Bending
Symbols Bank
• 𝑅𝐴 and 𝑅𝐵 are the reaction forces at points A
and B.
• L is the length of the beam.
• 𝜎𝑚𝑎𝑥 is the maximum bending stress.
• M is the internal bending moment.
• I is moment of Inertia.
• Y is the distance from NA to outer fiber.
• 𝛿𝑚𝑎𝑥 is the maximum deflection.
• 𝜃𝐴 and 𝜃𝐵 are the slope at the end of the beam.
• y is the equation of elastic curve.
• 𝑉𝑚𝑎𝑥 is the maximum shear load.
The Maximum
Stress Formula
𝜎𝑚𝑎𝑥
𝑀𝑌
=
≤ 𝜎𝑎𝑙𝑙
𝐼
The Flexural
Stiffness Formula
= 𝐸𝐼
Moment of
Inertia of a
Circle
𝜋𝑅4 𝜋𝐷4
𝐼=
=
4
64
Moment of
Inertia of a
Circular Tube
𝜋(𝑅4 − 𝑟 4 ) 𝜋(𝐷4 − 𝑑4 )
𝐼=
=
4
64
Moment of
Inertia of a
Rectangle
𝐵𝐻 3
𝐼=
12
Load Diagram
Deflection Diagram
Shear Diagram
Moment Diagram
Case 1 : Point load
at the midspan of a
simply supported
beam
Reaction
Forces
𝑃
𝑅𝐴 = 𝑅𝐵 =
2
Maximum
Moment
𝑀𝑚𝑎𝑥
Maximum
Deflection
𝑃𝐿
𝑃𝐿3
=
𝛿𝑚𝑎𝑥 =
4
48𝐸𝐼
Load Diagram
Deflection Diagram
Shear Diagram
Moment Diagram
Case 2 : Two
identical point
loads, symmetrically
places on a simply
supported beam.
Reaction
Forces
𝑅𝐴 = 𝑅𝐵
=𝑃
Maximum
Moment
𝑃𝐿
𝑀𝑚𝑎𝑥 =
8
Load Diagram
Deflection Diagram
Shear Diagram
Moment Diagram
Case 3 : Point load
at the end of a
cantilever beam.
Reaction
Forces
Maximum
Moment
Maximum
Deflection
𝑅𝐵 = 𝑃
𝑀𝑏 = 𝑀𝑚𝑎𝑥
= 𝑃𝐿
𝛿𝑚𝑎𝑥
𝑃𝐿3
=−
3𝐸𝐼
A cantilever beam is 5 m long and has a point load of 50 kN at the free
end. The deflection at the free end is 3 mm downwards. The modulus of
elasticity is 205 GPa. The beam has a solid rectangular section with a
depth 3 times the width. (D=3B).
Given:
𝐿 =5𝑚
𝐹 = 50000 𝑁
𝛿 = − 3 × 10−3 𝑚
𝐸 = 205 × 10−3 𝑀𝑃𝑎
𝐻 = 3𝐵
Required:
a- the flexural stiffness
b- The dimension of the section
c- The yield stress
F
5 mm
a- The
Flexural
Stiffness
𝐹𝐿3
50000 × 5
𝛿=−
= −0.003 = −
3𝐸𝐼
3𝐸𝐼
0.003 × 3
𝐸𝐼 =
50000 × 5
3
3
= 694.4 × 106 𝑁𝑚2
694.4 × 106 694.4 × 106
−3 𝑚4
𝐼=
=
=
3.38
×
10
𝐸
205 × 109
b- The
Dimension of
the Section.
𝐵𝐻 3 𝐵 3𝐵
𝐼=
=
12
12
3
27𝐵4
=
= 0.003387
12
𝐵 = 0.19 𝑚
𝐻 = 0.59 𝑚
𝑀𝑦
𝜎𝑦 =
𝐼𝑥
c- The Yield
Stress
𝐻 0.59
𝑦= =
= 0.295 𝑚
2
2
𝑀𝑦 = 𝐹𝐿 = 50000 5 = 25000 𝑁. 𝑚
347
𝐼=
𝑚
102500
𝜎𝑦 = 21.82 𝑀𝑃𝑎
A simply supported beam is made from a hollow tube 80 mm outer
diameter and 40 mm inner diameter. It is simply supported over a span
of 6 m. A point load of 900 N is placed at the middle. Find the
deflection at the middle if E=200 GPa.
Given:
𝑑𝑜 = 80 × 10−3 𝑚
𝑑𝑖 = 40 × 10−3 𝑚
𝐹 = 900 𝑁
𝐿 =6𝑚
𝐸 = 200 × 109 𝑀𝑃𝑎
Required:
a- Deflection
F
3 mm
3 mm
𝜋(𝐷4 − 𝑑4 ) 𝜋((80)4 − (40)4 )
𝐼=
=
64
64
𝐼 = 1.885 × 10−6 𝑚4
a- Deflection
𝐹𝐿3
900 × 6 3
𝛿=−
=−
48𝐸𝐼
48 200 × 109 1.885 × 10−6
𝛿 = −0.0107 𝑚
10.7 mm downward
Find the flexural stiffness of a simply supported beam which limits the
deflection to 1 mm at the middle. The span is 2 m and the point load is
200 kN at the middle.
Given:
𝐹 = 200000 𝑁
𝐿 =2𝑚
𝛿 = 1 × 10−3 𝑚
Required:
a- Flexural Stiffness
1 mm
1 mm
F
a- The
Flexural
Stiffness
𝐹𝐿3
200000 × 2
𝛿=−
=1=−
48𝐸𝐼
3𝐸𝐼
200000 × 2
𝐸𝐼 =
1 × 48
3
3
= 33.3 × 106 𝑁𝑚2
A cantilever 1.2 meters consisting of a steel tube with external and
internal diameters of 60 mm and 50mm respectively. Carries a
concentrated load (W) at the free end, neglecting the weight of the tube
it's required to find the value of (W) if the maximum bending stress is
200 Mpa and the factor of safety = 1.5
W
Given:
𝑑𝑜 = 60 × 10−3 𝑚
𝑑𝑖 = 50 × 10−3 𝑚
𝐿 = 1.2 𝑚
𝜎𝑚𝑎𝑥 = 200 𝑀𝑃𝑎
𝑁 = 1.5
Required:
a- The Point Load W
1.2 mm
W
60
𝑌=
= 30
2
1.2 mm
𝜎𝑚𝑎𝑥
𝑀𝑌 𝐹𝐿𝑌
=
=
𝐼
𝐼
𝜋(𝐷4 − 𝑑 4 )
𝐼=
64
𝜋((60)4 −(50)4 )
=
64
a- The Point
Load W
𝐹 × 1200 × 30
200 =
𝜋((60 × 10−3 )4 −(50 × 10−3 )4 )
64
𝐹 = 1829.869 𝑁
𝐹 1829.869
=
= 1219.91 𝑁
𝑁
1.5