎯⎯
𝐿
𝐿
𝜏=𝐺𝛾
𝑇𝐿
𝜙 = ⎯⎯⎯
𝐽𝐺
Chapters 4 and 5
(Bending of beams)
𝑀𝑦
𝑀𝑐
𝜎 = ⎯⎯⎯ , 𝜎
= ⎯⎯⎯
𝐼
𝐼
𝜎
𝜀 = ⎯⎯
𝐸
𝜎=𝐸𝜀
𝐸𝐼
𝜌 = ⎯⎯⎯
𝑀
c maximum distance
(maximum radius in the case of torsion)
(maximum distance from NA in the case of bending)
Centroidal area moment of inertia (I) is the resistance to rotation of an
area about an axis parallel to its plane and passing through the center or
centroid of the area:
1
𝐼 = ⎯⎯𝜋 𝑟
4
(circular area)
1
𝐼 = ⎯⎯𝜋 𝑟 − 𝑟
4
(hollow circular area)
radius of curvature of the NA (a beam under bending deforms into a
circular arc)
Stress is a function of load (P, T, M) and resistance to that load (A, J, I)
Strain is a function of load, total deflection ( , ), and length (L)
Deflection is a function of load, resistance to that load, length, and material
(E, G)
Ch4 Page 3
What is the significance of the moment of inertia I in bending?
From Statics, the area moment of inertia I is the resistance to rotation of an area about
an axis that is in the plane of the area. For example:
1
𝐼 = ⎯⎯⎯𝑏ℎ
12
1
𝐼 = ⎯⎯𝑏ℎ
3
When a beam bends, each cross section rotates about its centroidal axis:
Each vertical line on the left represents a cross section before bending. After
bending (right), each line has rotated so that it's no longer vertical.
Note that the centroidal axis of rotation of the cross section is perpendicular to the
neutral axis.
Ch4 Page 4
Applying the bending stress equation
1. If applicable, locate the centroid of the cross section. In the normal situation where
the beam is bent up or down, find 𝑦. This locates the neutral axis and is set as y = 0.
The maximum distances from the neutral axis (c values) can also be found.
2. Calculate the moment of inertia about the centroidal axis.
3. The formal equation for calculating the normal stress due to bending is:
𝑀𝑦
𝜎 = − ⎯⎯⎯
𝐼
Using accepted sign conventions for moment and distance from the NA, this
equation gives the correct sign for the stress (negative for compression and positive
for tension):
4. The stress varies linearly from 0 at the NA to the maximum values at the top and
bottom surfaces of the beam. These maxima are calculated according to:
𝜎
𝑀𝑐
= ⎯⎯⎯
𝐼
where c denotes the maximum distance from the NA. The sign (+/-) is assigned
manually.
Ch4 Page 5
Rectangular beams, I-beams, and H-beams
The most common beams are rectangular beams and I-beams.
Structural rectangular beams are traditionally made of wood and used in smaller
structures. Composites are becoming more prevalent.
1
𝐼 = ⎯⎯⎯(base)(height)
12
Rectangular beams are almost always oriented with the longer dimension being the
height (above left) because that results in a much larger moment of inertia or
resistance to bending.
Structural I-beams and H-beams are usually made of steel. They are used in larger
structures (e.g., large buildings and bridges). In construction, standard designs are used
(i.e., engineers do not design I-beams and H-beams from scratch). Geometric properties
of these standard designs can be found in tables (see the Appendix in the textbook).
Standard I-beams are referred to as "S shapes" (for American Standard). Standard Hbeams are referred to as "W shapes" (for Wide-flange).
I-beams and H-beams offer the advantage of large moment of inertia relative to area.
This efficiency in design allows for great resistance to bending while keeping the material
used (and weight) relatively low.
H-beams are more common in
buildings because of their higher
strength-to-weight ratio.
Ch4 Page 6
How efficient are H-beams?
Example: W12 x 40 H-beam
From the table of H-beams in the Appendix
W12 x 40
Area = 11.7 in2
Depth = 11.9 in
Flange width = 8.01 in
Flange thickness = 0.515 in
Web thickness = 0.295 in
Ix = 307 in4
Iy = 44.1 in4
Compare to a rectangular beam with dimensions equal to the depth and flange
width:
Area = (8.01 in)(11.9 in) = 95.3 in2
1
1
𝐼 = ⎯⎯⎯(base)(height) = ⎯⎯⎯(8.01 in)(11.9 in) = 1124.8 in
12
12
Ratio of Ix:
307/1124.8 = 0.2729
Ratio of areas:
11.7/95.3 = 0.1228
Ch4 Page 7
Example: The most basic type of bending problem
Given a beam and its cross section, find the maximum tensile and compressive stresses.
1. Locate the centroid of the cross section or the neutral axis of the beam.
The coordinate system of a horizontal beam is usually set up as shown below:
Note that the cross section is in the yz plane and bending is about the z axis. Hence,
we need Iz, or the moment of inertia about the centroidal z axis. However, we do
not know where the origin of the coordinate system is. We must find the location of
the centroid, or specifically, 𝑦.
Area (m2)
𝒚 (m)
𝒚A (m3)
1
(0.06)(0.06) = 0.0036 0.21 + ⎯⎯(0.06) = 0.23 0.000828
2
𝜋(0.04) = 0.005027 0.13 + 0.04 = 0.17
3
(0.12)(0.15) = 0.018 0.06 + 0.075 = 0.135 0.00243
4
⎯⎯𝜋 (0.06) = 0.005655

0.032281
( . )
0.06 − ⎯⎯⎯⎯⎯
= 0.0345
𝟎. 𝟎𝟎𝟒𝟑𝟎𝟖
𝒀 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 𝟎. 𝟏𝟑𝟑𝟒𝟒𝟓 𝐦
𝟎. 𝟎𝟑𝟐𝟐𝟖𝟏
Ch4 Page 8
0.000855
0.000195
0.004308
Example: The most basic type of bending problem (cont'd)
Given a beam and its cross section, find the maximum tensile and compressive stresses.
1. (cont'd) The location of the centroid determines the maximum distances from the
neutral axis.
