Mod 4 Assignment 4

AUE 8580 - ADVANCED VEHICLE STRUCTURE DESIGN
Module 4: BIW Topology Development-Structural Bending: Assignment 4
OCTOBER 5, 2021
Abhijeet Mordekar
amordek@clemson.edu
Body bending: Backbone structure

One topology for resisting bending loads is a backbone beam down the center of the vehicle. The
backbone can be idealized as a uniform beam supported at the axle position. The test load
condition is shown. The material is steel with E=207,000 N /mm2 andσ DESIGN =175 N /mm2. Use
the basic beam equations.
a) What is the minimum thickness to meet the strength requirement of supporting a 6670-N load?
b) What is the minimum thickness if the deflection at center span is to be at most 1mm?
c) Given the thickness computed in part a, what is the deflection in that case?
d) Given the thickness computed in part b, what is the maximum stress in that case?
e) If we desire to minimize mass (i.e., minimum t), what is the dominant requirement—strength
or stiffness?
f) How short does the wheelbase, L, need to be before the strength and stiffness requirements
result in the same thickness?
Solution:
Given:
2
The critical stress for design consideration , σ DESIGN =175 N /m m
Load at the center of the beam , F=6670 N
Height of the section , h=300 mm
Width of the section , w=200 mm
2
Young' s Modulus of the material , E=207,000 N /m m
h 300
The ratio of height ¿ width, r= = =1.5
w 200
Length of the backbone beam, L=2790 mm
Maximum deflection at the center of the beam, δ o=1 mm
1|Page
a) Minimum thickness to meet the strength requirement of supporting a 6670-N load
The stress for the beam section is given by:

Mc
σ Design = −−−−−(1)
I
Where,
2
σ Design =critical stress required for the design , N /mm
M =Bending moment of the section , Nmm
c=Vertical distance ¿ the point of interest ¿ the neutral axis ,mm
4
I =Moment of inertia of the section , mm
The bending moment for a simply supported beam at the center of the beam is given by:
FL
M= −−−−−( 2 )
4
6670∗2790
M=
4
M =4652325 Nmm
The vertical distance from the point of interest to the neutral axis is given by:
h
c= −−−−−(3)
2
300
c=
2
c=150 mm
Critical stress requirement for the design is given as:
2
σ Design =175 N /mm
The moment of inertia of the section is given by:
( 12 + 6r )−−−−−(4)
I =t 1 w 3 r 2∗
I =t 200 1.5 ∗( +
2 6 )
3 1 1.5
2
1
7 4
I =13500000 t=1.35∗10 t 1 mm
Substituting the values of M, c, σ Designand I in equation (1), we get,
2|Page
4652325∗150
175= 7
1.35∗10 ∗t 1
Therefore, we obtain the value of t 1 as,

t 1=0.295 mm
Thus, the minimum thickness to meet the strength requirement for supporting a force of
6670 N ist 1=0.295 mm
b) Minimum thickness if the deflection at center span is to be at most 1mm.
Deflection of the simply supported beam when the force is acting on the center of the beam is
given by:
3
FL
δ o= −−−−−(5)
48 EI
Where,
δ o=Maximum deflection , mm
F= Applied force /load , N
L=Length of thebeam , mm
' N
E=Youn g s modulus of thematerial , 2
mm
Substituting the given parameters and value of I in the earlier calculation in the equation (5), we
get,
3
6670∗2790
δ o= 7
48∗207000∗1.35∗10 t 2
Therefore, by solving the above equation we obtain the value of t 1 as,

t 2=1.079 mm
c) Given the thickness computed in part a, the deflection in that case:

