1. Elastic-equivalency of a Mansard-roof from a given Quonset (curved) roof
By Julio C. Banks, PE. CGC
05/03/2014
ABSTRACT
In general, the overall elastic properties of an arch cannot be simulated by a rectangular portal
frame made from axially-rigid beams. However, if the legs are at an equal and opposite slope, as
shown in Figure 1.21, then an elastically equivalent arch can be found. These legs are both of
length where and of bending stiffness, . The lintel (horizontal member) is of
length and bending stiffness . [1].
This study is intended to generate the properties of a Mansard frame elastically equivalent to a
given Quonset (curved) frame. The resulting Mansard frame consists of three (3) elastically
equivalent beams with essentially the same mean height.
While the ASCE 7-10 requires the curved roof to be ¼ for the windward and leeward (leg) portions
of the curved roof and ½ for the top (Mansard lintel), the work indicates that the resulting Mansard
roof will have approximately the values required by the ASC 7-10 for wind loads.
It is recommended that for the ASCE 7-10 snow load case, the legs be divided into two sub-
elements and the snow load be linearly reduced to zero on the member joining the vertical column
of frames.
Future MathCAD and FEA work
The Mansard roof will be placed on columns to create an elastically equivalent Mansard (effective)
frame from a given MQS (Modified Quonset Structure). Once the effective geometry generation
procedure is established, then the ASCE 7-10 loads can will be generated, applied and solved
using FEM (Finite Element Method). Initially, the effective frame will be exported to VA
(VisualAnalysi) and CAEFEM FEA (Finite Element Analysis) software. Once the solution is
validated in VA and CAEFEM environment, then a MathCAD-FEA solution will be utilized to obtain
the utilization structural check. Ultimately, the MathCAD-FEA solution will allow the automatic
generation of FEA solutions for quick-bid designs to generate sales quotes for MQS (Modified
Quonset Structure).
j L j h
2
k
2
 p EI
1 2 k( ) L q EI

2. MathCAD - Quonset-Mansard
Equivalent Frame, R-01.xmcd
Elastic-equivalency of a Mansard-roof from a given Quonset (curved) roof
by Julio C. Banks, PE. CGC
In general, the overall elastic properties of an arch cannot be simulated by a rectangular portal
frame made from axially-rigid beams. However, if the legs are at an equal and opposite slope,
as shown in Figure 1.21, then an elastically equivalent arch can be found. These legs are both
of length j L where j h
2
k
2
= and of bending stiffness, p EI . The lintel (horizontal member)
is of length 1 2 k( ) L and bending stiffness q EI
Reference
"Elastic Beams and Frames", 2nd Ed., John D. Renton, ISBN 1-898563-1, Pp. 11.26 - 11.27.1.
"Engineering Formulas", 8th Ed. Gieck. ISBN 0-07-145774-7. Page B12.
Example: Transform a Quonset (curved) frame. into an elastically equivalent Mansard frame.
Span: L 28 ft E 29 10
3
 ksi
Roof height: Δh 5.799 ft Imax 91.82 in
4
 A 0.7494 in
2

Solution
Calculate the radius of the curved roof:
Since L 2 w= then, w
L
2
14.00 ft xmid
w
ft
14.00
The aerodynamic factor is φ
Δh
w
0.4142
R
1
2
φ
1
φ






 w 19.80 ft
Julio C. Banks, P.E. page 1 of 6

3. MathCAD - Quonset-Mansard
Equivalent Frame, R-01.xmcd
The total angle is Φ 4 atan φ( ) 90.00 °
The half-angle is α
1
2
Φ 45.00 °
Calculate the area, A, and mean height, hM, of the segment
Λ
w
R
0.7071
AQ
1
2
Φ sin Φ( )( ) R
2
 111.9 ft
2

hM
A
L
0.00 in
Curve "Generate the nodes of a Curved Frame"
w
L
2

Λ
w
R

YG 1 Λ
2
  R
Δx 6 in
n
w
Δx
1
x
1
0 ft
Y
1
R
x
i
x
i 1( )
Δx
Y
i
1
x
i
R






2









R
i 2 nfor
Px1 x w
Py1 Y YG
Px2 x
Py2 reverse Py1 
Quonset
1 
stack Px2 Px1 
Quonset
2 
stack Py2 Py1 
Quonsetreturn
 x Curve
1 
 y Curve
2 

