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(9:30AM -1)
(5PM -2) Put # below.
Shear Flow, q: shear force per unit length. q is constant in an idealized, rectangular shear web.
Bucking of a uniform rectangular plate
under axial compression (per Megson)
σ cr = k
π E t
12(1 −ν 2 ) b
2
2
Shear buckling of a uniform rectangular plate with uniform
shear on all edges (per Roark):
E t
τ cr = K
(1 −ν 2 ) b
2
Simply Supported:
a/b
1.0
1.2
1.4
1.5
2.0
K
7.75 6.58 6.00 5.84 5.43
Clamped:
a/b
1.0
2
∞ .
K
12.7 9.5
7.38
Test results indicate K ≈ 4.1 for large a/b
3.0
5.02
∞ .
4.40
Crippling: Pcrippling = Pcr1+Pcr2+Pcr3+…
Sandwich Structures: Idealized analysis:
σ=
1
=
δ max
Pf
Pf
M
M
V
,σ 2 =
,τ c =
,
=
−
=
−
bt1 bht1
bt2
bht2
bh
kb Pl 3 k s Pl
[kb and ks will be provided if needed.]
+
D
S
Stress and strain transformation (planar)
γ
ε x = cos θ ε x + sin θ ε y + 2sin θ cos θ xy
2
γ
ε y = sin 2 θ ε x + cos 2 θ ε y − 2sin θ cos θ xy
2
2
γ xy
2
γ xy
=
− sin θ cos θ ε x + sin θ cos θ ε y + (cos 2 θ − sin 2 θ )
2
2
Principal Stresses and maximum shearing stress
Principal stresses follow from eigenvalues of
σ x τ xy τ zx
τ xy σ y τ yz
τ zx τ yz σ z
σ1 − σ 2 σ1 − σ 3 σ 2 − σ 3
,
,
2
2
2
τ max = MAX
Equations of Elasticity:
Equilibrium:
∂σ x ∂τ xy ∂τ xz
0
+
+
+X =
∂x
∂y
∂z
∂τ yx ∂σ y ∂τ yz
+
+
+Y =
0
∂x
∂y
∂z
∂τ zx ∂τ zy ∂σ z
+
+
+Z =
0
∂x
∂y
∂z
recall:
=
τ xy τ=
; τ xz τ=
; τ yz τ zy
yx
zx
Strain-displacement relations:
∂u
ε=
x
∂x
∂v
ε=
y
∂y
∂w
ε=
z
∂z
Generalized Hooke’s Law:
Orthotropic Plane Stress
Isotropic 3D:
1
(σ x −ν σ y −ν σ z )
E
1
ε y = (−ν σ x + σ y −ν σ z )
E
1
ε z = (−ν σ x −ν σ y + σ z )
E
1
γ yz = τ yz
G
1
γ zx = τ zx
G
1
γ xy = τ xy
G
εx =
=
εL
ν
1
σ L − LT σ T
EL
EL
ν
1
− TL σ L +
εT =
σT
ET
γ LT =
1
τ LT
GLT
ν LT
ν TL
EL
=
ET
ET
Isotropic 3D:
E
[(1 −ν )ε x +νε y +νε z ]
=
σx
∂u ∂v
(1 +ν )(1 − 2ν )
+
γ=
xy
∂y ∂x
E
[νε x + (1 −ν )ε y +νε z ]
=
σy
∂v ∂w
(1 +ν )(1 − 2ν )
+
γ=
yz
∂z ∂y
E
[νε x +νε y + (1 −ν )ε z ]
=
σz
(1 +ν )(1 − 2ν )
∂w ∂u
+
γ=
zx
τ yz = G γ yz
∂x ∂z
τ zx = G γ zx
τ xy = G γ xy
Limit Load, Ultimate Load, Ultimate Margin of Safety
Limit Load: Maximum load expected in Service
Ultimate Load: Limit Load multiplied by Factor of Safety
UMOS =
Actual load (stress) to cause failure − Design ultimate load (stress)
Design ultimate load (stress)
Load factor: n = 1+a/g or n =a/g
R. M. Rivello, Theory and Analysis of Flight Structures, McGraw-Hill, 1969.
Tresca (τmax) criterion:
σY
τ max =
SIF for central crack in infinite plate under tension:
KI = σ 0 π a
Incipient yielding
2
General SIF:
Von Mises criterion:
K I = Yσ 0 π a
σ VM = σ Y Incipient yielding
σ=
VM
1
2
(σ x − σ y ) 2 + (σ y − σ z ) 2 + (σ z − σ x ) 2 + 6(τ xy2 + τ yz2 + τ zx2 )
1
2
(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2
Formulas/charts for Y for various cases will be
provided in test booklet if needed.
