chapter 5 torsion and transverse shear of thin-walled...

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\1_transverseShear_advanced.docx p. 1 of 7

Chapter 5

Torsion and Transverse Shear of Thin-walled Beams

update this outline

o 3D FEA examples of thin-walled beams o Simplifying assumptions o Shearing stresses in open sections o Shear center o Shearing stresses, strains, and deformations in single-cell

closed sections o Analysis of multicell thin-walled sections in shear o Combined torsion and transverse shear of multicell thin-

walled beams

Coupled behavior: “C-beam” subjected to transverse load See more here: C:\W\whit\Classes\304\Notes\5_Torsion_and_transverseShear\Example_4.7\405_HW_4_C-beam


Shear in Advanced Beams Caveat: The discussion in this file assumes that the distributed moments my and mz =0, Iyz=0 and there are no gradients of thermal load in the x-direction. Although this is restrictive, it does simplify the equations a lot… and we still get to see some interesting behavior. If time permits, we will also consider the more complicated cases. Ref. A&H p. 205 Warning: A&H uses a “reference modulus” which I have chosen not to use. You have probably noticed that we tend to spend more time talking about normal stresses than shear stress. One reason is that for a typical non-optimized beam, the failure is due to excessive normal stress. However, once one decides that every ounce is important and you optimize the dimensions, shear stresses can become very important. Weight is obviously critical for aerospace structures, so we must understand how to evaluate the severity of the shear stresses. Earlier, we derived a formula for transverse shear stress in a beam. The formula is

1( )( ) b

y xxxy y

dy bdyb y dx

σσ = ∫ . We then converted this formula into a usable form for beams in

which there is no coupling between extension and bending. The formula was

( )b

x y

yV

b y EIy

b E y d yσ = − ∫ for the case of no thermal load. If you wish to include thermal

loads, you will need to go back to 1( )( ) b

y xxxy y

dy bdyb y dx

σσ = ∫ and include thermal effects in the

constitutive relation when expressing xxσ in terms of “beam” parameters. This formula can be further simplified for homogeneous cross-sections. When we derived this formula, we were only considering simple cross-section shapes, such as rectangular and we had not defined the quantity known as shear flow. (q). Now we want to consider more complicated shapes, such as that below.

We could start the derivation from scratch again (see AS&H for details) or we can extend what we already know. Let’s try using what we already know.

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\1_transverseShear_advanced.docx p. 3 of 7 Recall that we defined shear flow to be xsq tσ= when we were discussing torsion of thin-walled beams. That definition will not change. What does change is that the shear flow is not constant in the cross-section of a beam subjected to transverse loads. In fact, our goal is to determine how it changes. Using the formula above, we obtain

shear flow is

b

x y

yVq tEI

yE yt dyσ= = − ∫ , where I have replaced “b” with “t” just to be

consistent with our discussion of thin-walled cross-sections in which we used “t” for the thickness of the thin wall. Recall that by refers to the y-coordinate at the bottom of the beam where the shear stress is zero. We can generalize this formula for piecewise evaluation as follows.

0

0 0( ) ( ) where y is simply the starting point for the integration

yVq y q yEI

yE yt dy= − ∫

This is more useful, since when we do piecewise evaluation, the shear flow will not usually be zero at the lower limit of integration. This formula is also valid if the cross-section is more complicated, such as that shown above. We only need one minor change… replace “y” with “s”, where “s” is the curvilinear coordinate that goes along the middle of the thin wall. Hence, our formula is

0

0

( ) ( )

sVq s q sEI

s

E yt ds= − ∫ .

The presence of “y” in this formula is not a typo. It appeared in the formula originally because the axial strain varies linearly with “y”, and so when we replaced xxσ , it appeared in the integral. In performing the integration, we will need to express “y” in terms of “s”. This formula can be expressed as

0

0

( ) ( )

for transverse shear load in the y-direction

sVQEq s q s where QE

EIs

E ytds= − = ∫

This formula is for transverse shear load in the y-direction. If the transverse load is in the z-direction, the formula changes slightly, as you would expect.