Tension: c = 0.133445 m
Compression: c = 0.27 - 0.133445 = 0.136555 m
2. Calculate the moment of inertia about the centroidal axis.
Area (m2)
𝒚 (m)
𝒚A (m3)
1
0.0036
0.23
0.000828
2
0.005027
0.17
0.000855
3
0.018
0.135
0.00243
4
0.005655
0.0345
0.000195

0.032281
𝟎. 𝟎𝟎𝟒𝟑𝟎𝟖
𝒀 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 𝟎. 𝟏𝟑𝟑𝟒𝟒𝟓 𝐦
𝟎. 𝟎𝟑𝟐𝟐𝟖𝟏
0.004308
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
= ⎯⎯⎯(0.12)(0.06) + (0.0036)(0.23 − 0.133445)
36
1
+ ⎯⎯𝜋(0.04) + 0.005027(0.17 − 0.133445)
4
1
+ ⎯⎯⎯(0.12)(0.15) + (0.018)(0.135 − 0.133445)
12
1
8
+ ⎯⎯𝜋(0.06) − ⎯⎯⎯(0.06) + (0.005655)(0.0345 − 0.133445)
8
9𝜋
𝐼 = 1.33548 × 10
m
Ch4 Page 9
Example: The most basic type of bending problem (cont'd)
Given a beam and its cross section, find the maximum tensile and compressive stresses.
2. (cont'd) Be especially careful when calculating the moment of inertia of semicircles
and applying the parallel axis theorem.
Looking up the moment of inertia in a table:
𝐼 = 𝐼 ̅ = ⎯⎯𝜋𝑟
𝐶𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑖𝑠 𝑎𝑡 𝑦 = ⎯⎯⎯
In the table, Iy is centroidal but Ix is not. The parallel axis theorem uses the
centroidal moment of inertia as the starting point. The find the moment of inertia
about the centroidal x axis above, apply the parallel axis theorem in reverse:
𝐼 = 𝐼 ̅ + 𝐴𝑑
⎯⎯𝜋 𝑟
= 𝐼 ̅ + ⎯⎯𝜋𝑟
⎯⎯⎯
𝐼 ̅ = ⎯⎯𝜋𝑟 − ⎯⎯⎯𝑟
Ch4 Page 10
Example: The most basic type of bending problem (cont'd)
Given a beam and its cross section, find the maximum tensile and compressive stresses.
3. Calculate stresses
Tension:
𝜎
𝑀 𝑐 (20 × 10 N m)(0.133445 m)
= ⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 19.99 MPa
𝐼
1.33548 × 10 m
Compression:
𝜎
(20 × 10 N m)(0.136555 m)
𝑀𝑐
= − ⎯⎯⎯ = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −20.45 MPa
𝐼
1.33548 × 10 m
Ch4 Page 11
Beams of different materials
The materials will not behave the same because of different properties. How do we
calculate the maximum stresses in each material?
Ch4 Page 12
Expansion method
Method: convert a beam made of different materials into an equivalent beam made of a
single material. "Equivalent" refers to strains remaining the same.
1. We usually choose the material with the smallest E value. This means that the
components of the beam that are made of other materials must be converted to a
weaker or less stiff material. Their cross sections must expand to compensate. That
is, in order to replace a beam with a large Young's modulus with one that is weaker,
the new beam must be made larger.
2. The expansion must be done such that distances from the neutral axis are not
affected. For a horizontal beam that is bent up or down, cross sections can only
grow horizontally, not vertically.
3. The area of each component is expanded by an amount equal to the ratio of E values
(n). According to Hooke's law (i.e., stress and strain are linearly proportional), this
ensures that strains remain unchanged.
4. If applicable, find the centroid of the expanded shape.
5. Calculate the centroidal moment of inertia of the expanded shape.
6. Apply the bending stress equation to each material. For components that were
expanded, the calculated stress must be multiplied by the ratio of E values. This is
necessary because the load was applied to an artificially larger area.
Ch4 Page 13
Expansion method: Example 1
Find the maximum tensile and compressive stresses in each material of the composite
beam shown below.
1. Create an equivalent beam made of material 2 (smallest E).
2. Expand materials 1 and 3 horizontally.
3. Create the expanded cross section.
Ratios of E values used to expand widths:
n1 = E1/E2 = 150/100 = 1.5
n3 = E3/E2 = 200/100 = 2.0
4. Find the centroid of the expanded shape.
Because of symmetry, the centroid remains at the centerline of the section.
For both tension and compression in each component, c = 150 mm.
Ch4 Page 14
Expansion method: Example 1 (cont'd)
Find the maximum tensile and compressive stresses in each material of the composite
beam shown below.
5. Calculate the centroidal moment of inertia of the expanded shape.
1
1
𝐼 = ⎯⎯⎯𝑏ℎ = ⎯⎯⎯(0.45 m)(0.3 m) = 0.001013 m
12
12
6. Apply bending stress equation to find maximum tensile and compressive stresses in
each material.
Material 1:
𝜎
(20 × 10 N m)(0.15 m)
𝑀𝑐
= ± 𝑛 ⎯⎯⎯ = ±1.5 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= ± 4.444 MPa
𝐼
0.001013 m
Material 2:
𝜎
(20 × 10 N m)(0.15 m)
𝑀𝑐
= ± ⎯⎯⎯ = ± ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= ± 2.963 MPa
𝐼
0.001013 m
Material 3:
𝜎
(20 × 10 N m)(0.15 m)
𝑀𝑐
= ± 𝑛 ⎯⎯⎯ = ±2.0 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= ±5.926 MPa
𝐼
0.001013 m
Ch4 Page 15
Expansion method: Example 2
Find the maximum tensile and compressive stresses in each material of the composite
beam shown below.
1. Create an equivalent beam made of material 2 (smallest E).
2. Expand materials 1 and 3 horizontally.
3. Create the expanded cross section.
Ratios of E values used to expand widths:
n1 = E1/E2 = 150/100 = 1.5
n3 = E3/E2 = 200/100 = 2.0
The equivalent beam made of
material 2.
Ch4 Page 16
Expansion method: Example 2 (cont'd)
Find the maximum tensile and compressive stresses in each material of the composite
beam shown below.
