SKRIPTA RIJEÅ ENIH ZADATAKA IZ OTPORNOSTI MATERIJALA

23. Zadatak
�eli�na kocka brida a = 10 cm postavljena je bez zazora izme�u dviju krutih stijenki na podlogu i
na gornjoj plohi optere�ena s q = 60 MPa.
Odrediti : a) naprezanja i deformacije u tri okomita smjera i deformaciju dijagonale kocke
b) promjenu volumena kocke
c) normalno i posmi�no naprezanje u presjeku pod kutem od 45° prema stjenki.
E = 2 ⋅10
ν =
σ
ε
ε
ε
xx
yy
zz
1
=
E
1
=
E
1
=
E
d = a
ε
ε
d
d
y
z
0,
30
MPa
= −q
= −60MPa
σ = 0
= ε
=
xx
[ σ −ν
( σ + σ ) ] = 0 → σ = νσ = 0,
3⋅
( − 60)
1
[ σ −ν
( σ + σ ) ] = [ − 60 − 0,
3(
−18)
]
1
[ σ −ν
( σ + σ ) ] = − 0,
3(
−18
− 60)
3
∆d
= ε ⋅ d = −5,
2 ⋅10
d
5
x
y
z
cos
2
−5,
2 ⋅10
−5
z
x
z
α + ε
yy
y
y
x
cos
−5
2
⋅10
2 ⋅10
2 ⋅10
β + ε
2
⋅
zz
5
5
x
y
[ ]
2
cos γ = 0 − 27,
3⋅10
3 = −9
⋅10
Relativna i apsolutna promjena volumena :
∆V
ε V = = ε xx
V
+ ε yy + ε
∆V
= ε
−
⋅V
= −15,
6 ⋅10
V
⎡−18
ρ
n = [ σ ij ] ⋅ n =
⎢
⎢
0
⎢⎣
0
ρ = −12,
72MPa
ρ
ρ
n
nx
ny
nz
n
n
= −42,
42MPa
= 0
=
=
n
2
nx
2
n
+ ρ
−σ
2
ny
2
n
+ ρ
zz
0
0
2
nz
5
− 60
= 21MPa
= 0 −
⋅10
=
3
27,
3⋅10
−5
= −0,
15cm
0⎤⎡cos
45°
⎤
0
⎥⎢
cos 45
⎥
⎥⎢
°
⎥
0⎥⎦
⎢⎣
cos90°
⎥⎦
12,
72
2
+
−3
42,
42
mm
=
+ 11,
7 ⋅10
2
+ 0
2
−5
= −18MPa
−27,
3⋅10
= 11,
7 ⋅10
σ = ρ ⋅ n = −12,
72 ⋅ cos 45°
− 42,
42 ⋅ cos 45°
= −39MPa
ρ
τ
ρ
ρ
−5
=
−5
−5
1
⋅ + 11,
7 ⋅10
3
−15,
6 ⋅10
= 44,
28MPa
−5
−5
1
⋅
3
28