Deflection of the simply supported beam when the force is acting on the center of the beam is
given by:
3
FL
δ o=
48 EI
Where,
δ o=Maximum deflection , mm
F= Applied force /load , N
3|Page
L=Length of thebeam , mm
' N
E=Youn g s modulus of thematerial , 2
mm
Substituting the given parameters and value of t 1, obtained in part a in the above equation, we get
3
6670∗2790
δ o= 7
48∗207000∗1.35∗10 ∗0.295
δ o=3.66 mm
d) Given the thickness computed in part b, the maximum stress in that case;
The stress for the beam section is given by:
Mc
σ Design =
I
Substituting the values of M, c, I &t 2 in the above equation, we get,
4652325∗150
σ Max = 7
1.35∗10 ∗1.079
2
σ Max =47.907 N /mm
e) If we desire to minimize mass (i.e., minimum t), what is the dominant requirement
strength or stiffness.
As per the values of thickness calculated from the strength and stiffness requirements, the
thickness for strength is accounted to be 0.295 mm and the thickness for stiffness is accounted to
be 1.079 mm. If we choose the thickness for strength requirement as desired thickness, then the
beam will fail in the stiffness requirement test since the calculated required thickness for stiffness
is high. Thus, the stiffness requirement is dominant to minimize mass.
f) How short does the wheelbase, L, need to be before the strength and stiffness
requirements result in the same thickness.
In order to have the same thickness for the strength requirements and to find the shortest length
of the wheelbase L, we equate the formulas of thickness considered for both the cases.
From equations (1), (2), (3) & (4), the formula for thickness for strength requirement is given by:
4|Page
FL
∗h
4
2
t 1= −−−−−( 6 )
3 21 r
σ design w r ∗ +
2 6 ( )
From equations (4) & (5), the formula for thickness for stiffness requirement is given by:
3
FL
t 2= −−−−−(7)
1 r
3 2
48 E δ 0 w r ∗ +
2 6 ( )
Equating equations (6) & (7), we get,
FL
∗h
4
3
2 FL
=
3 21 r
σ design w r ∗ +
2 6 ( ) 48 E δ 0 w3 r 2∗ ( 12 + r6 )
2 6 hE δ 0
L= −−−−−(8)
σ design
Substituting the values of given parameters in equation (8), we get,

2 6∗300∗207000∗1
L=
175
L=1459.158 mm
Therefore, in order to have the same deflection to meet the strength and stiffness
requirements, the shortest length of the wheelbase should be 1459.158 mm.
5|Page
Load Paths for Bending

A three-wheel concept car is shown below. The structure is made of shear panels which can only
react loads within their plane. We are interested in the bending performance of the structure. The
car body is supported at three points as shown, and loaded with a bending load, F, which is
applied in the plane of the shear panel oriented along the center of the car. Using free body
6|Page
diagrams for each shear panel and bar element (labeled A through G), determine numerical
values for the loads on each structural element. Assume F = 6000 N.
Solution
F = 6000N
a = 400mm b = 600mm c = 500mm
RF RR
Fig 1. Free body diagram.

Using free body analysis,
Summation of forces in the y-direction.
2 RF
+ R R=F
2
R F + R R=6000 -----(1)
Summation of moments about the front axle,

F∗1000=R R∗1500
6000∗1000
R R=
1500
R R=4000 N
-----(2)
Substituting equation (2) in (1)

R F + 4000=6000
R F=2000 N
-----(3)
Considering each element in static equilibrium,

PA
7|Page
RF
RF 2
2
Let the unknown load be P A .
Consider element A
Let F L be acting on the top and bottom edge of the element A.
Summation of moments about the edge of this element A,
RF
∗a=P A∗h
2
2000∗400
PA=
2∗800
P A =500 N
Now, consider element B,

PB 2
From the figure given in the question we can see that,
P A =PB 1=500 N
Considering equilibrium along the horizontal direction we PB 1

get,
PB 1=PB 2=500 N
Now, consider element C, PC 2
From the figure given in the question we can see that,
P A =PC 1=500 N
PC 1
Considering equilibrium along the horizontal direction we
get,
PC 1=PC 2=500 N
Now, consider element D,
Summation of moments about an edge gives us, RF

2
8|Page
RF PPD
∗1200=P D∗600 D
2
2000∗1200
P D=
2∗600 RF
P D=2000 N 2
Now consider element E,

PE1
Summation of moments about an edge gives us, PE2
P E 2∗600=PE 1∗1200
PE1
P E 1=PB 2 =500 N
500∗1200
P E 2=
600
P E 2=1000 N
Now consider element F,
Summation of moments about an edge gives us,

P F 1∗1200=PF 2∗600 PF1
PF2
P F 1=PC 2=500 N
500∗1200 PF1
P F 2=
600
P F 2=1000 N
Now consider element G,

PG 1 RR
PG 1=P E 2=1000 N
PG 2=P D =2000 N
PG 2
F = 6000N F
PG 1
R R=4000 N
9|Page
Side frame deformation

