1. The beam is a cantilever 1.2 m long made of steel tube with an external diameter of 6 cm and internal diameter of 5 cm. 2. A concentrated load W is applied at the free end of the cantilever beam. 3. The maximum bending stress in the beam is not to exceed 1. The value of the load W that satisfies this condition is required.

1. Bending Test
Brittle materials
(Cast iron – Concrete)
Ductile materials (Steel)
• Types of bending tests:
Cold bend test Hot bend test
•ductile
1- Check the ductility
2
3
•bars &wires
•bar
Pin radius (R) = (1 –1.5) Bar diameter (D)
Bar diameter up to 25 mm (pin radius (R) = D)
Bar diameter more than 25 mm (R = 1.5 D)
1-1000ºC
2-
Quench bend test
quench
Nick bend test
• Signs of failure in cold bend test
1

2. Mp.l
. Y
I
Mmax
. Y
I
½ Pp.l
X ∆p.l
Ao
Lo
⅔ Pmax
X ∆max
Ao
Lo
Pp.l
. L3
48 ∆p.l
. I
Tensile stresses Compressive stresses Shear stresses Defects
• Loads causing bending stresses
Moment (M)
1. 3-Point bending
2. Center load
3. Simple beam
M =
Cantilever M = P L
Two point loading
Third points
4-Point bending M =
Quarter points
1bendingtwo point loading
2cast ironcenter load
σP.LBending stress at proportional limit
2
P L
4
P L
6
P
/2
P
/2
P
/2
P
/2
P
P

3. σmaxModulus of rupture
M.RModulus of resilience
M.T.Modulus of toughness
E
Modulus of elasticity (stiffness)
Area (A) Y Moment of inertia (I)
π
/4 D2 π
/64 D4
π
/4( D2
-d2
) π
/64 (D4
– d4
)
bh
• The effect of variables on the results:
1. Specimen dimensions
When the cross-section area (or) the length (decrease), the modulus of
rupture (Increase) and modulus of elasticity (decrease)
modulus of rupture
3
P
Δ
bh3
21
D
2
D
2
h
2
D
D
d
h
b

4. The modulus of rupture and the modulus of elasticity are lower for I section
extreme fiberStrength
2. Types of loading
modulus of rupture
Modulus of rupture from Cantilever > Modulus of rupture from Center
load (3-point) > Modulus of rupture from 4-point loading
Modulus of rupture for 4-point loading is less than that for 3-point loading
by (10 – 25%)
Rate of loading
The greater the test speed the higher the measured strength and hardness
1-To avoid shear failure L/d = (6 -12(
2- To avoid buckling L/b < 15 (b = width)
( ) A timber beam of circular cross-section simply supported over a span of 50
cm was tested in bending under a central load. If the modulus of elasticity of
this wood was 790 ton/cm2
and the loads versus mid span vertical deflection
were as follows:
P, ton 0 0.6 1 1.4 1.8 2.2 2.6 2.7 2.8
Δ, mm 0 0.5 0.8 1.2 1.5 2 3 4 5
Draw the load-deflection diagram and determine:
a- Diameter of the beam b- Elastic bending strength
c- Modulus of rupture d- Modulus of resilience
e- Modulus of toughness
4

5. Modulus of elasticity E = = = 790 ton/cm2
D = 5.3 cm
Elastic bending strength = = = 1.54 ton/cm2
Modulus of rupture = = = 2.4 ton/cm2
Modulus of resilience = = = 1.22 x 10-4
ton/cm2
Modulus of toughness = = = 8.46 x 10-4
ton/cm2
( ) A cast iron beam of circular cross section was simply supported over a span
of 50 cm and tested in bending under concentrated load at its mid-span. The
modulus of elasticity is 800 ton/cm2
. Find:
− The diameter of the beam if the resilience is 0.2 ton.cm and the load at the
proportional limit is 2 ton
− The maximum load on the beam if the maximum bending strength is 2.8
ton/cm2
− Explain the fracture behavior of the beam.
5
0
0.5
1
1.5
2
2.5
3
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5
P
Δ
W
D L = 50 cm I = π
/64
D4
Y = D/2M =
PL
4
½ Pp.l
X ∆p.l
Ao
Lo
⅔ Pmax
X ∆max
Ao
Lo
Pp.l
. L3
48 ∆p.l
. I
1.8 x (50)3
48 x 0.15 x π
/64
D4
Mp.l
Y
I π
/64
x (5.3)4
4
1.8 x 50
2
5.3
X
Mmax
Y
I π
/64
x (5.3)4
4
2.8 x 50
2
5.3
X
½ x 1.8 x 0.15
π
/4
(5.3)2
x 50
⅔ x 2.8 x 0.5
π
/4
(5.3)2
x 50
W
D L = 50 cm I = π
/64
D4
Y = D/2M =
PL
4