0

0

( ) ( )

for transverse shear load in the z-direction

sVQEq s q s where QE

EIs

E z tds= − = ∫

In each case you must choose the correct section stiffness and shear force. In the formulas below, I have added subscripts to clarify this.

0

0

0

0

( ) ( )

( ) ( )

y yy

zz

z zz

yy

sV QE

q s q s where QEEI

ss

V QEq s q s where QEEI

s

E ytds

E z tds

= − =

= − =

∫

∫

Note: In A&H, the subscripts for Q are different than the notation here. The normal procedure in using this formula is to start at a location where the shear flow is zero (i.e. 0( )q s =0) and start your piecewise integration. In this class, it is normally piecewise because the wall usually consists of a sequence of straight pieces, but they are not collinear. Generally,

the integrand is piecewise constant, so

0

y

s

QE

s

E ytds= ∫ can usually be expressed as

( )

( )

0

0

up to s

and

up to s

y c ii

z c ii

s

QE E y A

s

s

QE E z A

s

E ytds

E z tds

= =

= =

∑

∑

∫

∫

In this course, I will encourage you to evaluate the shear flow for one loading direction at a time. This is because I like to keep my formulas simple. Analyze the beam for transverse loading in the y-direction. Then consider the z-direction loading. Then add the contributions. Make sure the curvilinear coordinate “s” is defined in exactly the same way for both analyses.

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\1_transverseShear_advanced.docx p. 5 of 7 I should point out that you will often find these formulas specialized for homogeneous beams. For the case of homogeneous beams, we obtain

0

0

0

0

( ) ( )

( ) ( )

y yy

zz

z zz

yy

sV Q

q s q s where QI

ss

V Qq s q s where QI

s

y tds

z tds

= − =

= − =

∫

∫

The derivation in this file has shown you that

• if you know the shear flow at one location, you can integrate to obtain the shear flow at any other location

• to determine the shear stress, you must know how the stress xxσ varies in the x-direction. Fortunately, this is not difficult for many configurations

Appendix If you wish to start from scratch, here are the sketches you will need. The biggest difference between our original derivation for shear stress in a simple beam and the derivation using these sketches is that the original derivation considered a piece of the differential element that wen from the bottom to some ybar or from ybar to the top. The reason was that that the shear stress was zero at the top and the bottom, so it was convenient. Take a look at Allen and Haisler to go through the details.


Etc. Note that in the above sketch, if delta s had been taken very small, you would not even know the wall was curved.

1b_coupledCase_alt_2.mw p. 1 of 2

> >

(2)(2)

(1.1)(1.1)

(1 2)(1 2)

(4)(4)

> >

(1)(1)

> >

(3)(3)

> >

> >

coupledCase_alt_2.mw

In the derivation of the formula for shear stress, an intermediate result was

What is the derivative of sigmaXX for more complex cases?

restart : currentdir ; with LinearAlgebra :"C:\W\whit\Classes\304\Notes\5_Torsion_and_transverseShear"

Note that S is diagonal if EIyz=0.S dMatrix s11, 0, 0 , 0, s22, s23 , 0, s23, s33

S :=

s11 0 00 s22 s23

0 s23 s33

R = list of stress resultantse = list of generalized strains <epsXX, kz, ky>

R d F x , Mz x , My x C FT x , MTz x , MTy x ;e d S . R

R :=

F x CFT x

Mz x CMTz x

My x CMTy x

e :=

s11 F x CFT x

s22 Mz x CMTz x Cs23 My x CMTy x

s23 Mz x CMTz x Cs33 My x CMTy x

vxx := e[2]: wxx := -e[3]: sigma[xx] := E*(-y*vxx-z*wxx-alpha * deltaT(x)):dSigmaXX_dx := diff(sigma[xx],x);dSigmaXX_dx := eval(dSigmaXX_dx, {diff(My(x), x) = -my+Vz, diff(Mz(x), x) = -mz-Vy}):dSigmaXX_dx := eval(dSigmaXX_dx, {diff(MTy(x), x) = dMTy_dx, diff(MTz(x), x) = dMTz_dx});

dSigmaXX_dx := E Ky s22 ddx

Mz x Cddx

MTz x Cs23 ddx

My x Cddx

MTy x Kz

Ks23 ddx

Mz x Cddx

MTz x Ks33 ddx

My x Cddx

MTy x Kα ddx

deltaT x

dSigmaXX_dx := E Ky s22 Kmz KVyCdMTz_dx Cs23 KmyCVz CdMTy_dx Kz Ks23 KmzKVy

CdMTz_dx Ks33 KmyCVz CdMTy_dx Kα ddx

deltaT x

Simplifications for various casesIf EIyz=0 (i.e. s23=0)eval(dSigmaXX_dx, s23 = 0);