Data are provided for a side frame structure. The stiffness requirement is 7000 N/mm for the vehicle,
or 3500 N/mm for each side frame.
a) Compute the initial vehicle bending deflection, Δ, with rigid joints (a joint stiffness of
10
K j=1 x 10 Nmm/rad may be considered as rigid). Does the resulting stiffness meet the
requirement?
To improve bending stiffness, any of the sections may be increased in size (w, h) by up to 200% of
the initial dimensions, except the rocker which is restricted to an increase in size up to 125% due to
entry constraints. Thickness on all sections may be increased up to 3 mm. (Do not reduce the size of
any beams from the given initial size.)
b) Continuing with rigid joints, adjust the side frame beams to meet the stiffness requirement in
the most mass efficient way. Do at least two iterations of resizing. Which beams did you
adjust; why did you choose them; what are the final beam sizes; and what is the final
stiffness?
c) After doing part b, enter the joint stiffness values shown and determine the bending stiffness
with flexible joints. What is the new bending stiffness? What is the fraction of stiffness with
flexible joints to stiffness with rigid joints?
10 | P a g e
Solution:
11 | P a g e
a) Compute the initial vehicle bending deflection, Δ, with rigid joints,
Using the Don Malen sheet, we get the following solution,
Fig 1. Side Frame Geometry.
Fig. 2 Side Frame Section and Joint Properties.
12 | P a g e
Fig 3. Side Frame Restraints and Loads.
Fig 4. Side Frame Results.
13 | P a g e
From the above figure we can see that the maximum deflection occurs at node 8 with a deflection of
∆=3.73 mm.
Body bending stiffness is given by:

F
K=
∆
Where,
K=Body bending stiffness.
F= Applied force on each side of the frame=3337.5 N .
∆=Deflection of the frame .
3337.5
K=
3.73
K=894.77 N
The resulting stiffness does not meet the requirement that is 3500 N/mm.
In order to meet the stiffness requirement, the deflection of the beam should be less than:
F
Δ=
K
3337.5
Δ=
3500
Δ=0.953 mm
b)
Beam resizing
The fundamental concept behind selectively optimizing the sizing of the beams is increasing
the dimension of the cross section along the direction of bending force. This can be
explained by the bending formula shown below:
M σ E
= =
I y R
Where,
𝑀 = Bending moment created by the applied force, Nmm
𝐼 = Moment of inertia of the beam cross-section, m m4
𝜎 = Bending stress generated in the beam, N /m m2
𝑦 = distance of the extreme element from the neutral axis
𝐸 = Young’s modulus, N /m m2
14 | P a g e
𝑅 = Bending radius of the beam, mm
Thus, the resisting moment generated by the beam section is:

Mr=σz
z=section modulus=I / y
Thus, for a rectangular cross-section the second area moment of Inertia I =b d 3 /12 and y=d /2.
If we double the value of ‘d’ (the section dimension along the direction of application of bending
force), the max stress induced in the beam element at its extremes reduces and thus the
corresponding strain and consequently, deflection is also reduced. Therefore, we use this concept
in resizing the beams.
Therefore, the steps that were followed to carry out successive iterations are as follows,
 We first chose the beams with the highest strain energy as seen from the figure (4), and
since the rocker arm and the roof rail have the highest strain energy, these beams are
chosen to alter.
 By increasing the section modulus by increasing the dimension of the cross section in the
direction of application of force, we can increase the resisting moment generated by the
beam. This concept was used when resizing the beam.
 The first step in beam resizing was to increase the height of the rocker arm.
 The next step if the stiffness requirement is still not met is to increase the height of the
rocker arm and roof rail beam.
 Subsequently, if the stiffness target is still not met, the width of both the beams is
increased incrementally, starting with the rocker beam.
 Increasing the thickness of the beams is the last step in case the stiffness target is not met
even after the height and width of the cross-section have been increased to their max.
permissible limit.
The table below shows the iterations that were performed to come to final mass optimized side
frame design.
Iteration Rocker Roof Rail Force Deflection Stiffness Mass

H W T H W T Δ K
1 100 50 1 25 40 1 3337.5 3.72 897.177 12.61
2 125 50 1 25 40 1 3337.5 2.91 1146.907 13.40
3 125 50 1 50 40 1 3337.5 1.82 1833.791 13.88
4 125 62.5 1 50 40 1 3337.5 1.73 1929.192 14.28
5 125 62.5 1 50 80 1 3337.5 1.47 2270.408 15.06
6 125 62.5 1.5 50 80 1 3337.5 1.26 2648.809 18.00
7 125 62.5 1.5 50 80 1.5 3337.5 1.12 2979.910 19.27
8 125 62.5 2.0 50 80 1.5 3337.5 0.99 3371.212 22.20
9 125 62.5 2.0 50 80 1.8 3337.5 0.95 3513.158 22.97
15 | P a g e
10 125 62.5 1.8 50 80 2.1 3337.5 0.95 3513.158 22.56

