Compression

The reason for barrel shape:

كبير - 1 احتكاك يحدث العينة ضغط عندور نهايتىبين الماكينةى سأالعينة

تقاوم ىقو- 2 الناتجة التمدداالحتكاكنهايتيها عند للعينة وتسمح العرضى

عرض تمددب حيث ياالعينة المنتصف منلقو تاثير االحتكاكى اضعف

Types of failure:

Brittle material (Cast iron)

Semiductile materials (Brass)

Ductile materials(Steel)

Shear failureانضغاط يحدث بسيط

عل كسر زاوية يثمبين )60º-55(تتراوح

Shear failure

على الكسر يحدث50ºزاوية

يحدث ولكن failureاليحدث

ظهور مع كلى انضغاطشروخ

العينة سطح فى(θ = 45º+ Φ /2)

الكسر زاوية بين االحتكاك θالعالقة زاوية و

1

P

Δ L

55 - 60°

Δ LP

50°

P

Δ L

المستويات Φ الذرية بين

Limitations of compression test:

تطبيق- 1 بدقة load axialصعوبةالعينة للعينة unstable loadingحدوث- 2 على انحناء عزوم تولد و

ذات- 3 ماكينات استخدام يتطلب كبيرة مقاطع ذات عينات استخدام

مرتفعة قدرةراس- 4 بين الكبير و ىاالحتكاك العينة نهايتىالماكينة

Precautions االحتياطات على -1 للحصول دائرية مقاطع ذات اسطوانية المستخدمة العينات

لالجهادات منتظم .توزيع

قياسية -2 ابعاد ذات حدوث )Lо = (2-10( dо) العينات للعينة Bucklingلعدم

ت الضغط نتيجة اجهادات الى باالضافة عليها انحناء عزم ولدالعينة - 3 محور على ومتعامدان ومتوازيان مستويان العينة سطحاالعينة - 4 سطح على بزاوية يميل الماكينة رأس نجعل االحتكاك لتقليل

بين تتراوح هى و االحتكاك زاوية °)14-3(تسمى

The use of spherical bearing blocks

ميل - 1 فى انحراف اى تعديل على تعمل

عدم نتيجة العينة على المؤثر الحمل

العينة سطح انتظامانطباق - 2 على تعمل

محور مع العينة محور

االختبار وعندها ماكينة

محصلة وتكون لالجهادات منتظم توزيع على نحصل

عمودية ى القو العينة المؤثرة سطح على

2

Spherical block

Specimen

Machine head

Pp.l

Aо

Pmax

Aо

Aо Lо

½ Pp.l X Lp.l

Aо Lо

⅔ Pmax X Lmax

LD

LD

القياسية العينات انواع

UseSizeSpecimenFor bearing materialsL = 0.9DShort

Determination of compressive strengthL = (2- 3)DmediumDrawing the P- ΔL and determining the

Young's modulusL = (8-10)DLong

: االختبار نتائج على العينة أبعاد تأثير

اال .1 البد لكن و العينة على االحتكاك تاثير يقل نسبة زادت كلما

عن النسبة وبالتالى 10تزيد العينة على انحناء عزوم تولد لتجنب

لالجهادات المتجانس التوزيع عدمنسبة .2 تقل العينة 1.5عن عندما قاعدة الكسر مستوى يقطع

Strengthفتزيد

(L/D > 2 (to provide suitable middle region free from end restarints

القوانين

p.lElastic (proportional( limit stress

maxUltimate compressive strength

M.RModulus of resilience

M.T.Modulus of toughness

3

Lf

Lо

ε1

1

2

% Contraction

tan βSecant modulus at stress 1

Prove that (θ = 45º+ (

الكسر زاوية بين العالقة هى السابقة بين θالعالقة االحتكاك زاوية و المستويات

Friction force = عموديةقوة x االحتكاك معامل

االحتكاك μF = معاملاستبدال tan Φبالمعامل μتم

F= (Pcos θ) (tan Φ)τ = force/areaArea = A/ cos θ τ = (cos θ/A)[Psin θ - Pcos θ tan Φ]τ = (P/A)[ sin θ cos θ - cos 2θ tan Φ]τ = (P/A)[ ½ sin 2θ - cos 2θ tan Φ]

على تجعل θللحصول يمكن τالتى ما اكبرdτ/dθ = 0 cos 2θ +2 sin θ cos θ tan Φ = 0 cos 2θ + sin 2θ tan Φ = 0