E Ky s22 KmzKVyCdMTz_dx Cz s33 KmyCVz CdMTy_dx Kα ddx

deltaT x

If EIyz=0 and there are no distributed momentseval(dSigmaXX_dx, {my = 0, mz = 0, s23 = 0});

1b_coupledCase_alt_2.mw p. 2 of 2

> >

> >

> >

(1.2)(1.2)

(1.4)(1.4)

(1.3)(1.3)

> >

(2.1)(2.1)

E Ky s22 KVyCdMTz_dx Cz s33 Vz CdMTy_dx Kα ddx

deltaT x

If EIyz=0, there are no distributed moments, and thermal loads are constant in x-directioneval(dSigmaXX_dx, {my = 0, mz = 0, s23 = 0, dMTy_dx = 0, dMTz_dx = 0, diff(deltaT(x), x)=0});

E y s22 VyCz s33 Vz

If EIyz is non-zero, but there are no distributed moments and thermal loads are constant in x-directioneval(dSigmaXX_dx, {my = 0, mz = 0, dMTy_dx = 0, dMTz_dx = 0, diff(deltaT(x), x)=0});print(`Can rewrite as `,Vy * diff(%,Vy) + Vz * diff(%,Vz) );

E Ky Ks22 VyCs23 Vz Kz s23 VyKs33 VzCan rewrite as , Vy E y s22Kz s23 CVz E Ky s23Cz s33

What is in S?

C d Matrix EA, 0, 0 , 0, EIzz, KEIyz , 0, KEIyz, EIyy ; S d CK1;

C :=

EA 0 00 EIzz KEIyz

0 KEIyz EIyy

S :=

1EA

0 0

0 EIyy

EIzz EIyyKEIyz2EIyz

EIzz EIyyKEIyz2

0 EIyz

EIzz EIyyKEIyz2EIzz

EIzz EIyyKEIyz2

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\2_example_4.6.docx p. 1 of 4

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\3_shear_beams.docx p. 1 of 3

Shear in Advanced Beam: Summary

----------update formulas for inhomogeneous beams! Transverse shear stress in laminated beam

( )b

xy

x y

V dA

yV

b y EIy

b E y d y

σ

σ

=

= −

∫

∫

Euler-Bernoulli beam theory assume shear deformation due to transverse shear stress=0. Torsion of circular cylinder (homogeneous)

x x

x

x

x x

M r rdrd

dM GJdx

d rdx

θ

θ

θ θ

σ θ

φ

φε

σ ε

=

=

=

=

∫ ∫

Torsion of thin wall, closed cell beam (single cell)

2

1 2

x n xsM q r ds qA q t

q dsx A t G

σ

φ

= = =

∂=

∂

∫

∫

For single cells, q can be pulled out of integral 2 24 4 x

A AM GJds dsxt G t G

φ∂= => =

∂∫ ∫

Torsion of thin wall, closed cell beam (multi-cell)

2 relates moment due to shear flow to moment in beam

1 must impose all rotations are the same2

x i i

i i

x

M q A

q dsx A t G

MGJ

x

φ

φ

=

∂ = ∂

=∂∂

∑

∫


Combined torsion and transverse shear of thin wall, closed cell beams

0

0

M = moment due to shear flowMust integrate to obtain moment due to shear flow.

