1-4 解答 2. τ = 3. 反力は外向き法線方向を正とする (1) (ⅰ)40[kN] (ⅱ

1-4 解答
1.
σ=
P 85 × 9.8
=
= 28[Mpa]
A
5× 6
2. τ = P = 6000 = 34[Mpa]
A
π
×152
4
3. 反力は外向き法線方向を正とする
(1) (ⅰ)40[kN]
(ⅱ) σ 1 = 40000 = 354[Mpa]
π
×12 2
4
σ 2 = 40000 = 199[Mpa]
π
×16 2
4
(2)
(a) (ⅰ) R = 10[kN ]
(ⅱ) ① σ =
P 10[kN ]
=
= 1.0 ×10 3[MPa]
A 10[mm 2 ]
②σ =
P 15[kN ]
=
= 1.5 ×10 3[MPa]
A 10[mm 2 ]
(b) (ⅰ) R = 10[kN ]
(ⅱ)① σ =
P 10[kN ]
=
= 1.0 ×10 3[MPa]
A 10[mm 2 ]
②σ =
P 15[kN ]
=
= 1.5 ×10 3[MPa]
A 10[mm 2 ]
③σ =
P 10[kN ]
=
= 1.0 ×10 3[MPa]
A 10[mm 2 ]
(c) (ⅰ) R = −20[kN ]
(ⅱ)① σ =
P −20[kN ]
=
= −2.0 ×10 3[MPa]
A 10[mm 2 ]
②σ =
P −10[kN ]
=
= −1.0 ×10 3[MPa]
A 10[mm 2 ]
(d) (ⅰ) R = 10[kN ]
(ⅱ)① σ =
P 10[kN ]
=
= 1.0 ×10 3[MPa]
A 10[mm 2 ]
②σ =
P
0[kN ]
=
= 0[MPa]
A 10[mm 2 ]
③σ =
P 20[kN ]
=
= 2.0 ×10 3[MPa]
A 10[mm 2 ]
4. 変形後の棒の長さを l’とする
l '− l
l
l'
200[mm]
l=
=
= 199[mm]
ε +1 5 ×10 −3 +1
ε=

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