Julio C. Banks, P.E. page 2 of 6

4. MathCAD - Quonset-Mansard
Equivalent Frame, R-01.xmcd
0 10 20
2
4
y
ft
xmid
x
ft
Now define the elastically equivalent mansard frame geometry
C α( ) s sin α( )
c cos α( )
fMM
α
s

fPP
1 2 c
2
  α 3 s c
8 s
3


fMP
s α c
2 s
2


fSS
2 s
2.
 1  α s c
8 s
3


a
1
fMM
a
2
3 fPP
a
3
fMP
a
4
3fSS
areturn

α 45.00 °
f C α( )
1.1107
0.0751
0.1517
1.1358













Julio C. Banks, P.E. page 3 of 6

5. MathCAD - Quonset-Mansard
Equivalent Frame, R-01.xmcd
Given the set of equations 1 through 6
2 J K f
1
= 1( )
2 J 3 K( ) h
2
 f
2
= 2( )
J K( ) h f
3
= 3( )
2 k
2
 3 k 3  J k
2
k 1  K f
4
= 4( )
J
j
p
= 5( )
K
1 2 k
q
= 6( )
h 0.5 k 0.25 p 0.9 q 0.9 j h k( ) h
2
k
2

Let J h k p( )
j h k( )
p

and K k q( )
1 2 k
q

Given 2 J h k p( ) K k q( ) f
1
= h 0 1( )
2 J h k p( ) 3 K k q( )( ) h
2
 f
2
= k 0 2( )
J h k p( ) K k q( )( ) h f
3
= p 0 3( )
2 k
2
 3 k 3  J h k p( ) k
2
k 1  K k q( ) f
4
= q 0 4( )
h
k
p
q












Find h k p q( )
0.1892
0.2481
1.0103
1.0219













Julio C. Banks, P.E. page 4 of 6

6. MathCAD - Quonset-Mansard
Equivalent Frame, R-01.xmcd
Post-process Let Tsi 10
6
ksi
α 45.00 ° J h k p( ) 0.3089 K k q( ) 0.4929
EI E Imax 2.66 Tsi in
4
 EIλ p EI 2.69 Tsi in
4
 EIΛ q EI 2.72 Tsi in
4

Draw the resulting Mansard frame with curved frame. Let ORIGIN 1
L 28.0 ft Δx'
1
k L 6.95 ft Δx'
2
1 2 k( ) L 14.11 ft ΔhM h L 5.30 ft
x' 0 k 1 k( ) 1[ ]
T
L y' 0 ΔhM ΔhM 0 T

0 10 20
2
4
y
ft
y'
ft
xmid
x
ft
x'
ft

x'
0.00
6.95
21.05
28.00










ft y'
0.00
5.30
5.30
0.00










ft Imax 91.82 in
4

i 1 3
Δx'
i
x'
i 1( )
x'
i
 Δy'
i
y'
i 1( )
y'
i

Mansard-frame elements (MFE) Normalized MFE
λ Δx'
2
Δy'
2

 8.74
14.11
8.74








ft BM λ
 31.58 ft λN
λ
BM
27.7
44.7
27.7








%
Julio C. Banks, P.E. page 5 of 6

7. MathCAD - Quonset-Mansard
Equivalent Frame, R-01.xmcd
Shift the frame such that its mid-span position is located at the origin of the coordinate system
Quonset: xS x w Mansard: x's x' w
10 0 10
2
4
Quonset (curved) Roof
Mansard Roof
Curved Roof Made Ellastically Equaivalent to a Mansard Roof
AQ 111.9 ft
2
 α 45.00 ° AM
1
2
L Δx'
2
  ΔhM 111.6 ft
2

Compare the relative error of Quonset-Mansard frame solution.
Quonset Mansard
Area AQ 111.9 ft
2
 AM 111.6 ft
2
 %Δ AQ AM  0.3 %
Mean Height hM
AQ
L
4.00 ft hMM
AM
L
3.98 ft %Δ hM hMM  0.3 %
A 0.7494 in
2

