2.2.75 – 6.525 Problem Set 3: Solutions to ME problems
Fall 2013
Jacob Bayless
Problem 1: Bolted joint
a) If a bolt is over-tightened, which will fail first—the bolt, or the plastic?
The bolt is made of grade 5.8 steel (yield strength 420 MPa) and the flanges are made of polycarbonate
(yield strength 60 MPa). When the bolt is tightened, it puts the threads in tension and the flange in
compression. Given a root diameter of 4.8 mm, the cross-sectional area of the bolt is:
2
𝜋𝑑𝑟𝑜𝑜𝑡
𝐴𝑏𝑜𝑙𝑡 =
= 1.8 × 10−5 𝑚2
4
So for a clamping force 𝐹, the tensile stress in the bolt is
𝐹
4𝐹
𝜎𝑏𝑜𝑙𝑡 =
= 2
𝐴𝑏𝑜𝑙𝑡 𝜋𝑑𝑟𝑜𝑜𝑡
The area of the polycarbonate that is in compression follows a “stress cone”; it spreads into the
thickness of the plastic. The stress will be highest where the cone is narrowest, which is right under the
head of the screw. Assuming a clearance bore diameter of 6.4 𝑚𝑚, the area directly under the head of
the screw will be:
𝐴ℎ𝑒𝑎𝑑
2
2
𝜋𝑑ℎ𝑒𝑎𝑑
𝜋𝑑𝑏𝑜𝑟𝑒
=
−
= 4.6 × 10−5 𝑚2
4
4
𝜎ℎ𝑒𝑎𝑑 =
𝐹
𝐴ℎ𝑒𝑎𝑑
The flange and bolt both carry 𝐹 of equal magnitude (although one is tensile and the other is
compressive). To determine which will yield first, the stresses are compared with the yield strengths:
𝜎
(𝑠 𝑏𝑜𝑙𝑡 ) 𝐴
𝑦𝑠𝑡𝑒𝑒𝑙
ℎ𝑒𝑎𝑑 𝑠𝑦𝑝𝑜𝑙𝑦
=
= 0.366
𝜎ℎ𝑒𝑎𝑑
(𝑠
) 𝐴𝑏𝑜𝑙𝑡 𝑠𝑦𝑠𝑡𝑒𝑒𝑙
𝑦𝑝𝑜𝑙𝑦
This ratio is less than one, so the polycarbonate will yield first. (Polycarbonate is quite ductile so it won’t
break, but when the bolt is unscrewed, there will be a depressed area where the plastic was deformed).
b) The bolts are tightened until a tensile stress of 300 MPa is developed in the bolts. What is the
stiffness of the joint?
As long as the bolt is tightened, the stiffness can be determined without needing to know the exact
amount of stress in the bolt. The relevant equations are found on page 20 of chapter 9 of
FUNdaMENTALs:
𝜋𝐸𝑓𝑙𝑎𝑛𝑔𝑒
𝑘𝑓𝑙𝑎𝑛𝑔𝑒 =
4ℎ𝑓𝑙𝑎𝑛𝑔𝑒 (
1
𝑑ℎ𝑒𝑎𝑑 (𝑑ℎ𝑒𝑎𝑑 + 2ℎ𝑓𝑙𝑎𝑛𝑔𝑒 cos 𝛼𝑐𝑜𝑛𝑒 ) −
𝑘𝑏𝑜𝑙𝑡 =
2
𝑑𝑏𝑜𝑟𝑒
)+
(1 + 𝜈𝑓𝑙𝑎𝑛𝑔𝑒 ) ln 2
ℎ𝑓𝑙𝑎𝑛𝑔𝑒
𝜋𝐸𝑏𝑜𝑙𝑡
4𝐿𝑏𝑜𝑙𝑡
1
1
+ (1 + 𝜈𝑏𝑜𝑙𝑡 ) ln 2 (
+
)
2
ℎ𝑛𝑢𝑡 ℎℎ𝑒𝑎𝑑
𝑑𝑏𝑜𝑙𝑡
The Poisson’s ratio of steel is: 𝜈𝑏𝑜𝑙𝑡 = 0.3 and Young’s modulus 𝐸𝑏𝑜𝑙𝑡 = 200 𝐺𝑃𝑎
And of polycarbonate: 𝜈𝑓𝑙𝑎𝑛𝑔𝑒 = 0.37 and Young’s modulus 𝐸𝑓𝑙𝑎𝑛𝑔𝑒 = 2 𝐺𝑃𝑎
We can assume a strain angle 𝛼𝑐𝑜𝑛𝑒 of 30°.