4. Find the centroid of the expanded shape.
The neutral axis is horizontal. In order to locate it, 𝑦 is needed.
Area (m2)
𝒚 (m)
𝒚A (m3)
1
(0.45)(0.1) = 0.045 0.25
0.01125
2
(0.3)(0.1) = 0.03
0.15
0.0045
3
(0.6)(0.1) = 0.06
0.05
0.003

0.135
0.01875
𝟎. 𝟎𝟏𝟖𝟕𝟓
𝒀 = ⎯⎯⎯⎯⎯⎯⎯⎯= 𝟎. 𝟏𝟑𝟖𝟗 𝐦
𝟎. 𝟏𝟑𝟓
5. Calculate the centroidal moment of inertia of the expanded shape.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
= ⎯⎯⎯(0.45)(0.1) + (0.045)(0.25 − 0.1389)
12
1
+ ⎯⎯⎯(0.3)(0.1) + (0.03)(0.15 − 0.1389)
12
1
+ ⎯⎯⎯(0.6)(0.1) + (0.06)(0.05 − 0.1389)
12
𝐼 = 0.0011458 m
Ch4 Page 17
Expansion method: Example 2 (cont'd)
Find the maximum tensile and compressive stresses in each material of the composite
beam shown below.
6. Apply bending stress equation to find maximum tensile and compressive stresses in
each material.
The materials have different c values (i.e., different maximum distances from the
neutral axis):
Material 1 is in compression only (c = 0.3  0.1389 = 0.1611 m).
Material 2 is in both tension (c = 0.1389  0.1 = 0.0389 m)
and compression (c = 0.1611  0.1 = 0.0611 m).
Material 3 is in tension only (c = 0.1389 m).
Ch4 Page 18
Expansion method: Example 2 (cont'd)
Find the maximum tensile and compressive stresses in each material of the composite
beam shown below.
6. (cont'd) Apply bending stress equation to find maximum tensile and compressive
stresses in each material.
Material 1:
𝜎
,
(20 × 10 N m)(0.1611 m)
𝑀𝑐
= −𝑛 ⎯⎯⎯ = −1.5 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= − 4.218 MPa
𝐼
0.0011458 m
Material 2:
𝜎
𝜎
,
(20 × 10 N m)(0.0389 m)
𝑀𝑐
= ⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 0.679 MPa
𝐼
0.0011458 m
,
(20 × 10 N m)(0.0611 m)
𝑀𝑐
= − ⎯⎯⎯ = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −1.0665 MPa
𝐼
0.0011458 m
Material 3:
𝜎
,
(20 × 10 N m)(0.1389 m)
𝑀𝑐
= 𝑛 ⎯⎯⎯ = 2.0 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 4.849 MPa
𝐼
0.0011458 m
Ch4 Page 19
Expansion method: Example 3
Find the maximum tensile and/or compressive stresses in each material of the composite
beam shown below.
1. Create an equivalent beam made of material 1 (smallest E).
2. Expand materials 2 and 3 horizontally.
3. Create the expanded cross section.
Ratios of E values used to expand widths:
n2 = E2/E1 = 100/50 = 2.0
n3 = E3/E1 = 150/50 = 3.0
Ch4 Page 20
Expansion method: Example 3 (cont'd)
Find the maximum tensile and/or compressive stresses in each material of the composite
beam shown below.
4. Find the centroid of the expanded shape.
The neutral axis is horizontal. In order to locate it, 𝑦 is needed.
Area (m2)
𝒚 (m)
𝒚A (m3)
a
(0.25)(0.3) = 0.075
0.15
0.01125
b
(0.125)(0.05) = 0.00625 0.225
0.001406
c
(0.36)(0.05) = 0.018
0.125
0.00225
d
(0.15)(0.05) = 0.0075
0.025
0.000188

0.10675
𝟎. 𝟎𝟏𝟓𝟎𝟗𝟒
𝒀 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 𝟎. 𝟏𝟒𝟏𝟒 𝐦
𝟎. 𝟏𝟎𝟔𝟕𝟓
5. Calculate the centroidal moment of inertia of the expanded shape.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
= ⎯⎯⎯(0.25)(0.3) + (0.075)(0.15 − 0.1414)
12
1
+ ⎯⎯⎯(0.125)(0.05) + (0.00625)(0.225 − 0.1414)
12
1
+ ⎯⎯⎯(0.36)(0.05) + (0.018)(0.125 − 0.1414)
12
1
+ ⎯⎯⎯(0.15)(0.05) + (0.0075)(0.025 − 0.1414)
12
𝐼 = 7.25 × 10
m
Ch4 Page 21
0.015094
Expansion method: Example 3 (cont'd)
Find the maximum tensile and/or compressive stresses in each material of the composite
beam shown below.
6. Apply bending stress equation to find maximum tensile and compressive stresses in each
material.
The materials have different c values (i.e., different maximum distances from the neutral
axis):
Material 1, max. tension: c = 141.4 mm = 0.1414 m
Material 1, max. compression: c = 158.6 mm = 0.1586 m
Material 2, max. tension: c = 141.4 mm = 0.1414 m
Material 2, max. compression: c = 108.6 mm = 0.1086 m
Material 3, max. tension: c = 41.4 mm = 0.0414 m
Material 3, max. compression: c = 8.6 mm = 0.0086 m
Ch4 Page 22
Expansion method: Example 3 (cont'd)
Find the maximum tensile and/or compressive stresses in each material of the composite
beam shown below.
6. (cont'd) Apply bending stress equation to find maximum tensile and compressive stresses
in each material.