11 125 62.5 1.8 50 80 2.2 3337.5 0.93 3588.709 22.81
Observation (for iteration 1 to 8)

 Because of the vertical loading of the side frame, the maximum deflection is observed in
the horizontal beams namely the rocker beam and the upper side rail. Also, it can be seen
that the maximum bending energy is found in the rocker front beam and rocker rear
beam. This bending energy is nothing, but the strain energy absorbed by the rocker beam.
 Thus, the rocker beam dimensions (height first) is incrementally increased so that we can
reduce the maximum deflection.
 With respect to the loading condition and the deflection of the frame, it is observed that
the rocker beam absorbs the highest strain energy.
 It was consequently observed that, the rocker beam with its height increased to the max.
permissible limit, had a deflection (Iteration 2 deflection of 2.44 mm) which was lower
than the one calculated above. Thus, we decided to alter the rocker beam as well as the
roof rail beam.
The last two iteration are shown. These iterations are discussed in detail below.
Iteration 10
16 | P a g e
Fig 5: Iteration 10 for bending stiffness requirement- geometry.

Fig 6: Iteration 10 Results - Deflection at node 8.
Conclusion:
As can be seen from the above figure, the maximum deflection on node 8 is 0.95 mm. The excel
shows negative deflection because the deflection is in the negative y-axis direction.
The bending stiffness 𝐾 for iteration 10 is given by:
F
K=
Δ
3337.5
K=
0.9 5
N
K=3513.158
mm
Thus, by altering the rocker beam and roof rail, the computed bending stiffness (
K iterr 10=3 513.158 N /mm which meets the required bending stiffness K req =3500 N /mm .
The final beam sizes are as follows,
17 | P a g e
Beam 𝒉 𝒘 𝒕
Rocker 125 62.5 1.8
Hinge pillar 100 50 1
A-pillar 35 35 1
Roof Rail 50 80 2.1
B – pillar upper 50 25 1
B – Pillar lower 75 50 1
C – Pillar upper 75 25 1
C – Pillar lower 50 50 1
Table 2. Iteration 10 finalized section properties of key frame members.
Iteration 11
18 | P a g e
Fig 7: Iteration 11 for bending stiffness requirement- geometry.
Fig 8: Iteration 11 Results - Deflection at node 8.
Conclusion:
As can be seen from the above figure, the maximum deflection on node 8 is 0.93 mm. The excel
shows negative deflection because the deflection is in the negative y-axis direction.
The bending stiffness 𝐾 for iteration 11 is given by:
F
K=
Δ
3337.5
K=
0.9 3
N
K=3588.709
mm
Thus, by altering the rocker beam and roof rail, the computed bending stiffness (
K iterr 11=3588.709 N /mm which meets the required bending stiffness K req =3500 N /mm .
The final beam sizes are as follows,
19 | P a g e
Beam 𝒉 𝒘 𝒕
Rocker 125 62.5 1.8
Hinge pillar 100 50 1

A-pillar 35 35 1
Roof Rail 50 80 2.2
B – pillar upper 50 25 1
B – Pillar lower 75 50 1
C – Pillar upper 75 25 1
C – Pillar lower 50 50 1
Table 3. Iteration 11 finalized section properties of key frame members.
The mass of the side frame was 22.81 kg, thus achieving a satisfactory value of bending stiffness
to not much increase in mass of the frame.
c) Determine the bending stiffness with flexible joints.
Iteration 10
20 | P a g e
Fig 9: Iteration 10 for bending stiffness requirement- geometry. (flexible joints)
Fig 10: Iteration 10 Results - Deflection at node 3. (flexible joints)
From the above figure 10 we can see that the maximum deflection is 3.79mm. The deflection is
negative because the deflection is in negative y-axis direction.
The bending stiffness for iteration 10 is given by:

F
K=
∆
3337.5
K iterr 10 −fj =
3.79
K iterr 10 −fj =8 80.60 N /mm
The fraction of bending stiffness with flexible joints to bending stiffness with rigid joints is given
by:
K iterr 1 0− fj
K fraction =
K iterr 1 0
21 | P a g e
880.606
K fraction =
3513.158
K fraction =0. 2506∨25 . 06 %
The fraction of bending stiffness with flexible joints to bending stiffness with rigid joints
for iteration 10 is 25.06%.
Iteration 11
22 | P a g e
Fig 11: Iteration 11 for bending stiffness requirement- geometry. (flexible joints)
Fig 12: Iteration 11 Results - Deflection at node 3. (flexible joints)

