Multi-dimensional stress and strain relations (Strength of Materials - I ...

Problem :
Multi-dimensional stress and strain relations
(Strength of Materials - I, Final Exam-41-4)
1. A differential element on the side of a large steel pressure vessel
A 20-mm square was scribed on the side of a large steel pressure vessel. After pressurization, the
biaxial stress condition of the square is as shown. Using the data available E = 200 GPa and υ =0.29,
for structural steel, determine the change in length of (a) side AB, (b) side BC, (c) diagonal AC, (d)
the percent change in the slope of diagonal DB, (e) if the plate thickness is 5 mm the new volume of
this plate portion and the material dilatation.
Solution :
The normal strains (σzz =0, plane stress)
εxx = 1
E (σxx − υ (σyy + σzz))
εxx =
a)thechangeinlengthofsideAB
1
εxx =
(160 − 0.29 (80 + 0))
200 × 103 6. 84 × 10 −4
εyy = 1
E (σyy − υ (σxx + σzz))
εyy =
1
(80 − 0.29 (160 + 0))
200 × 103 εyy = 1. 68 × 10 −4
εzz = 1
E (σzz − υ (σxx + σyy))
εzz =
1
(0 − 0.29 (80 + 160))
200 × 103 εzz = −3. 48 × 10 −4
∆εAB = εxx × AB =6. 84 × 10 −4 × 20
∆εAB
b) the change in length of side BC
= 0.013 68 mm
∆εBC = εyy × BC =1. 68 × 10 −4 × 20
∆εBC = 0.003 36 mm
c) the change in length of diagonal AC
q
∆εAC = (20 + 0.013 68) 2 +(20+0.003 36) 2 − 20 √ 2
∆εAC = 0.012 05 mm
Dr. M. Kemal Apalak 1

d) the percent change in the slope of diagonal DB
tan θDB = 20
20 , θDB =45 ◦
tan θ 0
DB =
20 + 0.003 36
20 + 0.013 68
∆θDB =
44.985 − 45
× 100
45
∆θDB
e) the dilatation of material
= −0.033%
, θ0
DB =44.985 ◦
e = εxx + εyy + εzz =6. 84 × 10 −4 +1. 68 × 10 −4 − 3. 48 × 10 −4
e = 5. 04 × 10 −4
V = 20× 20 × 5 = 2000 mm 3
∆V = eV =5. 04 × 10 −4 × 2000
∆V = 1. 008 mm 3
V 0 = 2000 + 1. 008 = 2001.008 mm 3
Dr. M. Kemal Apalak 2