Some additional bolt dimensions are required. For these we can look up a typical M6 bolt on McMasterCarr to determine the measurements:
And for the nut:
From these drawings, ℎ𝑛𝑢𝑡 = 5 𝑚𝑚, ℎℎ𝑒𝑎𝑑 = 4 𝑚𝑚.
Note that much of the relevant information (such as the dimensions of an M6 screw, the diameter of the
clearance hole, and the strain cone angle) were deliberately omitted from the question. The reason is
that in a real design situation, engineers are not usually handed problems to solve with all of the
numbers already given, and so it’s important to know how to make educated guesses and research
appropriate numbers. In some settings, Before a design is finalized,
Now we have enough information to find the stiffness. Just plug in the numbers, and:
𝑘𝑓𝑙𝑎𝑛𝑔𝑒 = 2.43 × 107
𝑘𝑏𝑜𝑙𝑡 = 2.52 × 108
𝑁
𝑚
𝑁
𝑚
But there are four bolts, and these are the values from just one of them. The stiffness of the joint is the
stiffness of all four bolts in parallel (and in parallel with the stiffness of the flange – also times four). So,
𝑘𝑗𝑜𝑖𝑛𝑡 = 4(𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡 ) = 1.1 × 109
𝑁
𝑚
c) If a tensile load of 1000 N is applied across the joint, what is the stress in the bolts?
First of all, the bolts still have a 300 MPa tensile stress from being tightened, so the stress in the bolts
will include this as well as any stresses resulting from the load.
Secondly, when the applied load is taken up as an additional force across the bolts, the flange material
compressed underneath the bolt head has the opportunity to relax a little bit, reducing the stress
applied by the compressed flange on the bolt. This has the effect of the applied load being shared
between the flange and the bolts, so not all of the applied force results in additional stress in the bolts.
For 𝐹𝐿 = 1000 𝑁, and four bolts in the joint sharing the load equally,
𝐹𝑏𝑜𝑙𝑡 =
𝐹𝐿
𝑘𝑏𝑜𝑙𝑡
4 𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡
From the previous question, we know that:
𝑘𝑏𝑜𝑙𝑡
= 0.912;
𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡
So
𝐹𝑏𝑜𝑙𝑡 = 228 𝑁
This results in a total stress in each bolt of
𝜎𝑏𝑜𝑙𝑡 = 300 𝑀𝑃𝑎 +
𝐹𝑏𝑜𝑙𝑡
= 313 𝑀𝑃𝑎
𝐴𝑏𝑜𝑙𝑡
(This is not enough to yield the bolts).
d) What load will it take to separate the joint?
We start with the “Leak” equation in FUNdaMENTALs, which is given as:
𝐹𝑝 (𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡 )
𝐹𝑙𝑒𝑎𝑘 =
𝑘𝑓𝑙𝑒𝑎𝑘
Don’t forget that because there are four bolts in our joint, it will take four times this amount to
cause a leak. So,
𝐹𝑝 (𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡 )
𝐹𝑙𝑒𝑎𝑘 = 𝟒
𝑘𝑓𝑙𝑒𝑎𝑘
𝐹𝑝 is the preload force. We only know the preload stress, so if we want, we can convert that
back into a force.
𝜎𝑝 𝐴𝑏𝑜𝑙𝑡 (𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡 )
𝐹𝑙𝑒𝑎𝑘 = 𝟒
𝑘𝑓𝑙𝑎𝑛𝑔𝑒
Where 𝜎𝑝 = 300 𝑀𝑃𝑎
This results in 𝐹𝑙𝑒𝑎𝑘 = 250 𝑘𝑁
It is worth checking whether or not this would cause the bolts to yield, first. By repeating the
calculation in part c, we find the stress in the bolts to be:
𝜎𝑏𝑜𝑙𝑡 = 3420 𝑀𝑃𝑎
This is way higher than the yield strength of the bolts, so we can reasonably expect the joint to
break well before it would otherwise lose preload.
