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Corrections to Rocket Propulsion Elements, 9th Edition George P. Sutton, Oscar Biblarz ISBN: 978-1-118-75365-1 January 2017
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1
Bending of Square and Rectangular Box shaped members per AISC 8th and 9th Edition
TU8x8x3/16"
Member Size (TUhxbxt)
hw := 8 ⋅ in
Member height (in)
bf := 8 ⋅ in
Member width (in)
tf := 0.1875⋅ in
Nominal member flange thickness (in)
tw := 0.1875⋅ in
Nominal member web thickness (in)
2
Atot := 5.37⋅ in 4
Ix := 54.4⋅ in
Moment of Inertia about axis of bending (strong axis) (in^4)
4
Iy := 54.4⋅ in E := 29000 ⋅
Cross sectional area of section (in^2)
Moment of Inertia about weak axis (in^4)
kip
Modulus of Elasticity (ksi)
2
in Fy := 46
Yield strength of material (ksi)
Sy := 46⋅ ksi
Yield Strength with units in Mathcad (ksi
Mr := 275⋅ in⋅ kip
Maximum required moment for load case being considered (in-kip)
f a := 12
Computed axial stress (ksi)
f := 18
Computed compressive stress in the stiffened elements, based on the design properties as specified in Section C4 (ksi). If unstiffened elements are included in the total cross section, f for the stiffened element must bge such that the maximum compressive stress in the stiffened element does not exceed F a Qs or Fb Qs, as applicable.
M1 := 100⋅ in⋅ kip
Smaller bending moment at the ends of the unbraced length (in-kip)
M2 := 100⋅ in⋅ kip
Larger bending moment at the ends of the unbraced length (in-kip)
Lb := 50⋅ ft
Laterally unsupported length (ft)
2
Determine if HSS flange is: compact, non compact, or slender: b := bf − 3⋅ tf
For flanges of rectangular hollow structural sections (HSS), the width "b" is the clear distance between webs less the inside corner radius on each side. The inside radius shall be taken as 3 times the thickness. AISC 9th Ed. Pg 5-34 - Section B5.1d.
The lateral unsupported length of the compression flange of a box-shaped member of rectangular cross section shall not exceed the least of the following: Lc1 :=
Except that it need not be less than: Lc2 := 1200⋅
⎛ b ⎞ ⎝ Fy ⎠
Lc2 = 16.168⋅ ft
Lc := max( Lc1 , Lc2)
Lc = 42.442⋅ ft
Maximum laterally unsupported compression flange of a box shaped member
Also for compactness, a section with a depth that is not more than 6 times the width and For a section whose flange thickness is not more than 2 times the web thickness. AISCM 8th Ed. Pg 5-20 Section 1.5.1.4.1 Paragragh 6 HSSFlange :=
"Compact" if
b tf
≤ λ pf ∧ Lb ≤ Lc ∧ h w ≤ 6⋅ bf
"Non Compact" if
b t f
"Slender" otherwise HSSFlange = "Slender"
> λ pf ∧
b tf
≤ λ rf ∧ Lb > Lc
3
Determine if HSS web is: compact, non compact, or slender: For webs of rectangular hollow structural sections (HSS), the height "d" is the full nominal depth. AISC 9 Ed. Pg 5-35 - Section B5.1b
d := hw d = 8 ⋅ in
d tw
= 42.667
The depth thickness ratio of the web shall not exceed the following:
If flange is "Slender" then the allowable moment must be calculated using the calculated effective width and the resulting effective section modulus:
The AISCM says this is the compressive stress in the stiffened elements based on the section properties of the effective width calculated in Section C4.
be1 := 253 ⋅
tf f
⋅ ⎡1 −
⎢ ⎣
⎡
⎤⎤
⎛ b ⎞ ⎢ t ⋅ f ⎥⎥ ⎣ ⎝ f ⎠ ⎦⎦
be1 = 7.839⋅ in be :=
50.3
b e1 if b e1 < b b if b e1 > b
AISCM 8th Edition Pg 5-95 - Section C3 Is this value supposed to be the "actual" width or the "design" width?
be = 7.438⋅ in This is a situation the TU8x8x3/16" qualifies as a "Slender" tube and the section modulus needs to be reduced, but the "effective" width is larger than the "actual" width and thus ther is no difference in the section modulus.
4
Calculate Removed Area: brem := b − b e
b rem = 0 ⋅ in
Arem := b rem⋅ t f
Arem = 0 ⋅ in
Length of flange removed as a result of the effective width being used. Area of flange removed as a result of the effective width being used.
2
Calculate new neutral axis measured from the bottom of the section:
Calculate difference between "New" and "Old" neutral axis: d na :=
⎛ hw ⎞ −Y ⎝ 2 ⎠ eff
d na = 0 ⋅ in
Calculate the moment of inertia of the "Removed" area: 3
Irem := b rem⋅
tf
4
Irem = 0⋅ in
12
Calculate the distance from the "New" neutral axis to the centroid of the "Removed" area: d neg :=
tf ⎞ ⎛ − Yeff hw − 2 ⎠ ⎝
d neg = 3.906⋅ in
Calculate the effective moment of inertia of the "Total Area" minus the "Removed Area": Total area moment of inertia moving the axis from the old CG/N-A to the new CG/N-A (by means of parallel axis theorem): 2
4
Itotnew := Ix + Atot⋅ d na
Itotnew = 54.4⋅ in
Removed area moment of inertia moving the axis from the old CG/N-A to the new CG/N-A (by means of parallel axis theorem):
Iremnew := Irem + Arem ⋅ d neg
2
4
Iremnew = 0⋅ in
Is my rational on calculating the effective section modulus correct?
Is my rational on calculating the effective section modulus correct? (Continued from above)
5
Calculate the effective moment of inertia by subtracting the "Removed" area moment of inertia at the new CG/N-A from the "Total" area moment of inertia about the new CG/N-A: 4
Ieff := Itotnew − Iremnew
Ieff = 54.4⋅ in
Calcualte the effective seciton modulus:
Seff :=
Ieff
3
Seff = 13.6⋅ in
(hw − Yeff )
Calculate the nominal moment:
What allowable stress is supposed to be used with the effective section modulus (Slender flanges)? Is 0.60Fy correct? The AISC 8th and 9th edition are vague.
Mmax := 0.6Sy⋅ Seff
Mmax = 375.4 ⋅ in⋅ kip
⎛ Mmax ⎞ Fbs := ⎝ Seff ⎠
Fbs = 27.6⋅ ksi
Equivalent Allowable Stress based on allowable moment and section.
The allowable stress, Fb shall be the lowest value obtained according to : 1. Flange Local Buckling 2. Slenderness 3. Web Local Buckling Therefore: Fbf :=
Fbc if HSSFlange
=
Fbnc if HSSFlange Fbs if HSSFlange Fbw :=
Fbc if HSSWeb
=
=
=
"Compact"
Fbf = 27.6⋅ ksi
"Noncompact"
"Slender"
"Compact"
Fbw = 27.6⋅ ksi
Fbnc otherwise
(
)
Fb := min Fbf , Fbw
Fb = 27.6⋅ ksi
Note: Lateral Torsional Buckling does NOT apply to flexural members bent about their weak axis or HSS bent about either axis, per AISC Specification Sections F6, F7 and F8 (no mention is made of lateral torsional buckling). AISC 13th Edition, Pg 3-5.
