3 bending test

Bending Test

Brittle materials(Cast iron – Concrete) Ductile materials (Steel)

مقاومة ضعف نتيجة الكسر يحدثالشد الجهادات ظهور المادة يبدأ و

من عليها الشروخ التى الطبقةشد اجهادات

يتوازى حتى العينة تنثنىو فى طرفاها شروخ تظهر ال

نسبة حالة فى وخاصة العينةالمنخفضة الكربون

Types of bending tests:

Cold bend test Hot bend test ال المواد على اجرائه ductileيتم

بغرض:1- Check the ductility

و - 2 الكربون نسبة ارتفاع مدى تحديدالفوسفور

بعد -3 المطيلية فى الفاقد تحديدالحرارية المعالجات

على اجرائه شكل يتم على bars العينات&wires

ثنى على حادة barيعتمد بزاويةفى مدى ومالحظة الشروخ انتشار

العينةPin 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)

الحديد -1 على اجرائه يتمبتسخينه المطاوع

ثنيه 1000ºCلدرجة ومن -2 للتاكد اجرائه يتم

نسبة ارتفاع مدىالكبريت

Quench bend testمسامير على يجرى

بتسخينها الصلب البرشامعمل الماء quenchثم فى

الثنى ثم

فى االرتفاع لتحديد يجرىالكربون نسبة

Nick bend testالمادة عيوب لتحديد

1

ثم للعينة عنق عمل يتمالثنى

Signs of failure in cold bend test

Tensile stresses Compressive stresses Shear stresses Defects

Loads causing bending stresses لعزوم المولدة االحمالانحناء

Moment (M)

1. 3-Point bending2. Center load3. Simple beam

M =

Cantilever M = P L

2

P L4

P/2P/2

P/2P/2

P

P

Mp.l . YI

Mmax . YI

½ Pp.l X ∆p.lAo Lo

⅔ Pmax X ∆maxAo Lo

Pp.l . L3

48 ∆p.l . I

Two point loadingThird points

4-Point bendingM =

Quarter points

مالحظات:اختبار- 1 استخدام bendingفى يفضل الننا two point loadingللخرسانة

خالص انحناء عزم تأثير تحت بالكمرة منطقة على النوع هذا فى نحصلمثل- 2 القصفة المعدنية المواد اختبار استخدام cast ironفى يفضل

center load ولكن قص وقوة انحناء عزم الى الكمرة تعرض رغم لسهولته

بسبب فيها الكسر ويحدث القص اجهادات مقاومة فى قوية المواد هذه

االنحناء عزم من الناتجة الشد قوة

القوانين

P.LBending stress at proportional limit

maxModulus of rupture

M.RModulus of resilience

M.T.Modulus of toughness

EModulus of elasticity (stiffness)

3

P L6

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)

قيمة على المقطع شكل يؤثر modulus of كماrupture المساحة تساوى عند

The modulus of rupture and the modulus of elasticity are lower for I section

4

P

Δ

bh3

21

D 2

D 2

h 2

D

Dd

hb

من بالقرب المقطع مساحة زادت تقل extreme fiberكلما اى الكسر حمل يزدادStrength

2. Types of loading

قيمة التحميل modulus of rupture تختلف نوع باختالفModulus 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 2.8Δ, mm 0 0.5 0.8 1.2 1.5 2 3 4 5

5

Draw the load-deflection diagram and determine:a- Diameter of the beam b- Elastic bending strengthc- Modulus of rupture d- Modulus of resilience e- Modulus of toughness

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.

6

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

ΔW

D L = 50 cm I = π/64 D4 Y = D/2 M =PL4

½ Pp.l X ∆p.l

Ao Lo

⅔ Pmax X ∆max

Ao Lo

Pp.l . L3

48 ∆p.l . I1.8 x (50)3

48 x 0.15 x π/64 D4

Mp.l Y

I π/64 x (5.3)4

41.8 x 50

25.3

X

Mmax Y

I π/64 x (5.3)4

42.8 x 50

25.3

X

½ x 1.8 x 0.15 π/4 (5.3)2 x 50

⅔ x 2.8 x 0.5π/4 (5.3)2 x 50

W

D L = 50 cm I = π/64 D4 Y = D/2 M =PL4

Pp.l = 2 ton Resilience = ½ Pp.l X ∆p.l = ½ x 2 x ∆p.l = 0.2 ∆p.l = 0.2 cm


D = 5.1 cm


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 1.4B 60 800 2000 1.6Which one can be considered better than the other for strength property and toughness property?