Material 1:
𝜎
𝜎
,
(20 × 10 N m)(0.1414 m)
𝑀𝑐
= ⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 3.901 MPa
𝐼
7.25 × 10 m
,
(20 × 10 N m)(0.1586 m)
𝑀𝑐
= − ⎯⎯⎯ = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= − 4.375 MPa
𝐼
7.25 × 10 m
Material 2:
𝜎
𝜎
,
(20 × 10 N m)(0.1414 m)
𝑀𝑐
= 𝑛 ⎯⎯⎯ = 2.0 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 7.801 MPa
𝐼
7.25 × 10 m
,
(20 × 10 N m)(0.1086 m)
𝑀𝑐
= −𝑛 ⎯⎯⎯ = −2.0 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −5.992 MPa
𝐼
7.25 × 10 m
Material 3:
𝜎
𝜎
,
(20 × 10 N m)(0.0414 m)
𝑀𝑐
= 𝑛 ⎯⎯⎯ = 3.0 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 3.426 Mpa
𝐼
7.25 × 10 m
,
(20 × 10 N m)(0.0086 m)
𝑀𝑐
= −𝑛 ⎯⎯⎯ = −3.0 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −0.712 MPa
𝐼
7.25 × 10 m
Ch4 Page 23
Expansion method: Example 4
Draw the expanded cross section with all components converted to the material with the
smallest Young's modulus.
Ratios of E values used to expand widths:
n2 = E2/E1 = 100/50 = 2.0
n3 = E3/E1 = 150/50 = 3.0
n4 = E4/E1 = 200/50 = 4.0
Ch4 Page 24
Expansion method: Example 5
Draw the expanded cross section with all components converted to the material with the
smallest Young's modulus.
Ratios of E values used to expand widths:
n2 = E2/E1 = 150/50 = 3.0
The maximum distances from the neutral axis cannot be altered. Cut the circular
component (material 2) vertically down the middle and insert a rectangle such that
the new area is 3x the original area.
Expanded area = 3[(40 mm)2] = 15079.64 mm2
Area of inserted rectangle = 15079.64  (40 mm)2 = 10053.1 mm2
Width of inserted rectangle = 10053.1/80 = 125.6638 mm
Ch4 Page 25
4.33 and 4.34 A bar having the cross section shown has been
formed by securely bonding brass and aluminum stock. Using the
data given below, determine the largest permissible bending
moment when the composite bar is bent about a horizontal axis.
Aluminum
Brass
Modulus of elasticity
70 GPa
105 GPa
Allowable stress
100 MPa 160 MPa
Ch4 Page 26
Unsymmetric bending
Thus far, we have looked only at "regular bending" where the horizontal beam is aligned
in such a way that the neutral axis and centroidal moment of inertia are readily found.
The cross section in these situations usually have well-defined edges and/or axes of
symmetry that align with the coordinate axes. Under a bending moment, these beams
undergo symmetric bending. In the examples below, the bending occurs about the z axis.
There may be instances where the cross section of the beam is not aligned with the
coordinate axes. This is called unsymmetric bending.
Ch4 Page 27
Unsymmetric bending (cont'd)
The long (wrong) way to solve an unsymmetric bending problem: Treat the cross section
as a composite.
One way to treat this cross
section as a composite:
dividing it into several right
triangles and a rectangle.
The correct way to solve the problem is to treat the bending moment as a vector with
mutually perpendicular components.
Ch4 Page 28
Unsymmetric bending (cont'd)
By resolving the moment vector into two components, the loading is separated into two
simpler cases. Superpositioning is then applied.
Mz' causes bending about the z' axis, causing maximum tension and compression on
the top and bottom surfaces:
𝜎
𝑀 𝑐
= ⎯⎯⎯⎯⎯
𝐼
My' causes bending about the y' axis, causing maximum tension and compression on
the front and back surfaces:
𝜎
𝑀 𝑐
= ⎯⎯⎯⎯⎯
𝐼
Maximum overall tension is the point on the cross section that is under maximum
tension for both Mz' and My'. Likewise, maximum overall compression is the point on
the cross section that is under maximum compression for both Mz' and My'.
Ch4 Page 29
Unsymmetric bending (cont'd)
Interpreting moment components and their signs: the right-hand rule applies.
Ch4 Page 30
Unsymmetric bending: Example 1
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Separating into two different bending problems:
Point A is under max. tension due to both Mz and My. It is the point of max. tension.
Point C is the point of max. compression.
Due to Mz:
𝑀 𝑐 (20 × 10 cos 30° N m)(0.1 m)
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 9.743 MPa
1
𝐼
⎯⎯⎯(0.4 m)(0.2 m)
12
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −9.743 MPa
𝐼
Due to My:
𝑀 𝑐 (20 × 10 sin 30° N m)(0.2 m)
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 2.8125 MPa
1
𝐼
⎯⎯⎯
12 (0.2 m)(0.4 m)
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −2.8125 MPa
𝐼
Total:
𝜎 = 12.56 MPa,
𝜎 = −12.56 MPa
Ch4 Page 31
Unsymmetric bending: Example 2
4.136 The couple M acts in a
vertical plane and is applied to a
beam oriented as shown.
Determine the maximum tensile
and compressive stresses in the
beam and indicate the points where
they occur.
𝐼 = 14.77 × 10 mm
𝐼 = 53.6 × 10 mm
Separating into two different bending problems:
Point E is under max. tension due to
both Mz and My. It is the point of
max. tension.
Point B is the point of max.
compression.
Due to Mz:
(120 cos 20° N m)(0.016 m)
𝑀 𝑐
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 33.66 MPa
m
𝐼
(53.6 × 10 mm ) ⎯⎯⎯⎯⎯⎯⎯⎯
1000 mm
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −33.66 MPa
𝐼
Due to My:
(120 sin 20° N m)(0.01 m)
𝑀 𝑐
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 27.79 MPa
m
𝐼
(14.77 × 10 mm ) ⎯⎯⎯⎯⎯⎯⎯⎯
1000 mm
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −27.79 MPa
𝐼
Total:
𝜎 = 61.45 MPa,
𝜎 = −61.45 MPa
Ch4 Page 32
Unsymmetric bending: Example 3
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Separating into two different bending problems:
To find stresses due to Mz, we need 𝑦 and Iz. These were found in an earlier example to
be:
𝑦 = 0.133445 m
𝐼 = 1.33548 × 10
m
To find stresses due to My, we need 𝑧̅ and Iy. Because of symmetry, 𝑧̅ = 0 at the
vertical line of symmetry AF. Calculating the moment of inertia about this centroidal y
axis:
𝐼 = ∑ 𝐼 ̅ + 𝐴𝑑
1
= 2 ⎯⎯⎯(.06)(0.06)
12
+2
1
8
⎯⎯𝜋(0.04) − ⎯⎯⎯(0.04)
8
9𝜋
1
+ ⎯⎯𝜋(0.04)
2
1
+ ⎯⎯⎯(0.15)(0.12)
12
1
+ ⎯⎯𝜋(0.06)
8
𝐼 = 1.10529 × 10
m
Ch4 Page 33
4(0.04)
0.06 + ⎯⎯⎯⎯⎯⎯⎯
3𝜋
Unsymmetric bending: Example 3 (cont'd)
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Determining points of maximum tension and compression:
Due to Mz, point F is in max. tension and point A is in
max. compression.