6. Pp.l = 2 ton Resilience = ½ Pp.l X ∆p.l = ½ x 2 x ∆p.l = 0.2 ∆p.l = 0.2 cm
Modulus of elasticity E = = = 800 ton/cm2
D = 5.1 cm
Modulus of rupture = = = 2.8 ton/cm2
Pmax = 2.92 ton
( ) A bending test was carried out on two cast iron beams, A and B, by loading
each beam in the center of its span to fracture. The following readings were
recorded:
Beam Diameter (mm) Span (mm) Fracture load (kg) Central deflection (mm)
A 40 600 1500 1.4
B 60 800 2000 1.6
Which one can be considered better than the other for strength property
and toughness property?
Beam A Beam B
Y = D/2 = 40/2 = 20 mm
I= π
/64 D4
= π
/64 (40)4
Y = D/2 = 60/2 = 30 mm
I= π
/64 D4
= π
/64 (60)4
6
Pp.l
. L3
48 ∆p.l
. I
2 x (50)3
48 x 0.2 x π
/64
D4
Mmax
Y
I π
/64
x (5.1)4
4
Pmax
x 50
2
5.1
X

7. Strength properties
=
σ =0.036 ton/mm2
=
σ =0.019 ton/mm2
Toughness properties
T = ⅔ Pmax x ∆max = ⅔ x 1.5 x 1.4
T = 1.4 ton. mm
T = ⅔ Pmax x ∆max = ⅔ x 2 x 1.6
T = 2.13 ton. mm
Beam A is better than beam B for strength properties and lower in toughness
properties
( ) A beam of circular cross-section of 5 cm in diameter, simply supported over
a span of 40 cm, and was tested in bending under a central load. The loads
and corresponding deflections until rupture were as follows:
Load ( ton)
0 0.5 1 1.25 1.7 2 2.4
2.5
5
2.6
Deflection (mm) 0 0.4 0.8 1 1.5 2 3 4 4.8
(i) Draw the load-deflection diagram and determine:
• Stress at proportional limit
• Modulus of elasticity in bending
• Modulus of rupture
• Modulus of resilience
(ii) Draw the bending stress distribution on a section at 10 cm from the
support when the central load was 1 ton
7
π
/64
x (40)4
4
1.5 x 600
2
40
XMmax
Y
I
σ = π
/64
x (60)4
4
2 x 800
2
60
XMmax
Y
I
σ =
W
5 cm
L = 40 cm I = π
/64
(5)4
Y = 5/2 = 2.5M =
PL
4

8. Modulus of elasticity E = = = 543 ton/cm2
Elastic bending strength = = = 1.02 ton/cm2
Modulus of rupture = = = 2.12 ton/cm2
Modulus of resilience = = = 7.96 x 10-5
ton/cm2
( ) A cast iron beam of circular cross-section of 7 cm in diameter, simply
supported over a span of 60 cm was tested in bending under a central load.
If the loads and corresponding deflections until rupture were as follows:
Load ( ton) 0 1.2 2 2.8 3.6 4.4 5.2 5.5 5.6
Deflection (mm) 0 1 1.6 2.4 3 4 6 8 10
Draw the load-deflection diagram and determine:
i) Modulus of rupture.
ii) Modulus of elasticity, E, in bending.
iii) Draw fracture shape of the beam.
8
½ Pp.l
X ∆p.l
Ao
Lo
Pp.l
. L3
48 ∆p.l
. I
1.25 x (40)3
48 x 0.1 x π
/64
(5)4
Mp.l
Y
I π
/64
x (5)4
4
1.25 x
2
5
X
Mmax
Y
I π
/64
x (5)4
4
2.6 x 50
2
5
X
½ x 1.25 x 0.1
π
/4
(5)2
x 40
0
0.5
1
1.5
2
2.5
3
0 1 2 3 4 5 6
P
Δ
W
7 cm
L = 60 cm I = π
/64
(7)4
Y = 7/2 = 3.5M =
PL
4

9. Mmax
Y
I π
/64
x (7)4
X
Pp.l
. L3
48 ∆p.l
. I
3.6 x (60)3
48 x 0.3 x π
/64
(7)4
0
100
200
300
400
500
600
700
800
0 5 10 15 20 25 30 35 40 45
P
Δ
I = 1
/12
[2.5(5)3
] Y = 5/2 = 2.5
L = 90 cm M =
PL
4
W
2.5 cm
5cm
Modulus of rupture = = = 2.49 ton/cm2
Modulus of elasticity E = = = 458 ton/cm2
Fracture Shape
( ) A timber beam was centrally loaded; the distance between supports being 90
cm, the cross section of the beam was 2.5 cm breadth and 5 cm depth. If the
loads and corresponding deflections are as follows:
Load ( Kg) 100 200 300 400 500 600 700 750
Deflection (mm)
4.75 9.52 14.3 19.02
23.8
6
27.2
36.
1
Failure
Draw the load-deflection diagram and determine:
a) Extreme fiber stress at limit of proportionality
b) Modulus of rupture
c) Modulus of resilience
d) Modulus of toughness
e) Stiffness of material
9
0
1
2
3
4
5
6
0 2 4 6 8 10 12
P
Δ
4
5.6 x 60
2
7