( ) ( )

1 must impose all rotations are the same2

x

y yy

zz

i i

x

sV QE

q s q s where QEEI

s

q dsx A t G

MGJ

x

E ytds

φ

φ

= − =

∂ = ∂

=∂∂

∫

∫

Combined torsion and transverse shear of thin wall, open section beams

0

0

Cannot assume shear stress is constant across wall thickness

( ) ( )

?If the cross section is a collection of slender rectangles,

then the torsional stiffness i

xs

y yy

zz

sV QE

q s q s where QEEI

sGJ

E ytds

σ

= − =

=−

∫

31s approximately where "t" is the wall thickness3 i

ibt

∑

Shear center in thin wall beams The shear center is a special location. If the transverse load acts through the shear center, the beam does not twist. This gives us a clue as to how to calculate the location. Consider the following:

1. For any load system (including those that cause twist), the moment about the x-axis (i.e. the torque) due to the shear flow must equal the moment in the beam about the x-axis. For example, if the applied loading causes a twisting moment of 100 inch-lbs., then the moment due to the shear flow in the cross section must equal 100 inch-lbs.

2. The rate of rotation must equal zero. It is critical that you keep in mind that shear flow is a stress resultant. It is describing something inside the beam. It is not the applied load. The following assumes that the load is applied on the positive end of the beam. If you are applying the load on the negative end of the beam, there is a switch in sign


between what is applied and what is inside the beam. Hence, one solution strategy is

1. Determine the shear flows for the case that the rate of rotation is zero (i.e. there is no twist) 2. Determine the moment about the x-axis due to the shear flow. 3. Position the shear force such that the moment due to the shear force = moment due to the

shear flow. For example, if you sum moments about the origin when calculating the moment due to the shear flow, then the moment due to an applied shear force in the z-direction is Vz * y. Warning: If you wish to determine the y-coordinate of the shear center, you must only apply a shear force in the z-direction. If you wish to determine the z-coordinate of the shear center, you must only apply a shear force in the y-direction. If you have combined loads, you will be in trouble.

For an open section, (e.g. rectangular, C, S, or I cross-section), there is a bit of a subtlety.

• The shear stress due to an applied to an applied torque is not constant across the wall thickness. • The shear stress due to an applied shear force acting through the shear center is assumed to be

constant across the wall thickness. If you apply a shear force through the shear center, there will be a shear flow in the section. The moment due to the shear flow must equal the moment due to the shear force. You can sum moments about any point when calculating these moments, but it is particularly revealing to sum moments about the shear center. (I know we do not know where it is… so just say it is at y= ySC for the y-coordinate.) Assuming we sum moments about the shear center and the applied shear force in the z-direction acts through the shear center, then the moment due to the shear flow must equal zero. This requirement gives us an equation to determine the value of ySC. If we sum moments about any other point, we should still assume that the shear force acts through the shear center.

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\4a_summary_example4.7.docx p. 1 of 5

Summary of Maple Solutions for Example 4.7 in A&H (p. 213) The discussion of Example 4.7 in A&H needs some clarification. Here is the example.

If one was to apply the shear forces as indicated in the sketch, the shear flows would not be as calculated. You will note in the calculation that no mention is made about the z-location of the y-direction force or the y-location of the z-direction force… but it makes a difference. There are two ways to interpret this problem statement:

1) The calculation actually assumes that the shear forces are applied such that there is no twist. For that to be the case, the shear forces must act through the “shear center”. We have not discussed the shear center yet. If we allowed there to be twisting of the beam, the shear flows would be different. Let’s re-interpret the sketch. Assume the sketch simply shows that there are y-and z-direction shear forces. We will assume that they are being applied such that there is no twisting of the beam. After we analyze the beam, we will determine where the forces must have been acting to avoid twist.

2) The forces are applied such that there is twisting of the beam. However, the first approximation of the effect of the twisting is a shear stress distribution in each rectangular segment that is symmetric about the mid-line and hence the net shear flow is zero. Hence, the twist does not affect our calculation. Here is a schematic of the shear stress distribution in a rectangular region (by Felippa)

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\4a_summary_example4.7.docx p. 2 of 5 example_4.7.mw Solved like in A&H with the discontinuous path, but not using the reference modulus method. example_4.7_shearCenter.mw Solved using a continuous path. This worksheet also calculates the shear center, which is discussed in Example 4.8 in A&H example_4.7_path.mw This uses the same path as , xample_4.7_shearCenter.mw but it adds another approximation. Assuming the wall thickness is very small, the variation of the integrand is simplified. For example,

2 2is approximated asy EdA y E t ds∫ ∫

Making this approximation makes it much simpler to “automate” the calculation.

