Ain Shams University

Faculty of Engineering

New Program

4th assignment

Presented to: Dr. Nahed Abd El-Salam

Presented by: Ahmed Hassan Ibrahim

Mostafa sherif Ibrahim

S.MANF

Problem (1)

Given: Spherical, do=6m, thickness= 9mm= 0.009m, pressure= 500 KPa

Required: σhoop

σmax= 𝜏max = 𝑝𝑟𝑖

2𝑡=

(500𝐾𝑃𝑎)(6

2𝑚−(0.009𝑚))

2(0.009)= 83.083 𝑀𝑃𝑎

σmax=τmax=82.083 MPa Note: the hoop and axial stress are the same due to symmetry.

Problem (2)

Given : P=10 MPa, spherical, do= 200-mm, thickness= 6-mm, σu = 400 MPa

Required: factor of safety

σhoop= 𝑝𝑟𝑖

2𝑡=

(10𝑀𝑃𝑎)(0.200

2𝑚−(0.006𝑚))

2(0.006𝑚)= 78.33 𝑀𝑃𝑎

factor of safety (n)= σu

σℎ𝑜𝑜𝑝=

400 𝑀𝑃𝑎

78.33 𝑀𝑃𝑎= 5.106 n=5.106

Problem (3)

Given: do=240-mm, thickness= 3mm, pressure= 63 KPa

Required: normal stress

σhoop= σa = 𝑝𝑟𝑖

2𝑡=

(63𝐾𝑃𝑎)(0.240

2𝑚−0.003𝑚)

2(0.003𝑚)= 1.2285 𝑀𝑃𝑎 σa = 1.23 MPa

Note: An internal pressure p induces equal biaxial tangential tensile stresses in the walls.

Problem (4)

Given: Spherical, do= of 3m, thickness= 12mm, σall – 80MPa, E = 200GPa, ν = 0.29

Required : pressure, increase in the diameter

- σhoop= 𝑝𝑟𝑖

2𝑡 → p =

σℎ𝑜𝑜𝑝 ×2𝑡

𝑟𝑖=

(80 𝑀𝑃𝑎)(2)(0.012𝑚)3𝑚

2−(0.012𝑚)

= 1.29 𝑀𝑃𝑎

p=1.29 MPa

- 𝛿𝑑 = 2𝛿𝑟 =𝑝𝑟2

𝑡𝐸(1 −

𝑣

2)

= 2 ((1.29 𝑀𝑃𝑎)(1.5𝑚−0.012𝑚)2

(0.012)(200 ×103𝑀𝑃𝑎)(1 −

0.29

2)) = 2.035 × 10−3

δd=2.035 mm

Note: that the radial expansion is reduced by the Poisson term.

Problem (5)

Given: Spherical, do= 4.5m, thickness= 22mm, E= 200GPa, ν = 0.29, 1.7MPa.

Required : normal stress, increase in the diameter.

- σhoop= 𝑝𝑟𝑖

2𝑡=

(1.7 𝑀𝑃𝑎)(4.5𝑚

2−0.022𝑚)

2(0.022𝑚)= 86.0818 𝑀𝑃𝑎

- 𝛿𝑑 = 2𝛿𝑟 =𝑝𝑟2

𝑡𝐸(1 −

𝑣

2)

= 2 ((1.7 𝑀𝑃𝑎)(

4.5

2𝑚−0.022𝑚)

2

(0.022𝑚)(200 ×103𝑀𝑃𝑎)(1 −

0.29

2)) = 3.2796 × 10 3𝑚

δd=3.2796 mm

Problem (6)

Given: spherical, do= 750mm, σu = 400 MPa, factor of safety= 4, p= 4.2MPa

required: smallest thickness

Factor of safety =σ𝑢

σℎ𝑜𝑜𝑝→ σhoop =

σu

factor of saftey=

400 MPa

4= 100 MPa

σhoop= 𝑝𝑟

2𝑡 → 𝑡 =

𝑝𝑟𝑖

2σℎ𝑜𝑜𝑝=

(4.2 𝑀𝑃𝑎)(0.75

2𝑚−𝑡)

2(100 𝑀𝑃𝑎)= 7.875 × 10−3 − 1.65375 × 10−4𝑡

t = 7.713 mm

Problem (7)

Given: height= 14.6-m, thickness of 16-mm, Density= 1000-kg/m3, do=7.5m

Required: normal stress, shearing stress

p = 𝜌ℎ𝑔 = (1000 𝐾𝑔/𝑚3)(14.6 𝑚)(9.81 𝑚/𝑠2) = 143226 𝐾𝑔/𝑚𝑠2

= 143226 𝑁/𝑚2 = 143.226 𝐾𝑃𝑎

:

σmax= 𝑝𝑟

𝑡=

(143.226 𝐾𝑃𝑎)(𝟕.𝟓

𝟐𝒎−𝟎.𝟎𝟏𝟔𝒎)

(𝟎.𝟎𝟏𝟔𝒎)= 33.42536 𝑴𝑷𝒂

𝜏max= 𝑝𝑟

2𝑡=

(143.226 𝐾𝑃𝑎)(𝟕.𝟓

𝟐𝒎−𝟎.𝟎𝟏𝟔𝒎)

𝟐(𝟎.𝟎𝟏𝟔𝒎)= 16.71268 𝑴𝑷𝒂

Problem (8)

Given: r = 0.5 m, thickness t = 15 mm, P = 24 MPa, ν = 0.30, E = 200 GPa, cylindrical

Required : the circumferential (hoop) and longitudinal stresses. the circumferential,

longitudinal and through thickness strains

- Axial stress σa= 𝑝𝑟

2𝑡=

(24 𝑀𝑃𝑎)(0.5 𝑚)

2(0.015 𝑚)= 400 𝑀𝑃𝑎

- Circumferential σhoop=𝑝𝑟

𝑡=

(24 𝑀𝑃𝑎)(0.5 𝑚)

0.015𝑚= 800 𝑀𝑃𝑎

Note: the hoop stresses are twice the axial stresses. This result — different stresses in different directions — occurs more often than not in engineering structures, and shows one of the compelling advantages for engineered materials that can be made stronger in one direction than another (the property of anisotropy).