Due to My, point I is in max. tension and point C is in
max. compression.
The points of overall max. tension and compression
are not obvious. We must check several different
points (namely corners and transitions).
Ch4 Page 34
Eccentric loading
Eccentric loading (loading that is not centric or "through the center") is a tensile or
compressive force that is not applied at the centroid of the cross section of a rod or
column, leading to bending in addition to the axial loading.
There are two common types of eccentric loading problems. The first type of eccentric
load results in symmetric bending and axial loading. The second type results in
unsymmetric bending and axial loading.
Eccentric loading that leads to symmetric bending:
(Above) In this case, the applied forces are moved to the centroid. The moments
due to the forces when the forces were at their points of application must not be
neglected. As seen on the right, this leads to axial loading and symmetric bending.
In Statics, the process of relocating an eccentric force to the centroid is formally
justified by adding equal and opposite forces:
Ch4 Page 35
Eccentric loading (cont'd)
Eccentric loading that leads to unsymmetric bending:
In this case, the applied force is offset from the centroid by distance b from the x
axis, which causes a moment about that axis. It is also offset by distance a from the
y axis, which causes a moment about that axis. These moments produce
unsymmetric bending. As long as these moments are accounted for, the force can
be moved to the centroid.
Ch4 Page 36
Eccentric loading: Example 1
4.103 The vertical portion of the press shown
consists of a rectangular tube of wall thickness
t = 10 mm. Knowing that the press has been
tightened on wooden planks being glued
together until P = 20 kN, determine the stress
at (a) point A, (b) point B.
Moving the applied loads to the centroid:
𝐴 = (0.08 𝑚)(0.06 𝑚) − (0.06 𝑚)(0.04 𝑚) = 0.0024 m
1
1
𝐼 = ⎯⎯⎯(0.06)(0.08) − ⎯⎯⎯(0.04)(0.06) = 1.84 × 10
12
12
Stress at point A:
𝑀 𝑐 𝑃 (20 × 10 N)(0.24 m)(0.04 m) 20 × 10 N
𝜎 = ⎯⎯⎯ + ⎯⎯= ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 112.68 MPa
𝐼
𝐴
1.84 × 10 m
0.0024 m
Stress at point B:
(20 × 10 N)(0.24 m)(0.04 m) 20 × 10 N
𝑀𝑐 𝑃
𝜎 = − ⎯⎯⎯ + ⎯⎯= − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −96.01 MPa
𝐼
𝐴
1.84 × 10 m
0.0024 m
Ch4 Page 37
m
Eccentric loading: Example 2
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Find the location of the centroid. The force will be moved to the centroid and moments
about the x and y axes can be calculated.
Area (in2)
𝒚 (in)
𝒚A (in3)
1
(6)(4) = 24
5
120
2
(10)(3) = 30 1.5

54
45
𝟏𝟔𝟓
𝒀 = ⎯⎯⎯⎯= 𝟑. 𝟎𝟓 𝐢𝐧
𝟓𝟒
165
Calculating moment components Mx and My.
Mx = (50 kips)(3.05 in) = 152.5 kipin
My = (50 kips)(5 in) = 250 kipin
Ch4 Page 38
Eccentric loading: Example 2 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Area (in2)
𝒚 (in)
𝒚A (in3)
1
(6)(4) = 24
5
120
2
(10)(3) = 30 1.5

54
45
𝟏𝟔𝟓
𝒀 = ⎯⎯⎯⎯= 𝟑. 𝟎𝟓 𝐢𝐧
𝟓𝟒
165
Calculating moments of inertia Ix and Iy.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
= ⎯⎯⎯(6)(4) + (24)(5 − 3.05)
12
1
+ ⎯⎯⎯(10)(3) + (30)(1.5 − 3.05)
12
𝐼 = 217.835 in
𝐼 = ∑𝐼 ̅
1
1
= ⎯⎯⎯(4)(6) + ⎯⎯⎯(3)(10)
12
12
𝐼 = 322 in
Ch4 Page 39
Eccentric loading: Example 2 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Superposing the two bending problems and the axial loading problem:
Point B is under max. tension
due to Mx and under tension
due to My. It is the point of
max. tension.
Point F is the point of max.
compression.
Stress at B is due to Mx, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = ⎯⎯⎯⎯ + ⎯⎯⎯⎯ − ⎯⎯
𝐼
𝐼
𝐴
(152.5 × 10 lb in)(3.95 in) (250 × 10 lb in)(3 in) 50 × 10 lb
𝜎 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 4.169 ksi
217.835 in
322 in
54 in
Stress at F is due to Mx, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = − ⎯⎯⎯⎯ − ⎯⎯⎯⎯ − ⎯⎯
𝐼
𝐼
𝐴
(152.5 × 10 lb in)(3.05 in) (250 × 10 lb in)(3 in) 50 × 10 lb
𝜎 = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −5.39 ksi
217.835 in
322 in
54 in
Ch4 Page 40
Eccentric loading: Example 3
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Due to symmetry about both the y and z axis, the centroid is at the center of the cross
section. Calculating moment components Mz and My:
Mz = (50 kips)(4 in) = 200 kipin (bend down)
My = (50 kips)(8 in) = 400 kipin (bend forward)
Calculating area and moments of inertia Iz and Iy.