From the above figure 12 we can see that the maximum deflection is 3.82mm. The deflection is
negative because the deflection is in negative y-axis direction.
The bending stiffness for iteration 10 is given by:

F
K=
∆
3337.5
K iterr 11− fj=
3. 82
K iterr 11− fj=873.69 N /mm
The fraction of bending stiffness with flexible joints to bending stiffness with rigid joints is given
by:
K iterr1 1−fj
K fraction =
K iterr1 1
23 | P a g e
873.69
K fraction =
3588.709
K fraction =0. 2434∨2 4.34 %
The fraction of bending stiffness with flexible joints to bending stiffness with rigid joints
for iteration 11 is 24.34%.
Torsional stiffness of van with side frame

The right and left side frames of the van now consist of four steel pillars, 40 mm × 40 mm × 1.8
mm thick, and rigid roof and rocker beams. All other panels are as in Exercise 5.4. The pillars
form three door openings. The doors in this case do not contribute to torsional stiffness and can
be neglected. Consider the side frame to be isolated, restrained at the bottom, and loaded by a
shear load, Q, at the top. The resulting horizontal deflection is Δ.
Because the roof beam and rocker are rigid, each pillar acts as two cantilever beams of length
h/2, as shown. Each cantilever reacts a load, Q/4, and deflects by an amount Δ/2.
a) What is the shear stiffness, Q/Δ, of each side frame?
b) What is the effective shear rigidity, ( ¿ ) EFF of each side frame?
c) Determine the van torsional stiffness with a right and left side frame replacing the flat steel
sides of Exercise 5.4 (all other panels are flat panels).
24 | P a g e
d) Write a formula for the maximum bending stress in each pillar as a function of Q, the applied
shear force. (Useσ MAX =c M MAX /I on the cantilever illustrated.)
25 | P a g e

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AUE 8580 - ADVANCED VEHICLE STRUCTURE DESIGN

Module 4: BIW Topology Development-Structural Bending: Assignment 4

Body bending: Backbone structure

a) Minimum thickness to meet the strength requirement of supporting a 6670-N load

The stress for the beam section is given by:

The moment of inertia of the section is given by:

Substituting the values of M, c, σ Designand I in equation (1), we get,

Therefore, we obtain the value of t 1 as,

Therefore, by solving the above equation we obtain the value of t 1 as,

c) Given the thickness computed in part a, the deflection in that case:

Substituting the values of given parameters in equation (8), we get,

Load Paths for Bending

a = 400mm b = 600mm c = 500mm

Fig 1. Free body diagram.

Summation of moments about the front axle,

Substituting equation (2) in (1)

Considering each element in static equilibrium,

Let the unknown load be P A .

Let F L be acting on the top and bottom edge of the element A.

Summation of moments about the edge of this element A,

Now, consider element B,

Considering equilibrium along the horizontal direction we PB 1

Now, consider element C, PC 2

From the figure given in the question we can see that,

Now, consider element D,

Summation of moments about an edge gives us, RF

Now consider element E,

Now consider element F,

Summation of moments about an edge gives us,

Now consider element G,

Side frame deformation

a) Compute the initial vehicle bending deflection, Δ, with rigid joints,

Using the Don Malen sheet, we get the following solution,

Fig 1. Side Frame Geometry.

Fig. 2 Side Frame Section and Joint Properties.

Fig 3. Side Frame Restraints and Loads.

Fig 4. Side Frame Results.

Body bending stiffness is given by:

K=Body bending stiffness.

F= Applied force on each side of the frame=3337.5 N .

∆=Deflection of the frame .

𝑅 = Bending radius of the beam, mm

Thus, the resisting moment generated by the beam section is:

Iteration Rocker Roof Rail Force Deflection Stiffness Mass

10 125 62.5 1.8 50 80 2.1 3337.5 0.95 3513.158 22.56

Observation (for iteration 1 to 8)

Fig 5: Iteration 10 for bending stiffness requirement- geometry.

The bending stiffness 𝐾 for iteration 10 is given by:

The final beam sizes are as follows,

Rocker 125 62.5 1.8

Hinge pillar 100 50 1

Roof Rail 50 80 2.1

Fig 7: Iteration 11 for bending stiffness requirement- geometry.

Fig 8: Iteration 11 Results - Deflection at node 8.

The bending stiffness 𝐾 for iteration 11 is given by:

The final beam sizes are as follows,

Rocker 125 62.5 1.8

Hinge pillar 100 50 1

c) Determine the bending stiffness with flexible joints.