- ehoop direction=𝑝𝑟

𝑡𝐸(1 −

𝑣

2) =

(24 𝑀𝑃𝑎)(0.5𝑚)

(0.015 𝑚)(200×103)(1 −

0.3

2) = 0.0034

- eaxial direction=𝑝𝑟

𝑡𝐸(

1

2− 𝑣) =

(24 𝑀𝑃𝑎)(0.5𝑚)

(0.015 𝑚)(200×103)(

1

2− 0.3) = 0.00008

- eradial direction= −3𝑣𝑃𝑟

2𝑡𝐸= −

3(0.3)(24 𝑀𝑃𝑎)(0.5 𝑚)

2(0.015 𝑚)(200×103𝑀𝑃𝑎)= −0.018

Problem (9)

Given: cylindrical, do=1.75m, t=1.6mm, ultimate normal stress=450MPa, n=5 .

Required: pressure

Factor of safety =σ𝑢

σ𝑎→ σa =

σu

factor of saftey=

450 MPa

5= 90 MPa

Axial stress σa =𝑝𝑟

2𝑡→ 𝑝 =

2tσ𝑎

𝑟𝑖=

2(0.0016 𝑚)(90 𝑀𝑃𝑎)1.75

2𝑚−0.0016𝑚

= 0.329745 𝑀𝑃𝑎

p = 329.74 KPa

Problem (10)

Given : cylindrical, pressure of 1.5 MPa, do =320-mm, thickness = 3-mm

Required: maximum normal stress, maximum shearing stress

σmax = 𝑝𝑟𝑖

2𝑡=

(𝟏.𝟓 𝑴𝑷𝒂)(𝟎.𝟑𝟐

𝟐𝒎−𝟎.𝟎𝟎𝟑𝒎)

(𝟎.𝟎𝟎𝟑𝒎)= 𝟕𝟖. 𝟓 𝑴𝑷𝒂

𝜏max = 𝑝𝑟

2𝑡=

(𝟏.𝟓 𝑴𝑷𝒂)(𝟎.𝟑𝟐

𝟐𝒎−𝟎.𝟎𝟎𝟑𝒎)

𝟐(𝟎.𝟎𝟎𝟑𝒎)= 𝟑𝟗. 𝟐𝟓 𝑴𝑷𝒂

Note: One can now take as a characteristic radius the dimension r. This could be the inner radius, the outer radius, or the average of the two – results for all three should be close.

Problem (11)

Given: D= 200 mm, dimensions 400x400x20 mm, σx= 42 MPa, σy= 14 MPa,

E = 100 GPa, ν = 0.34

2D

- ac: 𝑒𝑥 =σ𝑥

𝐸−

𝑣σ𝑦

𝐸=

42 𝑀𝑃𝑎

100×103 𝑀𝑃𝑎−

0.34(14 𝑀𝑃𝑎)

100×103 𝑀𝑃𝑎= 3.724 × 10−4

= 0.0003724

𝑒𝑥 =ac

𝑎𝑐→ ac = (𝑒𝑥)(𝑎𝑐) = (0.0003724)(200 𝑚𝑚) = 0.07448𝑚𝑚

- bd : 𝑒𝑦 =σ𝑦

𝐸−

𝑣σ𝑥

𝐸=

(14 𝑀𝑃𝑎)

100×103 𝑀𝑃𝑎−

0.34(42 𝑀𝑃𝑎)

100×103 𝑀𝑃𝑎= −2.8 × 10−6

= −0.0000028

𝑒𝑦 =bd

𝑏𝑑→ bd = (𝑒𝑦)(𝑏𝑑) = (−2.8 × 10−6)(200𝑚𝑚) = −0.00056 𝑚𝑚

- t : 𝑒𝑧 = −𝑣σ𝑥

𝐸−

𝑣σ𝑦

𝐸= −

0.34(42 𝑀𝑃𝑎)

100×103 𝑀𝑃𝑎−

0.34(14 𝑀𝑃𝑎)

100×103 𝑀𝑃𝑎= −0.0001904

Note: that σx and σy are principal stresses and remember that the third principal stress σz= 0. However, a state of plane stress is not a state of plane strain. The sheet will experience a strain in the z direction equal to the Poisson strain contributed by the x and y stresses.

𝑒𝑧 =𝑡

𝑡→ t = (ez)(t) = (−0.0001904)(200 mm) = −0.003808 mm

- 𝑣 = 𝑣𝑜(𝑒𝑥 + 𝑒𝑦 + 𝑒𝑧)

= (400 × 400 × 20)(0.0003724 − 0.0001904 − 0.0000028) = (32 × 105 𝑚𝑚3)(1.792 × 10−4) = 573.44 𝑚𝑚3