Comparison of predictions example_4.7.mw example_4.7_shearCenter.mw example_4.7_path.mw

zc .8125 .800 .55+.25= .80

EIzz 2.629E9 2.757E9 2.756E9

EIyy 2.971E8 2.984E8 2.952E8

qA 1017.85 1041.5 1057.7

qB -1212.36 -1201.95 -1202.38

Shear center Z = -1.060 Z=-.8777 Z=-.8781

Twist Not done yet -66.7 degrees -66.7 degrees

Coordinates are given in terms of the original coordinate system.

The largest difference is for the the shear center. The differences will reduce if the wall is made thinner. The configuration specified above is marginally “thin-wall”.

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\4a_summary_example4.7.docx p. 3 of 5 I created another Maple worksheet that is more flexible. It has the option of making the “very thin” assumption or not, so it replaces the last two sheets mentioned above. It is named example_4.7_pathOptionsNew.mw . It was used for the following.

Now let’s cut the wall thickness to .2 and the load by a factor of 100.

w/o the very thin assumption

Very thin asumption

zc .800 .800

EIzz 1.1026E9 1.1025E9

EIyy 1.183E8 1.181E8

qA 10.551 10.577

qB -12.02 -12.02

Shear center Z = -1.028 Z= -1.028

Twist -10.43 -10.43 degrees


Appendix

This example is from A&H. For inhomogeneous beams, A&H have used something called a reference modulus. That is what E1 and all the starred terms are about. I have chosen not to introduce this approach. This difference in notation means that it will probably be confusing for you to use A&H on this example.

4b_example_4.7.mw p. 1 of 5

4b_example_4.7.mw

This worksheet uses the path used in A&H.

Let's determine the shear flow at points "A" and "B" for this "C" channel.

Note: Refer to the sketch on the right to identify the original coordinate system. Original means the coordinate system before we shift to the weighted centroid. Because of the limits on what Maple can doin terms of formatting, we will use "xyz" for the original coordinate system and the shifted one. (i.e. ignore the primes).

It is important to recognize that when one analyzes thin wall sections, there can be disagreement as to the idealization. It is a good idea to be consistent. In particular, the discussion in A&H, which this worksheet follows, does not use a continuous path when defining the integration path. Referring to the the sketch above, the path starts at the top left and goes for 5 inches. It then starts at the top middle of part 2. (I have shown "s" it to the side because of lack of space, but it is actually going down the middle of each section.) The sketch below shows what a continuous path would look like. Ifyou choose this second option, consistency would suggest that you would approximate the geometry and you will obtain a slightly different location for the centroid and section stiffnesses than what is given in this worksheet.

In this configuration, the subregions are rectangular, so the integrals become quite simple. For more


(1.1)(1.1)

(1.3)(1.3)

(1.2)(1.2)

(1)(1)

(1.4)(1.4)

complcated shapes, you might need to actually perform the required integrals rather using the summation formulas. (... or you would use formulas from a table)

Solving this problem requires manipulation of many dimensions and coordinates. You will help yourself a lot if you get a good sketch before you start. Below is an example. The z-coordinate of the centroid of the cross-section is unknown at this point. Once you know it, you can save yourself time (and mistakes) by updating the information. It will be much easier to keep track of your "s" coordinate if you restart it at zero for each section

restart : currentdir ; with linalg : Vy d 1.0e4; Vz d K1.0e4; #Vz d 0:

"C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear"10000.K10000.