or
σ VM
=
Constant amplitude fatigue:
R = Smin/ Smax
Maximum Normal Stress Criterion:
Power law curve fit of high-cycle S-N diagram:
Incipient Fracture:
S a = AN Bf
σ max = σ ult ,t
or
Goodman:
σ min = σ ult ,c
Maximum stress criterion for orthotropic material:
2
S
S
Gerber: a + m =
1
S a ,0 Sult
Failure does not occur if:
Xc < σ L < Xt
Yc < σ T < Yt
Palmgren-Miner Rule: Fatigue failure if:
− S < τ LT < S
Paris Law:
dθ
T
=
d z GJ
Tb
; J = 13 ab3 or
J
Neglect shear stress in stiffeners
Neglect axial stress in skins/web
a
∫ t (s) ds
3
Shear flow constant in skin/web segments between
stiffeners
0
Single-cell thin-walled closed section: q =
J=
T
2 Aenclosed
2
4 Aenclosed
ds
∫ t (s)
Multi-cell thin-walled closed sections:
# cells
T = ∑ 2 Ai qi ;
i =1
da
n
= C( ∆K )
dN
Idealized semimonocoque structure:
Thin walled open section: τ max =
J=
nj
∑ N =1
j
Torsion:
1
3
Sa
S
1
+ m =
S a ,0 Sult
1
q( s) d s
= constant
∫
2 Ai G i t ( s )
;
Normal stress is constant within an individual
stiffener
Symmetric bending:
M y
Iz
σx =
− z +
Shear in idealized box beam sections:
Myz
Joint Equilibrium:
or
Iy
+ (V I − V I ) y + (V I − V I ) z
−
∑(q − q ) =
I I −I
Af
out
M y
I xx
σz =
+ x +
∑T
internal
M z I y + M y I yz
Centroid and moments of inertia using
composite body approach (for idealized
cross-sections, neglect I0 terms):
i
i
i
i i
i
∑(I + A d )
=
I
∑(I + A d )
=
I
∑(I + Ad d )
=
Iz
z 0,i
i
2
yi
y
y 0,i
i
2
zi
yz
yz 0,i
i
yi
q
dθ ∫ t ds
=0
=
dx 2 AG
zi
σ xε x + σ yε y + σ z ε z + τ yz γ yz + τ zxγ zx + τ xyγ xy dV
U 0= 12 σ xε x + σ yε y + σ z ε z + τ yz γ yz + τ zxγ zx + τ xyγ xy
U 0 dV
1 F2
2 k
U N = 12 ∫
N2
dx
EA
M2
UM = ∫
dx
EI
1
1
σ x2 + σ y2 + σ z2 + 2ν (σ xσ y + σ xσ z + σ yσ z ) +
τ yz2 + τ zx2 + τ xy2
2E
2G
∂U
∂qi
Castigliano’s Theorem on Deflections: qi =
=
kq 2
1
2
1
2
For linear elastic isotropic material
Castigliano’s First Theorem: Pi =
=
U
Utotal = UN + UM + UV + UT
1
2
Volume
If Linear Elastic, U = UC.
For slender structures, strain energy can be
evaluated for each static resultant
∂U
∂U
∂U
δU =
δ q1 +
δ q2 +
δ q3 + ...
∂q1
∂q2
∂q3
∫∫∫
Strain Energy in Simple Structures
Hookean Spring:
δV = Pδ q
; U
dU U=
=
0 dV
= ∑ Tapplied
If Transverse shear loading is applied at the shear center:
Point force moving through a virtual displacement:
U0
=
f
q
dθ ∫ t ds
=
dx 2 AG
Principle of Virtual Work: δU = δV
dU=
y yz
Twisting of a cross-section:
+ ( M x I yy − M y I xy ) y + ( M y I xx − M x I xy ) x
I xx I yy − I xy2
∑Ay
∑A
∑Az
z=
∑A
z z
1 q
ds = constant for all cells
Ai ∫i t
M y I z + M z I yz
1
y=
f
Equal Twist (assuming homogeneous material):
or
σz
=
z yz
Torque Equivalence:
I yy
1
− ( M z I y + M y I yz ) y + ( M y I z + M z I yz ) z
I y I z − I yz2
tan ϕ =
y y
2
yz
y z
Myx
Unsymmetric bending
σ=
x
in
k = 1.2 rectangular
kV 2
UV = ∫
dx k = 1.33 solid circle
GA
k = 2.0 thin circular tube
1
2
∂U C
∂Pi
Dummy load method: apply fictitious load at location of
interest, later set it equal to zero.