𝐴 = (16)(8) − 𝜋(2) − (6)(3) = 97.4336 in
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
= ⎯⎯⎯(16)(8)
12
1
8
−2 ⎯⎯𝜋(2) − ⎯⎯⎯(2)
8
9𝜋
1
−4 ⎯⎯⎯(3)(3)
12
𝐼 = 1099.4 in
1
4(2)
+ ⎯⎯𝜋(2) 1 + ⎯⎯⎯⎯
2
3𝜋
Treated as 4 triangles, all
with their bases on the z
axis.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
1
= ⎯⎯⎯(8)(16) − ⎯⎯𝜋(2)
12
4
1
1
−2 ⎯⎯⎯(6)(3) + ⎯⎯(6)(3)(5)
36
2
𝐼 = 2259.1 in
Ch4 Page 41
Eccentric loading: Example 3 (cont'd)
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Superposing the two bending problems and the axial loading problem:
Stress at B (max. tension) due to Mz, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = ⎯⎯⎯⎯+ ⎯⎯⎯⎯ + ⎯⎯
𝐼
𝐼
𝐴
[(50 × 10 lb)(4 in)](4 in) [(50 × 10 lb)(8 in)](8 in) 50 × 10 lb
𝜎 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
1099.4 in
2259.1 in
97.4336 in
= 2303.2 psi
Stress at D (max. compression) due to Mz, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = − ⎯⎯⎯⎯− ⎯⎯⎯⎯ − ⎯⎯
𝐼
𝐼
𝐴
[(50 × 10 lb)(4 in)](4 in) [(50 × 10 lb)(8 in)](8 in) 50 × 10 lb
𝜎 = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
1099.4 in
2259.1 in
97.4336 in
= −1276.9 psi
Ch4 Page 42
Eccentric loading: Example 4
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Find the location of the centroid. The force will be moved to the centroid and moments
about the x and y axes can be calculated.
Area (in2)
𝒚 (in)
1
⎯⎯𝜋 (4) = 25.1327
( )
12 − ⎯⎯⎯
= 13.698 344.26
2
(12)(6) = 72
9
648
3
(6)(6) = 36
4
144

133.1327
𝒚A (in3)
1136.26
𝟏𝟏𝟑𝟔. 𝟐𝟔
𝒀 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 𝟖. 𝟓𝟑𝟒𝟖 𝐢𝐧
𝟏𝟑𝟑. 𝟏𝟑𝟐𝟕
Calculating moment components Mx and My.
Mx = (20 kips)(3.4652 in) = 69.304 kipin
My = (20 kips)(6 in) = 120 kipin
Ch4 Page 43
Eccentric loading: Example 4 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Area (in2)
𝒚 (in)
1
⎯⎯𝜋 (4) = 25.1327
( )
12 − ⎯⎯⎯
= 13.698 344.26
2
(12)(6) = 72
9
648
3
(6)(6) = 36
4
144

133.1327
𝒚A (in3)
𝒀 = 𝟖. 𝟓𝟑𝟒𝟖 𝐢𝐧
1136.26
Calculating moments of inertia Ix and Iy.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
8
= ⎯⎯𝜋(4) − ⎯⎯⎯(4)
8
9𝜋
+ (25.1327)(13.698 − 8.5348)
1
+ ⎯⎯⎯(12)(6) + (72)(9 − 8.5348)
12
1
+ ⎯⎯⎯(12)(6) + (36)(4 − 8.5348)
36
𝐼 = 1741.91 in
𝐼 = ∑𝐼 ̅
1
1
1
= ⎯⎯𝜋(4) + ⎯⎯⎯(6)(12) + 2 ⎯⎯⎯(6)(6)
8
12
12
𝐼 = 1180.53 in
Ch4 Page 44
Eccentric loading: Example 4 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Superposing the two bending problems and the axial loading problem:
Axial loading:
𝑃
20 × 10 lb
𝜎 = − ⎯⎯= − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = −150.226 psi
𝐴
133.1327 in
Ch4 Page 45
Sample exam problems
Ch4 Page 46
Ch4 Page 47
Ch4 Page 48
Ch4 Page 49
Ch4 Page 50
Ch4 Page 51
Ch4 Page 52
Unsymmetric bending
Thus far, we have looked only at "regular bending" where the horizontal beam is aligned
in such a way that the neutral axis and centroidal moment of inertia are readily found.
The cross section in these situations usually have well-defined edges and/or axes of
symmetry that align with the coordinate axes. Under a bending moment, these beams
undergo symmetric bending. In the examples below, the bending occurs about the z axis.
There may be instances where the cross section of the beam is not aligned with the
coordinate axes. This is called unsymmetric bending.
Ch4 Page 53
Unsymmetric bending (cont'd)
The long (wrong) way to solve an unsymmetric bending problem: Treat the cross section
as a composite.
One way to treat this cross
section as a composite:
dividing it into several right
triangles and a rectangle.
The correct way to solve the problem is to treat the bending moment as a vector with
mutually perpendicular components.
Ch4 Page 54
Unsymmetric bending (cont'd)
By resolving the moment vector into two components, the loading is separated into two
simpler cases. Superpositioning is then applied.
Mz' causes bending about the z' axis, causing maximum tension and compression on
the top and bottom surfaces:
𝜎
𝑀 𝑐
= ⎯⎯⎯⎯⎯
𝐼
My' causes bending about the y' axis, causing maximum tension and compression on
the front and back surfaces:
𝜎
𝑀 𝑐
= ⎯⎯⎯⎯⎯
𝐼
Maximum overall tension is the point on the cross section that is under maximum
tension for both Mz' and My'. Likewise, maximum overall compression is the point on
the cross section that is under maximum compression for both Mz' and My'.
Ch4 Page 55
Unsymmetric bending (cont'd)
Interpreting moment components and their signs: the right-hand rule applies.
Ch4 Page 56
Unsymmetric bending: Example 1
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Separating into two different bending problems:
Point A is under max. tension due to both Mz and My. It is the point of max. tension.
Point C is the point of max. compression.
Due to Mz:
𝑀 𝑐 (30 × 10 cos 30° N m)(0.1 m)
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 9.743 MPa
1
𝐼
⎯⎯⎯(0.4 m)(0.2 m)
12
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −9.743 MPa
𝐼
Due to My:
𝑀 𝑐 (30 × 10 sin 30° N m)(0.2 m)
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 2.8125 MPa
1
𝐼
⎯⎯⎯
12 (0.2 m)(0.4 m)
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −2.8125 MPa
𝐼
Total:
𝜎 = 12.56 MPa,
𝜎 = −12.56 MPa
Ch4 Page 57
Unsymmetric bending: Example 2
4.136 The couple M acts in a
vertical plane and is applied to a
beam oriented as shown.