4

N

μ N

θ

θ

P

P

P sin θ

P cosθ tan

P cos θ

2

cos 2θ / sin 2θ + tan Φ = 0 -tan Φ = cot 2θ-tan Φ = tan (90 - 2θ)

θ = 45º+

Φ المعدن تزداد جزيئات حجم زاد كلما

( ( Determine the relative displacement of points A and D of a steel rod of variable cross section area shown in Fig. when subjected to 4 concentrated load forces P1, P2, P3, and P4 . (Es = 200 GPa(. Cross sectional area of AB and CD = 0.001 m2, and BC = 0.002 m2

ΔL =

A1 = 0.001 m2 A2 = 0.002 m2 A3 = 0.001 m2

Tension Compression Tension

ΔL1 = ΔL2 = ΔL3 =

5

P1 = 100 KN

2 m

P 4NK 001 = P 3NK 001 =

P 2NK 001 = A B C D

1 m 5.1 m

L PA E

P 1L 1

A E 1

P 2L 2

A E 2

P 3L 3

A E 3

01 x 2 x 001 3

002 01 x 9100.0 x 051 01 x 1 x 3

01 x 002 9200.0 x 01 x 5.1 x 05 3

002 01 x 9100.0 x

NK001

2 m

NK 05NK 051NK 051NK001

1 m5.1

m NK 05

ΔL1 = ΔL2 = ΔL3 =

ΔL1 = 0.001 m ΔL2 = 0.000375 ΔL3 = 0.000375

ΔLtot = ΔL1 + ΔL3 – ΔL2 = 0.001 m (Tension)

( ( A brass bar having cross-sectional area of 10 cm2 is subjected to axial forces as shown in Fig. 1. Find the total change in length of the bar. Eb = 1.05 x 106 Kg/cm2

ΔL =

Tension Compression Compression

ΔL1 = ΔL2 = ΔL3 =

6

5000 Kg

60 cm

1000 Kg2000 Kg8000 Kg

100 mc

021 mc

L PA E

P 1L 1

A E 1

P 2L 2

A E 2

P 3L 3

A E 3

06 x 000550.1 01 x 601 x

0003 001 x50.1 01 x 601 x

021 x 000150.1 01 x 601 x

gK 0005

06 mc

gK 0001gK 0003gK 0003

gK0005

001 mc

021 mc

gK 0001

ΔL1 = ΔL2 = ΔL3 =

ΔL1 = 0.0286 cm ΔL2 = 0.0286 cm ΔL3 = 0.0114 cm

ΔLtot = ΔL1 + ΔL3 – ΔL2 = 0.0114 cm

( ( A member formed by connecting a steel bar to an aluminum bar is shown in the Fig. (3(. Assuming that the bars are presented from buckling sidewise, calculate the magnitude of force P that will cause the total length of the member to decrease 0.25 mm. The values of elastic modulus for steel and aluminum are 210 KN/mm2 and 70 KN/mm2 respectively.

ΔL =

ΔLtot = ΔLs + ΔLAl = 0.25

+ = 0.12

7

P LE A

300 P210 x 05 x

05

P 08307 001 x 001 x

rab munimulA001 001 x

rab leetS05 05 x

P

083 mm

mm 003

P = 224.4 kg

( ( A bar ABC shown in the Fig. (2( consists of two parts AB and BC, each part being 1 meter long and having cross-section area of 2 cm2 and 3 cm2 respectively. The bar is suspended from A and there is a rigid horizontal support at 2.001 m from A. A force of 10 tons acting vertically downwards is applied at B. Determine (i( the reaction produced by the rigid horizontal support, (ii( the stresses in parts AB and BC of the bar. Take E = 2 x 106 kg/cm2

(i) ΔL =

ΔLtot = ΔLAB - ΔLBC = 0.001

+ = 0.1

8

A

A1 = 2 cm2

10 not

1 m

m 1

1 mmA 2mc 3 = 2

B

C

L PA E

001 x (R – 01) 2 01 x 62 x

001 x R 2 01 x 63 x

noisserpmoC

R

R

noisneT

01 - R

01 - R

PBA

A о

6.3 – 01 2

PCB

A о

6.3 3

R = 3.6 kg

(ii)