U T = 12 ∫
T2
dx
GJ
1. (15 pts) An idealized box beam structure has a single-cell, closed cross-section, square shape. There are
four stiffeners. Note that the cross-sectional areas for the stiffeners are not all the same, as noted in the box
beside the figure. The thickness, t, of each web is the same. The location of the centroid has been computed
for you and is included in the figure.
a) (5 pts) Determine the area moments of inertia Iy, Iz and Iyz. Express them in terms of L, A, and/or t.
b) (10 pts) The location of the horizontal position of the shear center is defined by the parameter ‘d’ as
shown in the sketch below. Set up a system of equations that will permit solution of d, in terms of the given
geometry L, A, and/or t. These equations may involve only the geometric constants L, A, and/or t as well as
the vertical shear resultant V and the shear flows q1, q2, q3, and q4 (use nomenclature and sign conventions for
the shear flows given in the sketch). Simplify the equations appropriately. [Hint: you will need a system
of five equations.]
Note: you are not asked to solve the system of equation above.
A
2
A2 = A
A1 =
x
A3 = A
L/2
A4 =
L/3
2L/3
d: describes position
of shear center “X”
V
A
2
2. (10 points) The stiffener has the shape shown in the figure. Note that the location of the centroid, C, is
given, as are the moments and product of inertia (you do not need to compute these!). The beam is loaded by
a bending moment resultant about the horizontal axis, Mz, of 50 kip-in, as shown in the sketch. There is no
moment resultant about the vertical axis.
a) (4 pts) Determine the orientation of the neutral axis, ϕ, and draw the neutral axis on the sketch below.
b) (6 pts) Determine the maximum compression stress on the cross-section. Report both a numerical value
of the stress, and state clearly where this maximum stress is found.
3. (11 pts) A sandwich beam is to be designed to support a design load P of 3.6 kN (this number includes
the relevant factor of safety) at a the midpoint between two roller supports one meter apart as shown in the
sketch. The following design choices have already been made:
• The beam width, b, is fixed at 50 mm.
• The core thickness is limited to a maximum of tc = 25 mm.
• Both faces are to have the same thickness.
• The core material with be HRH-10 Honeycomb with 1/8 inch cell size (Properties as given in the
following table). Assume these values are valid for any core thickness.
• Note that the relative weight of the core is indicated by “Density” in the honeycomb designation.
Thus, the cores are listed in order from lightest to heaviest.
a) (3 pts) Sketch V-M diagrams for the beam, and determine the maximum bending resultant and the
maximum shear resultant that the beam must support.
b) (4 pts) If the faces are to be made of Aluminum 2024-T4 with a yield strength of 325 MPa, determine
the minimum face thickness necessary based on yielding of the faces. Indicate the core thickness used.
c) (4 pts) Considering the shear strength of the core only, determine the lightest core material (give its
designation) acceptable for the case considered in part b. Justify your choice.
Please start your work on the following page
4. (6 pts –2 points each)
During flight of a small airplane the structure of the wing failed causing the wings to collapse and
the plane to crash. Which of the following is a possible cause of such an accident
(select all correct answers, there is more than one).
The pilot improperly applied the controls to command an angle of attack that the structure was
not designed to support at the given flight speed.
The pilot executed a stall at low flight speed
Undetected fatigue cracks grew in the wing due to poor maintenance practices
The aircraft encountered a gust more severe than what the wing was designed to tolerate
A semimonocoque wing structure is shown in the following figure. Which parts of the structure are
most responsible for carrying the bending loading on the wing? Choose one answer.
a)Transverse Ribs and Longitudinal Stringers
b) Cover Skin and Spar Webs
c) Spar Caps and Longitudinal Stringers
d) Spar Webs and Longitudinal Stringers
Acoustic loads may be a significant factor in the loading of a spacecraft. During which stage(s)
are acoustic loads greatest?
a) During ground transportation and factory testing
b) During lift-off and transonic flight
c) During staging operations, e.g. due to pyrotechnic bolts
d) Immediately prior to orbital insertion
5. (6 points)
a) (2 pts) Explain in words, what is a Stress Concentration Factor
b) (2 pts) In SI, what are the units of shear flow q:
c) (2 pts) Here is the graph of fixity coefficient for a plate under axial compression with simply supported
boundary conditions on all four edges. Sketch the buckled shape of such a plate if it has an aspect ratio
a/b = 2.
6. (12 pts) A rectangular rosette is placed on the surface of a part that can be assumed to be in a state of plane
stress. The material is isotropic with E = 70 GPa, ν = 0.3.
Given: the gages in the rosette read εa = 1100 μstrain, εb = 700 μstrain, εc = –300 μstrain, determine the full
six component states of stress and strain. Show work and put the results in the following table.
σx =
εx =
σy =
εy =
σz =
εz =
τxy =
iE =
τyz =
γxy =
τzx =
γzx =
γyz =

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