10. W
D cm
L = 80 mm I = π
/64
(D)4
Y = D/2M =
PL
4
1000
1200P
Stress at proportional limit = = = 1.08 ton/cm2
Modulus of rupture = = = 1.62 ton/cm2
Modulus of resilience = = = 0.53 x 10-3
ton/cm2
Modulus of toughness = = = 1.87 x 10-3
ton/cm2
Modulus of elasticity E = = = 122 ton/cm2
( ) A cast iron beam of circular cross-section was tested in bending under a
central load and on support 80 mm apart. If the modulus of resilience of this
grade of cast iron equals 0.196 Kg/cm2
and the loads and corresponding
deflections measured during the test are as follows:
Load ( Kg)
150 300 450 600 750
90
0
1050 1100
Deflection (mm) 1.05 2.1 3.15 4.2 5.25 6.8 8.88 Failure
Draw the load-deflection diagram and determine:
a) Diameter of cross section
b) Modulus of rupture
10
½ Pp.l
X ∆p.l
Ao
Lo
⅔ Pmax
X ∆max
Ao
Lo
Mp.l
Y
I 1
/12
[2.5(5)3
]
4
0.5 x
2
5
X
Mmax
Y
I 1
/12
[2.5(5)3
]
4
0.75 x
2
5
X
½ x 0.5 x 2.386
2.5 x 5 x 90
⅔ x 0.75 x 4.2
2.5 x 5 x 90
Pp.l
. L3
48 ∆p.l
. I
0.5 x (90)3
48 x 2.386 x 1
/12
[2.5(5)3
]

11. 0
200
400
600
800
0 1 2 3 4 5 6 7 8 9 10
Δ
c) Modulus of elasticity
Modulus of resilience = = = 1.96 x 10-3
kg/mm2
D = 126 mm
Modulus of rupture = = = 0.112 kg/mm2
Modulus of elasticity E = = = 0.123 kg/mm2
( ) A simply supported cast iron beam of square cross-section 12cm x 12cm was
centrally loaded by 24 tons, if the bending strength of the beam is 25
kg/mm2
. Calculate the span of the beam.
= = 25 kg/mm2
L = 1200 mm
11
Mmax
Y
I π
/64
x (126)4
4
1100 x 80
2
126
X
½ Pp.l
X ∆p.l
Ao
Lo
½ x 750 x 5.25
π
/4
(D)2
x 80
Pp.l
. L3
48 ∆p.l
. I
750 x (80)3
48 x 5.25 x π
/64
(126)4
I = 1
/12
[120(120)3
] Y = 120/2 = 60mm
L =??? M =
PL
4
P
12 cm
12cm
1
/12
[120(120)3
]
4
24000 L
2
120
XM . Y
I
σ =

12. ( ) A cantilever 1.2 m consisting of a steel tube with external and internal
diameters of 6 and 5 cm respectively, carries a concentrated load of W (Kg)
at the free end. Neglecting the weight of the tube, it is required to find the
value of W if the maximum bending stress is not to exceed 1.3 ton/cm2
.
= = 1.3 ton/cm2
W = 0.119 ton = 119 kg
12
π
/64
[(6)4
- (5)4
]
W x 120 x 3M .Y
I
σ =
M = P L
P
L = 120 cm I = π
/64
[(6)4
- (5)4
] Y = D/2 = 6/2 = 3 cm
6
5

13. I = 1
/12
[10(15)3
] Y = 15/2 = 7.5cm
L = 240 cm
M =
W L
4
=
240W
4
I = 1
/12
[10(h)3
] Y = h/2
M =
W x 0.6 x 100
2
=
( ) A simply supported beam of 2.4 m span
has a constant width of 10 cm throughout
its length with varying depth of 15 cm at
the centre to minimum at the ends as
shown in fig. the beam is carrying a point
load W at its mid span. Find the
minimum depth of the beam at a section
0.6 m from the left hand support. Such that the maximum bending stress at
this section is equal to that at the mid span of the beam.
At mid span
= = 0.16 W
At 0.6 m from left support
30 W
13
2.4m
W
BCA
10
15
W
W/2W/2
0.6 m
10
10
15
1
/12
[10(15)3
]
4
240 W
2
15
XM . Y
I
σ1
=

14. 2 = =
σ1 = σ2
0.16 W = 18 W / h2
h = 10.61 cm
14
h
10
1
/12
[10(h)3
]
30 W XM . Y
I
σ = 2
h
h2
18W