Both the flange and bolts will yield well below such a high applied load. Yielding of the flange
will not necessarily cause the joint to fail, but the bolts breaking definitely would. So if there’s a
chance that the joint will be subjected to very high loads, it would be worth re-calculating the
maximum force that the joint can bear before the bolts break:
𝐹𝐿
𝑘𝑏𝑜𝑙𝑡
𝜎𝑏𝑜𝑙𝑡 = 𝜎𝑝 +
4𝐴𝑏𝑜𝑙𝑡 𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡
𝜎𝑏𝑜𝑙𝑡 = 𝑠𝑡𝑒𝑛𝑠𝑖𝑙𝑒 ≅ 500 𝑀𝑃𝑎
𝐹𝐿 = 4𝐴𝑏𝑜𝑙𝑡 (𝑠𝑡𝑒𝑛𝑠𝑖𝑙𝑒 − 𝜎𝑝 )
𝑘𝑓𝑙𝑎𝑛𝑔𝑒 + 𝑘𝑏𝑜𝑙𝑡
= 16 𝑘𝑁
𝑘𝑏𝑜𝑙𝑡
This is a more reasonable load limit.
Problem 2. Cantilever Beam
A) At each location x, what is the maximum tensile stress along the top surface of the beam?
We will let 𝑥 = 0 at the wall. (You could also use 𝑥 = 0 at the end of the beam if you wanted). The
moment in the beam is the product of the force 𝐹 and its distance from 𝑥, so 𝑀 = 𝐹(𝐿 − 𝑥).
The tensile stress in a beam is given by:
𝑀𝑦
𝐼
Where 𝑦 is the distance from the neutral axis (ie, the midplane) of the beam. In this case then, 𝑦 = 𝑡/2.
𝐼 is the second moment of area (also called the area moment of inertia), which for a rectangular profile
𝜎=
is
𝑤𝑡 3
.
12
Therefore,
𝜎=
12𝐹(𝐿 − 𝑥)
𝑤𝑡 2
b) How can we remove material to change the profile of the beam so that it is stressed equally along its
length when 𝐹 is applied?
Clearly, 𝑤 and 𝐿 cannot be constants with 𝑥 in our modified beam. Since it is done only by removing
material from the existing beam, we can say that at 𝑥 = 0, we will retain the original beam profile.
12𝐹(𝐿 − 𝑥)
12𝐹𝐿
=𝜎=
2
𝑤(𝑥)𝑡(𝑥)
𝑤(0)𝑡(0)2
With both 𝑤(𝑥) and 𝑡(𝑥) being free functions of x, there is an unlimited number of valid solutions. One
solution would be to taper 𝑤(𝑥), so that the thickness is kept constant 𝑡(𝑥) = 𝑡(0) and
𝐿−𝑥
= 𝑐𝑜𝑛𝑠𝑡
𝑤(𝑥)
Then
𝐿−𝑥
𝐿
=
𝑤(𝑥) 𝑤0
𝑥
𝑤(𝑥) = 𝑤0 (1 − )
𝐿
The width decreases linearly along the length, creating a triangular-tapered beam. Does this seem
familiar? An example was shown in FUNdaMENTALs:
From FUNdaMENTALs of Design,Chapter 9: Structures, by Alex Slocum
Another option would be to allow 𝑡 to vary, holding 𝑤 constant. In that case,
𝐿−𝑥
= 𝑐𝑜𝑛𝑠𝑡
𝑡(𝑥)2
𝑥
𝑡(𝑥) = 𝑡0 √1 −
𝐿
This creates a beam which tapers with a parabolic thickness profile.
For bearing the point-force load, both of these tapered beams could be considered more “structurally
efficient” than the conventional, uniform-profile beam, especially if a lightweight structure were needed,
or if the materials were expensive. Sometimes it’s a good idea to go beyond the simple textbook
designs!
c) If the force 𝐹 is changed to a moment 𝑀, which design of beam is more efficient and why?
With an applied moment, the maximum tensile stress in the beam is now
12𝑀
𝜎=
𝑤𝑡 2
In other words, the stress no longer varies with 𝑥; thanks to the moment loading, the stress is already
uniform along the beam when 𝑤 and 𝑡 are constant. In that case, the uniform-profile beam is the most
efficient structure.
Note: Structural efficiency typically refers to a structure’s ability to support a load with the least amount
of material being wasted. The efficiency is generally highest when all of the materials are stressed to the
limit. If there are parts of the beam that aren’t highly stressed while other areas are highly stressed, it
suggests that the material in the un-stressed areas is being wasted, and that same material would serve
its purpose better if it were relocated to reinforce the highly-stressed areas. But as shown above, it
depends strongly on the type of loading being considered.