Determine the maximum tensile
and compressive stresses in the
beam and indicate the points where
they occur.
𝐼 = 14.77 × 10 mm
𝐼 = 53.6 × 10 mm
Separating into two different bending problems:
Point E is under max. tension due to
both Mz and My. It is the point of
max. tension.
Point B is the point of max.
compression.
Due to Mz:
(120 cos 20° N m)(0.016 m)
𝑀 𝑐
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 33.66 MPa
m
𝐼
(53.6 × 10 mm ) ⎯⎯⎯⎯⎯⎯⎯⎯
1000 mm
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −33.66 MPa
𝐼
Due to My:
(120 sin 20° N m)(0.01 m)
𝑀 𝑐
𝜎 = ⎯⎯⎯⎯ = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 27.79 MPa
m
𝐼
(14.77 × 10 mm ) ⎯⎯⎯⎯⎯⎯⎯⎯
1000 mm
𝑀 𝑐
𝜎 = − ⎯⎯⎯⎯ = −27.79 MPa
𝐼
Total:
𝜎 = 61.45 MPa,
𝜎 = −61.45 MPa
Ch4 Page 58
Unsymmetric bending: Example 3
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Separating into two different bending problems:
To find stresses due to Mz, we need 𝑦 and Iz. These were found in an earlier example to
be:
𝑦 = 0.133445 m
𝐼 = 1.33548 × 10
m
To find stresses due to My, we need 𝑧̅ and Iy. Because of symmetry, 𝑧̅ = 0 at the
vertical line of symmetry AF. Calculating the moment of inertia about this centroidal y
axis:
𝐼 = ∑ 𝐼 ̅ + 𝐴𝑑
1
= 2 ⎯⎯⎯(.06)(0.06)
12
+2
1
8
⎯⎯𝜋(0.04) − ⎯⎯⎯(0.04)
8
9𝜋
1
+ ⎯⎯𝜋(0.04)
2
1
+ ⎯⎯⎯(0.15)(0.12)
12
1
+ ⎯⎯𝜋(0.06)
8
𝐼 = w1.10529 × 10
m
Ch4 Page 59
4(0.04)
0.04 + ⎯⎯⎯⎯⎯⎯⎯
3𝜋
Unsymmetric bending: Example 3 (cont'd)
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Determining points of maximum tension and compression:
Due to Mz, point F is in max. tension and point A is in
max. compression.
Due to My, point I is in max. tension and point C is in
max. compression.
The points of overall max. tension and compression
are not obvious. We must check several different
points (namely corners and transitions).
Ch4 Page 60
Eccentric loading
Eccentric loading (loading that is not centric or "through the center") is a tensile or
compressive force that is not applied at the centroid of the cross section of a rod or
column, leading to bending in addition to the axial loading.
There are two common types of eccentric loading problems. The first type of eccentric
load results in symmetric bending and axial loading. The second type results in
unsymmetric bending and axial loading.
Eccentric loading that leads to symmetric bending:
(Above) In this case, the applied forces are moved to the centroid. The moments
due to the forces when the forces were at their points of application must not be
neglected. As seen on the right, this leads to axial loading and symmetric bending.
In Statics, the process of relocating an eccentric force to the centroid is formally
justified by adding equal and opposite forces:
Ch4 Page 61
Eccentric loading (cont'd)
Eccentric loading that leads to unsymmetric bending:
In this case, the applied force is offset from the centroid by distance b from the x
axis, which causes a moment about that axis. It is also offset by distance a from the
y axis, which causes a moment about that axis. These moments produce
unsymmetric bending. As long as these moments are accounted for, the force can
be moved to the centroid.
Ch4 Page 62
Eccentric loading: Example 1
4.103 The vertical portion of the press shown
consists of a rectangular tube of wall thickness
t = 10 mm. Knowing that the press has been
tightened on wooden planks being glued
together until P = 20 kN, determine the stress
at (a) point A, (b) point B.
Moving the applied loads to the centroid:
𝐴 = (0.08 𝑚)(0.06 𝑚) − (0.06 𝑚)(0.04 𝑚) = 0.0024 m
1
1
𝐼 = ⎯⎯⎯(0.06)(0.08) − ⎯⎯⎯(0.04)(0.06) = 1.84 × 10
12
12
Stress at point A:
𝑀 𝑐 𝑃 (20 × 10 N)(0.24 m)(0.04 m) 20 × 10 N
𝜎 = ⎯⎯⎯ + ⎯⎯= ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 112.68 MPa
𝐼
𝐴
1.84 × 10 m
0.0024 m
Stress at point B:
(20 × 10 N)(0.24 m)(0.04 m) 20 × 10 N
𝑀𝑐 𝑃
𝜎 = − ⎯⎯⎯ + ⎯⎯= − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −96.01 MPa
𝐼
𝐴
1.84 × 10 m
0.0024 m
Ch4 Page 63
m
Eccentric loading: Example 2
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Find the location of the centroid. The force will be moved to the centroid and moments
about the x and y axes can be calculated.
Area (in2)
𝒚 (in)
𝒚A (in3)
1
(6)(4) = 24
5
120
2
(10)(3) = 30 1.5

54
45
𝟏𝟔𝟓
𝒀 = ⎯⎯⎯⎯= 𝟑. 𝟎𝟓 𝐢𝐧
𝟓𝟒
165
Calculating moment components Mx and My.
