課題2の解答

H26 年度基礎力学及び演習担当岩城・子田提出日：2014 年 5 月 29 日
学生番号氏名演習問題：モーメント，数力の合力，部材力１． (1)
O 点回りのモーメント力を求めよ．
P=5kN
M = P ⋅ l = 5kN × 3m = 15kN ⋅ m
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O
(2)
M = P1 ⋅ l1 + P2 ⋅ l2 − P3 ⋅ l3
= 4kN × 6m + 5kN × 3m − 8kN × 9m
= −33kN ⋅ m
P2=5kN
P1=4kN
O
€
P3=8kN
(3)
P1=2kN
P3=2kN
l1=6m
M = −P1 ⋅ l1 − P2 ⋅ l2 + P3 ⋅ l3
= −2kN × 6m − 3kN × 4m + 2kN × 9m
= −6kN ⋅ m
l3=9m
O
l2=4m
P2=3kN
€
H26 年度基礎力学及び演習担当岩城・子田提出日：2014 年 5 月 29 日
２． (1)
合力 R とその作用位置を求め，図示せよ．
R
P1=10kN
R
R = −P1 + P2 = −10kN + 3kN = −7kN (下向き)
l=3m
€O'
P2=3kN
O
M o = −P2 ⋅ l = 7kN ⋅ x
−P ⋅ l −3kN ⋅ 3m
x= 2 =
= −1.3m
7kN
7kN
€
(2)
R
P1=10kNR
l1=7m
R = −P1 + P2 = −10kN + 2kN = −8kN (下向き)
l2=8m
M o = P1 ⋅ l1 − P2 ⋅ ( l1 + l2 ) = 8kN ⋅ x
€
O
O''
P2=2kN
O'
x=
P1 ⋅ l1 − P2 ⋅ ( l1 + l2 ) 10 ⋅ 7 − 2 ⋅ ( 7 + 8)
=
= 5m
8kN
8
€
(3)
R
P1=10kN
P2=8kN R
l1=6m
O
R = −P1 − P2 + P3 = −9kN (下向き)
l2=6m
M o = P2 ⋅ l1 − P3 ⋅ ( l1 + l2 ) = 9kN ⋅ x
€ O'
P3=9kN
€
x=
P2 ⋅ l1 − P3 ⋅ ( l1 + l2 ) 8 ⋅ 6 − 9 ⋅ (6 + 6)
=
= −6.7m
9kN
9
H26 年度基礎力学及び演習担当岩城・子田提出日：2014 年 5 月 29 日
1 点に作用していない数力の合力 R とその作用位置を求め，図示せよ．
３． (1)
y
P1=50kN
P2=30kN
(2,2)
(-1,1)
(x0,y0)=(2,1)
!H
"V
H•y
P1
0kN
-50kN
0#2m=0
P2
-30kN
0kN
$
-30kN
-50kN
-30kN#1m
=-30kN•m
-30kN•m
V•x
50kN#2m
=100kN•m
0#1m=0
100kN•m
x
R
R = (ΣH ) 2 + (ΣV) 2 = (−30) 2 + (−50) 2 = 58.3kN
% ΣV (
−1 % −50 (

θ = tan−1'
* = tan '
* = 59
& ΣH )
& −30 )
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− ΣH ⋅ y 0 = Σ( H ⋅ y )
ΣV ⋅ x0 = Σ(V ⋅ x)
−30 × y 0 = −30
y 0 = 1m
50 × x0 = 100
x0 = 2m
€
€
(2)
R
y
P2=30kN
P1=50kN
(x0,y0)=(1.1, 4.8)
(2,2)
(-1,1)
!H
x
"V
P1
P 1 cos45°=35.4kN P 1 sin45°=35.4kN
P2
-P2 cos30°=-26kN
P 2 sin30°=15kN
$
9.4kN
50.4kN
H•y
35.4kN#2m
=70.8kN•m
-26kN#1m
=-26kN•m
44.8kN•m
R = (ΣH) 2 + (ΣV ) 2 = (9.4) 2 + (50.4) 2 = 51.3kN
% ΣV (
−1% 50.4 (

θ = tan−1'
* = tan '
* = 79.4
& ΣH )
& 9.4 )
ΣH ⋅ y0 = Σ( H ⋅ y) €
− ΣV ⋅ x 0 = Σ(V ⋅ x )
−50.4 × x 0 = −55.8
x 0 = 1.1m
9.4 × y0 = 44.8
y0 = 4.8m
€
€
V•x
-35.4kN#2m
=-70.8kN•m
15kN#1m
=15kN•m
55.8kN•m
H26 年度基礎力学及び演習担当岩城・子田提出日：2014 年 5 月 29 日
４． (1)
部材力 T1，T2 を求めよ．
鉛直方向と水平方向の釣り合いより
T1
A
ΣV = −T2 sin 30° − P = 0
C
T2 = −
T2
P
10kN
=−
= −20kN
sin 30°
sin 30°
ΣH = −T1 − T2 cos 30° = 0
T1 = −T2 cos 30° = −(−20kN) × cos 30°
= 17.3kN
P=10kN
B
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モーメントの釣り合いより
ΣM A = P ×  AC + T2 ⋅  AC sin 30° = 0
T2 = −
ΣM B = P ⋅  BC cos 30° − T1 ⋅  BC sin 30° = 0
P
10kN
=−
= −20kN
sin 30°
sin 30°
T1 =
P ⋅ cos 30° 10kN ⋅ cos 30°
=
= 17.3kN
sin 30°
sin 30°
(2)
A€
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鉛直方向と水平方向の釣り合いより
ΣV = T1 sin 60° − P = 0
T1 =
T1
P
10kN
=
= 11.5kN
sin 60° sin 60°
ΣH = −T1 cos60° − T2 = 0
T2 = −T1 cos60° = −(11.5kN) × cos60°
= −5.8kN
B
C€
T2
モーメントの釣り合いより
ΣM A = P ×  AC cos60° + T2 ⋅  AC sin 60° = 0
T2 = −
P=10kN
P ⋅ cos60°
10kN ⋅ cos60°
=−
= −5.8kN
sin 60°
sin 60°
ΣM B = P ×  BC − T1 ⋅  BC sin 60° = 0
€
T1 =
P
10kN
=
= 11.5kN
sin 60° sin 60°
(3)
T1 €
C
モーメントの釣り合いより
A
ΣM A = P ×  AC cos15° + T2 ⋅  AC sin 30° = 0
T2 = −
T2
P=10kN
€
B
€
P ⋅ cos15°
10kN ⋅ cos15°
=−
= −19.3kN
sin 30°
sin 30°
ΣM B = P ⋅  BC sin 45° − T1 ⋅  BC sin 30° = 0
T1 =
P ⋅ sin 45° 10kN ⋅ sin 45°
=
= 14.1kN
sin 30°
sin 30°

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