7

Pp.l . L3

48 ∆p.l . I2 x (50)3

48 x 0.2 x π/64 D4

Mmax Y

I π/64 x (5.1)44

Pmaxx 50 2

5.1 X

Beam A Beam BY = 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

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 2.55 2.6Deflection (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

8

π/64 x (40)44

1.5 x 600 2

40 X Mmax Y

Iσ = π/64 x (60)4

42 x 800

260

X Mmax Y

Iσ =


Elastic bending strength = = = 1.02 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 5.6Deflection (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.

9

½ Pp.l X ∆p.l

Ao Lo

Pp.l . L3

48 ∆p.l . I1.25 x (40)3

48 x 0.1 x π/64 (5)4

Mp.l Y

I π/64 x (5)44

1.25 x 40 2

5 X

Mmax Y

I π/64 x (5)44

2.6 x 50 2

5 X

½ x 1.25 x 0.1 π/4 (5)2 x 40

W5 cm L = 40 cm I = π/64 (5)4 Y = 5/2 = 2.5 M =

PL4

0

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5 6

P

Δ

W7 cm L = 60 cm I = π/64 (7)4 Y = 7/2 = 3.5 M =

PL4

Mmax Y

I π/64 x (7)4

X

Pp.l . L3

48 ∆p.l . I3.6 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 =PL4

W

2.5 cm

5 cm



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 750Deflection (mm) 4.75 9.52 14.3 19.02 23.86 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 resilienced) Modulus of toughnesse) Stiffness of material

10

0

1

2

3

4

5

6

0 2 4 6 8 10 12

P

Δ

45.6 x 60

2 7

Stress at proportional limit = = = 1.08 ton/cm2



Modulus of toughness = = = 1.87 x 10-3 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) 150 300 450 600 750 900 1050 1100Deflection (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 sectionb) Modulus of rupture

11

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 90 2

5 X

Mmax Y

I 1/12 [2.5(5)3]4

0.75 x 90 2

5 X

½ x 0.5 x 2.386 2.5 x 5 x 90

⅔ x 0.75 x 4.22.5 x 5 x 90

Pp.l . L3

48 ∆p.l . I0.5 x (90)3

48 x 2.386 x 1/12 [2.5(5)3]

WD cm L = 80 mm I = π/64 (D)4 Y = D/2 M =

PL4

0

200

400

600

800

1000

1200

0 1 2 3 4 5 6 7 8 9 10

P

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.

12

Mmax Y

I π/64 x (126)4

4 1100 x 80

2126

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 . I750 x (80)3

48 x 5.25 x π/64(126)4

I = 1/12 [120(120)3] Y = 120/2 = 60mm

L =??? M =PL4

P

12 cm

12 c

m

1/12 [120(120)3]4

24000 L 2

120 X M . Y

Iσ =

= = 25 kg/mm2

L = 1200 mm

( ) 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.

13

M = P LP

L = 120 cm I = π/64 [(6)4 - (5)4] Y = D/2 = 6/2 = 3 cm 65

L = 240 cm

= = 1.3 ton/cm2

W = 0.119 ton = 119 kg

( ) 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.

14

2.4m

W

B C A

π/64 [(6)4 - (5)4]W x 120 x 3 M . Y

Iσ =

I = 1/12 [10(15)3] Y = 15/2 = 7.5cm

M =W L

4=

240W 4

I = 1/12 [10(h)3] Y = h/2

M = W x 0.6 x 100

2=

At mid span

= = 0.16 W

At 0.6 m from left support

30 W

2 = =

σ1 = σ2

0.16 W = 18 W / h2

h = 10.61 cm

15

10

15

h

10

W

W/2W/2

0.6 m

10

10 15

1/12 [10(15)3]4

240 W 2

15 X M . Y

Iσ1=

1/12 [10(h) 3]30 W X M . Y

Iσ = 2

h

h2

18W

3 bending test

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