AB = = = 3.2 kg/cm2

BC = = = 1.2 kg/cm2

( ( A reinforced concrete column of cross-section is shown in the figure. The section is reinforced by four steel bars of 20 mm diameter each. If the column carry a load of 80 ton, find the load carried by the steel and concrete, the contraction in the column is 0.5 mm. Find also the stress on the steel bars and on the concrete. Es = 206 GPa, Ec = 14 GPa

Ptot = 80 ton = PS + PC ΔL = 0.5 mm

ΔL =

ΔLtot = ΔLS = ΔLC

AS = 4 x (/4)(20)2 = 1257 mm2

AC = (400 x 600) – 1257 = 238743 mm2

9

40 mc

06 mc

P S

A S

P sL 602 01 x 3 x

7521

P cL

41 01 x 3347832 x

L PA E

7.5521

7P C

A C

3.477832

34

Pmax

Aо

250177

0.05 x 10025

∆Lf

Lо%

4550.0075/25

0.5 = =

PS L = 129.5 x 106

PC L = 1671.2 x 106

PS = 0.07749 PC

PC = 74.3 ton

PS = 5.7 ton

S = = = 4.5 x 10-3 ton / mm2

C = = = 3.11 x 10-4 ton / mm2

( ( A concrete cylinder 15 cm in diameter with a gauge length of 25 cm is used to obtain the stress-strain diagram in compression. If the specimen failed at a load of 250 KN and a total contraction of 0.05 cm, determine the following:(i( The ultimate strength (ii( The percentage contraction(iii( The secant modulus at a stress of 455 N/cm2 and contraction of 0.0075cm

D = 15 cm L = 25 cm Pmax = 250 KN ΔLmax = 0.05 cm

Ao = (/4) (15)2 = 177 cm2

i) Ultimate strength max = = = 1.41 KN/cm2

ii) % Contraction = = = 0.2%

iii) tan α = = = 1516.7 KN/ /cm2

10

( ( The following data were obtained in the compression test of a cast iron cylindrical specimen 35 mm diameter and 250 mm height. The load (P( and the corresponding decrease in height (h( were as follows:

P (KN( 0 27 68 102 136 203 272h(mm( 0 0.056 0.14 0.208 0.28 0.457 0.71

Draw the ordinary stress-strain diagram and determine the following:(i( The crushing stress (ii( The initial tangent modulus(iii( The secant modulus at a stress of 2100 kg/cm2

(iv( Modulus of toughness

Lо = 250 mm Aо = /4 (35(2 = 962 mm2

(N/mm2 ( 0 28 71 106 141.4 211 282.7 x 10-4 0 2.24 5.6 8.3 11.2 18.3 28.4

110

50

100

150

200

250

300

0 5 10 15 20 25 30

Pmax

Aо

272962

p.l

p.l

∆Lp.l x Aо

Pp.l x Lо

0.056 x 96227 x 250

21018.75 x 10-4

i) The crushing stress max = = = 0.283 KN/mm2

ii) Modulus of elasticity E = = = = 125.3 KN/mm2

iii) The secant modulus

= 2100 kg/cm2 = 210 N/mm2

From drawing () = 18.75 x 10-4

tan α = = = 11.2 x 104 N/ /mm2 ( ( A compression test was carried out on a cast iron cylindrical specimen of 3.4

cm diameter and 25 cm length. The loads and corresponding decrease in height were as follow:

Load (KN( 0 27 68 102 136 203 272Decrease in height (mm( 0 0.056 0.14 0.208 0.28 0.457 0.71

Draw the ordinary stress-strain diagram and determine the following:(i( The fracture stress (ii( The modulus of elasticity(iii( The secant modulus at a stress of 21 KN/cm2

Lо = 25 cm Aо = /4 (3.4(2 = 9.1 cm2

(KN/cm2 ( 0 2.97 7.47 11.2 14.94 22.3 29.89 x 10-3 0 2.24 5.6 8.3 11.2 18.3 28.4

12

0

5

10

15

20

25

30

35

0 5 10 15 20 25 30

Pmax

Aо

2729.1

p.l

p.l

∆Lp.l x Aо

Pp.l x Lо

0.056 x 9.127 x 25

2117.25 x 10-3

0

5

10

15

20

25

30

0 5 10 15 20 25 30 35 40

i) The fracture stress max = = = 29.9 KN/cm2

ii) Modulus of elasticity E = = = = 1324.6 KN/cm2

iii) The secant modulus

= 2100 kg/cm2 = 21 KN/cm2

From drawing () = 17.25 x 10-3

tan α = = = 1217.4 KN/cm2

( ( A compressive test was carried out on a brass cylinder of 10 cm2 cross-section area and gauge length of 200 mm. The load and corresponding decrease in height were as follow:

