哈工大2009年材料力学期末考试题及答案

d GPa a= mm qmax
BC
σpl = . MPa [σ ] =
MPa
MPa
b=
MPa
b h q B d C
A 60◦
D
l
-
l
(
)
A
.C
.D
.A
.B.Bຫໍສະໝຸດ .D.C.B
.A
F
P
A
C
F
O B F
F P ⋅ ( L) − EI AB C F ⋅ ( L) , EI
D
A
δ
C vAB
C
C vCD
O B
2F L
B
2P L P 2P L
A
F
FL D
T
2F L
A
2F L
bending moment by P
bending moment and torque by F
A A M = ( FL) + ( PL) = T = FL = kN ⋅ m √ σr = √ →d≥ M +T π [σ ] dmin = . cm M +T ≤ [σ ] πd √ kN ⋅ m
A
D
( )P P
θ2
R
B
R θ1 C
X1
MA = NC ⋅ R − PR = C NC = P NC = P
NC
P
⎧ P ⎪ ⎪ ( − cos θ )R , ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪P M (θ ) = ⎨ [( − cos θ )R + R ], ⎪ ⎪ ⎪ ⎪ P ⎪ ⎪ ⎪ ( + cos θ )R , ⎪ ⎪ ⎩ CB BD DA , , , π ≤θ ≤π ≤θ ≤ π
√ RB [nst ] = ( ) RB λ= KL ×l l = = i d d N RB = ( + )P [σ ]
σcr =
λ > λp ( )
π E π Ed = λ l NRB σcr ≤ πd [nst ] d≥ . cm l = λ = d RB
d≥
NRB [nst ] l πE λp = π E = σpl .
A
: . A. B. C. D. (
( )
,
,
)
. A. B. C. D.
(
)
. τmax A. ≤ . [σ ] . A. B. C. D. τmax B. ≤√ [σ ] τmax C. ≤ . [σ ] (
( τmax D. ≤ [σ ] )
)
(
)
:
:
:
. A.
( B.
) C. D.
. A. B. C. D. σ σ E
bending moment and axial force by F
A A √ M= + T= N⋅m N=F= N = N⋅m
N M + = . + . = . MPa σ = σM + σN = πd πd T τ= = . MPa πd √ σr = σ + τ = MPa < [σ ]
( )
(
)
. A. B. C. D.
(
)
. A. B. C. D.
(
)
. A. B. C. D.
(
)
(
)
:
:
:
A (δ ≪ L) CD C EI = kL D AB
AB B
CD k CD O C
EI δ
P
P
AB
k
δ
P
A
4L L
O B
-
δ
C
3L
D
(
)
:
:
:
A L= . m E= GPa [ ∆] = .
O vCD
B vAB
D
CD C O
P
⋅ L ⋅ FL ⋅ L EI ,
O = vCD
⋅ L ⋅ L ⋅ PL CD F ⋅ L ⋅ L ⋅ FL EI C − B
C = vCD
F ⋅ ( L) EI
CD
B v AB =
C = v AB
EI F k
C C vCD − v AB = δ,
O B vCD − v AB =
(
)
. A. B. C. D.
(
)
. A. B. C. D.
(
)
. A. B. C. D.
(
)
. A. B. C. D.
(
)
(
)
:
:
:
-
OC AB OB O x C
O
EI P
B
x
A
x
O
P 4L
-
α
B
L
C
(
)
:
:
:
A D
d F = kN [σ ] = P = kN MPa
ABCD
xz L= . m
EI
v B = PL (
() MA = NB = (
√ + )P
P
√
+P
C
−P
P R
+(1 −
3 )P 2
D
1 − √ P 2 3
A
1 − √ P 2 3
1 + √ P 2 3
1 −(1 + √ )P 2 3
√ 1 3 )P +( + 2 4
+(
1 1 + √ )P 2 4 3
B
NB √ 1 3 )P =( + 4 2
CB BA
, ,
≤θ ≤π ≤θ ≤π
δ =
∫ ∫
π
∫
∫
∫
δ X +∆F = C
→X = NC = X =
P π P P π
()
BD
ABC √ MA = : NBD × L − ql × L = → NBD = √ ql
ABC
q
A
NBD
B
C
A
M
B
C
ql2 2
( ) B Mmax = ql √ B σmax ≤ [σ ] ( ) BD × l l KL = = = i d d E λp = π = σpl BD πE σcr = = . MPa < [σ ] λ σcr NBD ≤ πd [nst ] λ= q≤ σmax = . ql + ql bh bh kN m
A − BCD D b= . m G= mm . P= kN P F =
d= F kN
mm
GPa
[σ ] =
MPa
D
P
A
4L
C
-
D
F
L+b
B
L
(
)
:
:
:
A, C EI C D
AB
BC P
B
A
D
R
B
R C
P
-
(
)
:
:
:
-
ABC q
BD E= σs = l= m b= [nst ] = MPa mm h= mm d=
PL − EI PL − EI EI kL = P= EI δ= L kδ
FL − EI FL − EI F=
FL =δ EI FL F = EI k P
() P
F
T
A
+
300 N
·m
P
-
N
A
2000
N
+
D C
+
D
600
F
+
800 N · m
B
N·m
+
300 N · m
B
C
600 N
·m
bending moment and torque by P
P D
C
P
A
-
B
(
)
B
.D
.B
.C
.C
.B
.D
.B
.C
.D
L<x≤ L
≤x≤ L
M x
P x(1 −
x ) 4L
M
mx
mC C
P
x
mx
x x(1 − ) 4L
M = Px ( − vx =
x ) L
2x/3 (4L + 2x)/3
x ⋅ x( − ) EI L x M Lx ( − ) → vx = EI L ⋅ L⋅M
≤θ ≤π
( ) C ⎧ ⎪ ⎪R sin θ , M (θ ) = ⎨ ⎪ ⎪ ⎩−R sin θ , ( )
π R sin θ R sin θ πR R dθ + R dθ = EI EI EI PR PR ( − cos θ )R sin θ ( − cos θ )(−R sin θ ) π π ∆F= R dθ + R dθ EI EI PR ( + cos θ )(−R sin θ ) π + R dθ EI π PR PR = ( − − )=− EI EI
t
A
0.3 N
·m
1N
-
D C
+
D
D
+
bending moment and torque by unit load
× . × +
−
0.8 N · m
× . ×
B
0.3 N · m
× . + × . × . JG