x's
14.00
7.05
7.05
14.00












ft y'
0.00
5.30
5.30
0.00












ft
Imax 91.82 in
4

Define a relative error function: %Δ a b( )
b a
a

Julio C. Banks, P.E. page 6 of 6

8. Page 11.26 Matrix Analysis of Frames [Ch. 11]
From §7.6, it then follows that the end response ofthe cantilever to a moment M and a shear
force S is characterised completely by lIlEk and the coefficients/MM./MS andI lIS' Matching
coefficients are now sought using the compound cantilever shown in Figure 11.18b. This is also
of length I, but has a constant flexural stifthess Ell over a length al and a constant fleXUral
stiflhess Elzover the remaining length. The complementary strain energy induced by the same free
end loading as before is given by
al I
[j = J(M - SX)2 dx + J(M - Sx)2 dx
2EII 2EI2o al
(11.92)
=M21[~ + l-UJ-MSI2[~ + l-U2J+S2/3[~ + l-U3J
2EI) 2El2 2El1 2El2 6EI) 6EI2
By comparing the terms in (11.92) with those in (11.91),
the two systems are found to be elastically equivalent if
­k(3/ss - 2IMS )
f--.­
~ =--------=-~=------
3/~ss + 2/~MS -~ - 4f!s
k(31s8 +IMM - 4fMS ) ~
12 = 2 ' (11.93)
3/~ss-4fMS
Figure 11.19 Elastically equivalent
3/ss - 21MS
iii ~ --=:::.--= structural components.
2/MS - IMM
Figure 11.19 shows the equivalent system for a beam with a flexural stifthess which varies linearly
fromElo at the left-hand end to 2EIo at the right-hand end. For the equivalent system,
II = 1.140/0 12 = 1.80710 al ~ 0.43131. (11.94)
The-stiflhess distribution in the equivalent system is
shownsuperimposed on that for the non-uniform beam.
A similar process can be used to replace a
curved component by elasticaJly equivalent prismatic (
components. The circular arch shown in Figure 11.20
will be taken as axially rigid and only in-plane
1.. - ­behaviour will be considered. The complementary
IL "­ -;: i'w'Xenergy induced by a moment M and forces P and S
Figure 11.20 A circular arch.shown in the figure is then given by
U= f(M+Py-Sxi ds = f41 [M+PR(coscf>-cosa)+SR(sincf>-sinaW Rd'" )
2EI 2EI 't' (11.95
-41
Expressing this in terms ofthe span I ofthe arch ( equal to 2Rsina ),
- M2/IMM MSI"iMs SlJyss Pl(%p MPI"iMP SPIYsp
U = - - - - - - - + - - + - - + --=-.::::... (11.96)
2EI EI 2EI 2EI EI EI
where
Reference
Non-prismatic Members[Sect. 11-8] Page 11.27
u f, ~ u + 2ucos2u - 3sinacosa
IMM ~ 21MS =-.­
Slnu 'PP 8sin3a
(11.97)
= 2usin2 u + u - sinucosaIMP =21sp = sinu - ucosu .f,
2sin2 cx
ss 8sin3 u
Ingeneral, the overall elastic properties ofsuch an
arch cannot be simulated by a rectangular portal
frame made from axially-rigid beams. However, if
y,Sthe legs are at an equal and opposite slope, as
shown in Figure 11.21, then an e1astically­
equivalent arch can be found. These legs are both (~
oflengthjl (j2 = h2+k2 ) and ofbending stiffitess M~ I ~I
pEI. The lintel is of length (1-2k)/ and bending
stifthess qEI. The complementary energy induced Figure 11.21 An equivalent arch.
by the moment M and forces P and S are then
fl r ] HI
U= f2~l(M -7S + 7 Pt+(M -SI +7S +7P?ds + f ~I(M -Ss +Phl?ds (11.98)
o ~
This gives an expression ofthe same form as (11.96) where
IMM ~ 2/MS = 2J + K 31pp = hl(2J + 3K) ,
IMP = 2isp = h(J + K) 31ss =(2k1 - 3k + 3)J + (l - k + kl)K (11.99)
and J = L K = (I - lk) .
P q
Comparing this with (11.97) gives the conditions from which the four parameters h, k, P and q,
describing the elastica1ly-equivalent portal frame, can be found in terms ofa. Some results are
given in the following table.
a h k qP
2.9030
0.01589 0.4683 0.9998 1.0051
300 0.1223 0.2600 l.0055 1.0090 I
I
450
0.1893 0.2481 1.0103 1.0219
60· 0.2644 0.2298 1.0131 1.0440
90· 0.4626 0.1626 1.0008 1.1419
120' 0.8257 0.0058 0.9451 1.4726
The first line ofthe table is for the smallest angle a for which the values ofthe four parameters
are real. As before, the forms ofthe substitute portals are similar to those ofthe original arches.
Values for other angles IX CIII!.be found using TRAPARCH.EXE.
The expression for U can be expanded to include the axial strain energy, for example.
Then the axial compressibility ofthe arch can be allowed for in its equivalent. Note that the
overall flexibilities are additive so that any extra flexibilities ofthe original can be expressed in
terms ofextra flexibilities in the equivalent.}
John D. Renton, "Elastic Beams and Frames", Second Edition (' J
ISBN 1-898563-8'::, Pages 11.26 through 11.27., -I!. ;r 12--c:- ~ cp +- )~ W
-(.WC-2.-1 p-c ~_, fz4,b;;l~)-dX"-2..':±') v' <p
f.J..J