Mx = (50 kips)(3.05 in) = 152.5 kipin
My = (50 kips)(5 in) = 250 kipin
Ch4 Page 64
Eccentric loading: Example 2 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Area (in2)
𝒚 (in)
𝒚A (in3)
1
(6)(4) = 24
5
120
2
(10)(3) = 30 1.5

54
45
𝟏𝟔𝟓
𝒀 = ⎯⎯⎯⎯= 𝟑. 𝟎𝟓 𝐢𝐧
𝟓𝟒
165
Calculating moments of inertia Ix and Iy.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
= ⎯⎯⎯(6)(4) + (24)(5 − 3.05)
12
1
+ ⎯⎯⎯(10)(3) + (30)(1.5 − 3.05)
12
𝐼 = 217.835 in
𝐼 = ∑𝐼 ̅
1
1
= ⎯⎯⎯(4)(6) + ⎯⎯⎯(3)(10)
12
12
𝐼 = 322 in
Ch4 Page 65
Eccentric loading: Example 2 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Superposing the two bending problems and the axial loading problem:
Point B is under max. tension
due to Mx and under tension
due to My. It is the point of
max. tension.
Point F is the point of max.
compression.
Stress at B is due to Mx, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = ⎯⎯⎯⎯ + ⎯⎯⎯⎯ − ⎯⎯
𝐼
𝐼
𝐴
(152.5 × 10 lb in)(3.95 in) (250 × 10 lb in)(3 in) 50 × 10 lb
𝜎 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= 4.169 ksi
217.835 in
322 in
54 in
Stress at F is due to Mx, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = − ⎯⎯⎯⎯ − ⎯⎯⎯⎯ − ⎯⎯
𝐼
𝐼
𝐴
(152.5 × 10 lb in)(3.05 in) (250 × 10 lb in)(5 in) 50 × 10 lb
𝜎 = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯= −6.94 ksi
217.835 in
322 in
54 in
Ch4 Page 66
Eccentric loading: Example 3
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Due to symmetry about both the y and z axis, the centroid is at the center of the cross
section. Calculating moment components Mz and My:
Mz = (50 kips)(4 in) = 200 kipin (bend down)
My = (50 kips)(8 in) = 400 kipin (bend forward)
Calculating area and moments of inertia Iz and Iy.
𝐴 = (16)(8) − 𝜋(2) − (6)(3) = 97.4336 in
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
= ⎯⎯⎯(16)(8)
12
1
8
−2 ⎯⎯𝜋(2) − ⎯⎯⎯(2)
8
9𝜋
1
−4 ⎯⎯⎯(3)(3)
12
𝐼 = 609.2 in
1
4(2)
+ ⎯⎯𝜋(2) 1 + ⎯⎯⎯⎯
2
3𝜋
Treated as 4 triangles, all
with their bases on the z
axis.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
1
= ⎯⎯⎯(8)(16) − ⎯⎯𝜋(2)
12
4
1
1
−2 ⎯⎯⎯(6)(3) + ⎯⎯(6)(3)(5)
36
2
𝐼 = 2259.1 in
Ch4 Page 67
Eccentric loading: Example 3 (cont'd)
For the beam and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Superposing the two bending problems and the axial loading problem:
Stress at B (max. tension) due to Mz, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = ⎯⎯⎯⎯+ ⎯⎯⎯⎯ + ⎯⎯
𝐼
𝐼
𝐴
[(50 × 10 lb)(4 in)](4 in) [(50 × 10 lb)(8 in)](8 in) 50 × 10 lb
𝜎 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
609.2 in
2259.1 in
97.4336 in
= 3242.9 psi
Stress at D (max. compression) due to Mz, My, and axial load:
𝑀 𝑐 𝑀 𝑐 𝑃
𝜎 = − ⎯⎯⎯⎯− ⎯⎯⎯⎯ + ⎯⎯
𝐼
𝐼
𝐴
[(50 × 10 lb)(4 in)](4 in) [(50 × 10 lb)(8 in)](8 in) 50 × 10 lb
𝜎 = − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯− ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯+ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
609.2 in
2259.1 in
97.4336 in
= −2216.5 psi
Ch4 Page 68
Eccentric loading: Example 4
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Find the location of the centroid. The force will be moved to the centroid and moments
about the x and y axes can be calculated.
Area (in2)
𝒚 (in)
1
⎯⎯𝜋 (4) = 25.1327
( )
12 − ⎯⎯⎯
= 13.698 344.26
2
(12)(6) = 72
9
648
3
(6)(6) = 36
4
144

133.1327
𝒚A (in3)
1136.26
𝟏𝟏𝟑𝟔. 𝟐𝟔
𝒀 = ⎯⎯⎯⎯⎯⎯⎯⎯⎯ = 𝟖. 𝟓𝟑𝟒𝟖 𝐢𝐧
𝟏𝟑𝟑. 𝟏𝟑𝟐𝟕
Calculating moment components Mx and My.
Mx = (20 kips)(3.4652 in) = 69.304 kipin
My = (20 kips)(6 in) = 120 kipin
Ch4 Page 69
Eccentric loading: Example 4 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Area (in2)
𝒚 (in)
1
⎯⎯𝜋 (4) = 25.1327
( )
12 − ⎯⎯⎯
= 13.698 344.26
2
(12)(6) = 72
9
648
3
(6)(6) = 36
4
144

133.1327
𝒚A (in3)
𝒀 = 𝟖. 𝟓𝟑𝟒𝟖 𝐢𝐧
1136.26
Calculating moments of inertia Ix and Iy.
𝐼 = ∑(𝐼 ̅ + 𝐴𝑑 )
1
8
= ⎯⎯𝜋(4) − ⎯⎯⎯(4)
8
9𝜋
+ (25.1327)(13.698 − 8.5348)
1
+ ⎯⎯⎯(12)(6) + (72)(9 − 8.5348)
12
1
+ ⎯⎯⎯(12)(6) + (36)(4 − 8.5348)
36
𝐼 = 1741.91 in
𝐼 = ∑𝐼 ̅
1
1
1
= ⎯⎯𝜋(4) + ⎯⎯⎯(6)(12) + 2 ⎯⎯⎯(6)(6)
8
12
12
𝐼 = 1180.53 in
Ch4 Page 70
Eccentric loading: Example 4 (cont'd)
For the column and loading shown, find the maximum tensile stress and the maximum
compressive stress as well as the points on the cross section at which they occur.
Superposing the two bending problems and the axial loading problem:
Axial loading:
𝑃
20 × 10 lb
𝜎 = − ⎯⎯= − ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = −150.226 psi
𝐴
133.1327 in
Ch4 Page 71
Ch4 Page 72
Ch4 Page 73