Load (Kg( 0 2700 6750 10100 12500 17500 20000 27000Decrease in height (mm( 0 0.055 0.1375 0.205 0.255 0.365 0.445 0.700

Draw the stress-strain diagram and find:(a( Ultimate compressive strength(b( Modulus of resilience (c( Elastic compressive strength(d( The modulus of elasticity(e( Modulus of toughness(f( Contraction %(g( Secant modulus at a stress of 2400 Kg/cm2

Lо = 200 mm Aо = 1000 mm2

(Kg/mm2 ( 0 2.7 6.75 10.1 12.5 17.5 20 27 x 10-4 0 2.75 6.875 10.25 12.75 18.25 22.25 35

13

240029 x 10-4

0.7 x 100200

∆Lf

Lо%

Aо x Lо

⅔ Pmax x ∆Lmax

1000 x 200

⅔ x 27000 x 0.7

Pmax

Aо

27000100

0

p.l

p.l

∆Lp.l x Aо

Pp.l x Lо

0.365 x 1000

17500 x 200

Pp.l

Aо

17500100

0

Aо x Lо

½Pp.l x ∆Lp.l

1000 x 200

½ x 17500 x 0.365

(a) Ultimate compressive strength max = = = 27 Kg/mm2

(b) Modulus of resilience (M.R.) = = = 0.016 Kg/mm2

(c) Elastic compressive strength P.l = = = 17.5 Kg/mm2

(d) Modulus of elasticity E = = = = 9.6 ton/mm2

(e) Modulus of toughness = = = 0.063 Kg/mm2

(f) % Contraction = = = 0.35%

(g) The secant modulus

= 2400 kg/cm2 = 24 kg/mm2 From drawing () = 29 x 10-4

tan α = = = 8.3 x 105 Kg/cm2

14

( ( A compression test was carried out on a concrete cylinder of 20 cm in diameter and gage length of 25 cm. The following data load and were recorded:Load, ton 0 1.8 5.4 9 12.6 14.4 18 21.5 23.4 25.2 26 27

∆L, mm x 10-2 0 0.66 2.05 4 6.6 8.3 15.4 23.4 29.7 40 47 59.5Draw the stress-strain diagram and find:a- Ultimate compressive strengthb- The initial modulus of elasticityc- The modulus of toughnessd- Secant modulus at a stress of 45.5 Kg/cm2

e- Contraction %f- Explain the fracture characteristic of test specimen

Lо = 25 cm Aо = (/4( (Do(2 = 314 cm2

(Kg/cm2 ( 0 5.7 17.2 28.7 40.1 45.9 57.3 68.5 74.5 80.3 82.8 86 x 10-3

0 2.64 8.2 16 26.4 33.2 61.6 93.6 118.8 160 188 238

15

010

20

30

40

5060

70

80

90100

0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255

Pmax

Aо

27000314

p.l

p.l

∆Lp.l x Aо

Pp.l x Lо

2.64 5.7 x 1000

45.534.5 x 10-3

59.5 x 100

250 x 100

∆Lf

Lо%

Aо x Lо

⅔ Pmax x ∆Lmax

314 x 25 x 1000

⅔ x 27000 x 59.5

(a) Ultimate compressive strength max = = = 86 kg/cm2

(b) Modulus of elasticity E = = = = 2159 kg/cm2

(c) Modulus of toughness = = = 0.136 kg/cm2

(e) % Contraction = = = 0.238 %

(d) The secant modulus = 45.5 kg/cm2 From drawing () = 34.5 x 10-3

tan α = = = 1319 Kg/cm2

( ( A compression test was carried out on a brass cylinder of 1000 mm2 cross section area and gauge length of 200 mm.

Based on the shown stress-strain diagram calculate the following:

1- The proportional limit stress2- The modulus of elasticity3- The modulus of resilience 4- The modulus of toughness5- Secant modulus at stress 210 MPa6- Ultimate compressive strength7- Contraction %8-Explain the fracture characteristic of

test specimen

16

55 - 60°

0

50

100

150

200

250

300

0 0.002 0.004

Stre

ss, M

Pa

Strain