1. Designing for Stiffness
For beam which is supported at both ends, suffers max deflection, y in its middle, when subjected to central load, L.
Can be represented as,
y = L x l3
48 x E x I
l – length of beam, E – Young’s Modulus, I – moment of inertia of the cross section with respect to the neutral axis.
The deflection of the beam under load can be take as a measure of stiffness.
Stiffness component can be define as its resistance to elastic deflection is a function of both shape and modulus of elasticity of the material

2. Designing for Stiffness
Stiffness of material increase by increasing its second moment of area, which computed from cross-sectional dimension and selecting a high-modulus material for its manufacture.
An important characteristic of metallic materials is that their elastics modulus are very difficult to change by changing the composition or heat treatment.
On the other hand, the elastic modulus of composite can be changed over wide range by changing the volume fraction and orientation of the constituents.
When a metallic component is loaded in tension, compression, or bending the Young’s Modulus E, is used in computing its stiffness.

3. Designing for Stiffness
When the loading is in shear or torsion the modulus of rigidity, G, is used in computing its stiffness. The relation between these two elastics constant is given by :
μ is Poisson’s ratio
When initially straight beam is loaded, it becomes curved as a result of its deflection.
As the deflection at a given point increase, the radius of curvature at this point decreases.
The radius of curvature, r, is given by the relationship :
M = Weight/load applied along the beam length

4. The importance of stiffness arises in complex assemblies where differences in stiffness could lead to incompatibilities and misalignment between various components, thus delay their efficiency or even causing failure.
Using high strength materials in attempts to reduces weight usually comes at the expense of reduces cross-sectional area and reduced second moment of area.
This could be adversely affect stiffness of the component if the elastic constant of the new strong material does not compensate for the reduced second moment of area.
Another solution to the problem of reduced stiffness is to change the shape of the component cross section to achieve higher second moment of area, I.
Figure 4.3 gives the formula for calculating I for some commonly used shape and the value of I for constant cross-sectional area.

6. In application where both stiffness and weight of structure are important, it becomes necessary to consider the stiffness /weight or specific stiffness of the structure.
In the simple case of structural member under tensile or compressive load, the specific stiffness is given by E/P, where E is the Young’s Modulus of material and P is density.
In such case, the weight of a beam is given stiffness can easily shown to be proportional to P/E.
This shown that the weight of the component cab be reduced equally by selecting a material with lower density or higher elastic modulus.
The weight of the beam for square cross sectional area, w, can be shown to be :

7. This shown that for given deflection y under load L, the weight of the beam is proportional to (P/E1/2).
For solid round bar diameter D the second moment of area, I is given as :
The weight of the round bar is given by :
From the both weight equation shows that the weight of an axisymmetric bar can be reduced by reduction p or increasing E of the material.
However, reducing p is more effective as E is present as square root.

8. Cost per unit property method
In the simplest cases of optimizing the selection of materials, one property stands out as the most critical service requirement.
In such simple case, the cost per unit property can be used as a criteria for selecting the optimum material.
Considering the case of a bar of a given length, L, to support a tensile force, F. the cross sectional area, A, of the bar is given by :
Where S=working stress of the material, which is related to its yield strength by an appropriates factor of safety.
The cost of the bar (C’) is given by :
Where C=cost of the material per unit mass
p=density of the material

9. In comparing the different candidate materials, only the quality [(C
In comparing the different candidate materials, only the quality [(C*p)/S], which is the cost of unit strength, need to be compared as F and L are constant
The material with the lowest cost per unit strength is the optimum material.

10. Table 1 : Candidate materials for suspension bridge tie rods
Example 1
It is required to select a structural material for the manufacture of the tie rod of a suspension bridge. A representative rod is 10 m long and should carry a tensile load 500 kN without yielding. The maximum extension should not exceed 18 mm. Which one of the steels listed in table 1 will give the lightest tie rod.
Table 1 : Candidate materials for suspension bridge tie rods
Material YS(MPa) E (GPa) Specific density (kg/m3)
ASTM A675 Grade
ASTM A572 Grade
ASTM A717 Grade
Maraging steel Grade
Al 5052-H
Catridge brass 70 % hard
temper

11. Table 1 : Candidate materials for suspension bridge tie rods
Solution
Table 1 : Candidate materials for suspension bridge tie rods
Material YS(MPa) E (GPa) Specific Area based on yield Mass
density (kg/m3) strength (x 10-3 m2) (g)
ASTM A675 Grade
ASTM A572 Grade
ASTM A717 Grade
Maraging steel Grade
Al 5052-H
Catridge brass 70 % hard
temper
For the present case, calculate the area based on YS :
A can also be calculate based on deflection= (load x length)/(E x deflection)
Multiply A with length of the bar and specific density, get mass (kg)

12. Table 1 : Candidate materials for suspension bridge tie rods
Solution
Table 1 : Candidate materials for suspension bridge tie rods
Material YS(MPa) E (GPa) Specific Area based on Mass
density (kg/m3) deflection (x 10-3 m2) (g)
ASTM A675 Grade
ASTM A572 Grade
ASTM A717 Grade
Maraging steel Grade
Al 5052-H
Catridge brass 70 % hard
temper

13. The larger of the area will be taken as the design area and will be used to calculate the mass.
The result of the calculation shown in table 2 and show that steel A717 Grade 70 and maraging steel Grade 200 give the least mass.
As the former steel is more ductile and less expensive will be selected

14. Example 2
What design change are required when substituting HDPE for stainless steel in making a fork for picnic set while maintaining similar stiffness. The narrowest cross section of the original stainless steel fork was rectangular of 0.6 x 5 mm.
E for stainless steel = 210 GPa
E for HDPE = 1.1 GPa

15. Solution E for stainless steel = 210 GPa E for HDPE = 1.1 GPa
I for stainless steel section = 5x(0.6)3 = 0.09 mm4 12
EI should be kept constant for equal deflection under load :
EI for stainless steel = 210 x 0.09 = 18.9
EI for HDPE design = 1.1 x I
I for HDPE design = 17.2 mm4

16. Taking a channel section of thickness 0
Taking a channel section of thickness 0.5 mm, web height 4 mm and width 8 mm,
I = [8 x (4)3 – 7 x (3.5)3]/12 = 17.7 mm4
which means the required value.
Area of stainless steel section = 3 mm2
Area of the HDPE section = 7.5 mm2
The specific density of stainless steel is 7.8 and that of HDPE is Thus :
Relative weight of HDPE/stainless steel
= (7.5 x 0.96)/(3 x 7.8) = 0.3

19. Example 3
Given a 2 m long aluminum alloy tube of outer diameter 3 cm and inner diameter 2.6 cm, determine its second moment of inertia.
a) What would the diameter of a solid rod have to be to have the same flexural stiffness ?
b) How do the weight of the two structures compare ?
(p=2700 kg/m3)
c) What would the dimensions be if a square section was used to give the same second moment of inertia.