Determine the location of the weighted centroid We only need to calculate the z centroid, since symmetry tells us that yc = 0.0.E := 1.00 107, 3.00 107, 1.00 107 ;

1.000000000 107, 3.000000000 107, 1.000000000 107

Ad 2.5, 5.0, 2.52.5, 5.0, 2.5

ySubCentroid d 5.25, 0, K5.255.25, 0, K5.25

zSubCentroidd 2.5, 0.25, 2.52.5, 0.25, 2.5

zcd>i = 1

3

Ei A

i zSubCentroid

i

sum E i $A i , i = 1 ..3;

print `zc = ,̀ zc ; zSubCentroid d evalm zSubCentroidKzc ;


(2.2.3)(2.2.3)

(2.2.1)(2.2.1)

(2.1.2)(2.1.2)

(2.2.2)(2.2.2)

(1.5)(1.5)

(2.1.4)(2.1.4)

(2.1.3)(2.1.3)

(2.1.1)(2.1.1)

0.8125000000zc = , 0.8125000000

1.687500000 K0.5625000000 1.687500000

Calculate the weighted moments of inertia using the extended parallel axis theorem

EIzzbase d vector 3, 5, .5, 5

5 0.5 5height d vector 3, .5, 10, .5

0.5 10 0.5Calculate the weighted moments of inertia relative to the centroid of each regionEIzz0 d vector 3 : for i from 1 to 3 do

EIzz0 i d 112

$E i $ base i $ height i 3;

od: print ÈIzz0 =`, EIzz0

EIzz0 =, 5.208333333 105 1.250000000 109 5.208333333 105

Now use the parallel axis theorem to calculate the weighted moment of inertia for the section#EIzz := sum( EIzz0[i]+ ySubCentroid[i]^2 * E[i]*A[i], i = 1..3);EIzz d 0 :for i from 1 to 3 do EIzz d EIzz C EIzz0 i CySubCentroid i ^2 * E i * A i ; od:print ÈIzz = ,̀ EIzz ;

EIzz = , 2.629166666 109

EIyyReverse the roles of base and heightheight d vector 3, 5, .5, 5

5 0.5 5base d vector 3, .5, 10, .5

0.5 10 0.5Calculate the weighted moments of inertia relative to the centroid of each regionEIyy0 d vector 3 : for i from 1 to 3 do

EIyy0 i d 112


od: print ÈIyy0 =`, EIyy0

EIyy0 =, 5.208333333 107 3.125000000 106 5.208333333 107

Now use the parallel axis theorem to calculate the weighted moment of inertia for the sectionEIyy d 0 :for i from 1 to 3 do


(2.2.4)(2.2.4)

(4.3)(4.3)

(3.3)(3.3)

(3.1)(3.1)

(4.2)(4.2)

(3.2)(3.2)

(4.1)(4.1)

EIyy d EIyy C EIyy0 i CzSubCentroid i ^2 * E i * A i ; od:print `EIyy= `, EIyy ;

EIyy= , 2.971354166 108

Calculate the shear flow at loction "A"

#sbar d 4.5:QEy d E 1 $ 5.25 $ sbar$.5QEz d E 1 $ .5 C 2.25Kzc $ sbar$.5

9.687500000 106 sbarDue to the shear Vy, we obtainq0 d 0 :

qshearY

d q0 K Vy $ QEyEIzz

K99.84152142 sbarDue to the shear Vz, we obtainq0 d 0 :

qshearZ

d q0 K Vz $ QEzEIyy

326.0297985 sbarq12 d q

shearYCq

shearZ = 226.1882771 sbar

Therefore, the shear flow at "A" = eval qshearY

CqshearZ

, sbar = 4.5 = 1017.847247

Calculate the shear flow at loction "B"

We will do this using the path used in A&H.We repeat the calculation we performed for point "A", but extend the integration to the end, i.e. 5 inchessbar d 5.0 :QEy d E 1 $ 5.25 $ sbar$.5

1.312500000 108

QEz d E 1 $ 2.5Kzc $ sbar$.54.218750000 107

Due to the shear Vy, we obtainq0 d 0 :

qshearY

d q0 K Vy $ QEyEIzz

K499.2076071Due to the shear Vz, we obtainq0 d 0 :


(4.5)(4.5)

(5.2)(5.2)

(4.4)(4.4)

> >

(5.3)(5.3)

(4.9)(4.9)

(4.8)(4.8)

(5.4)(5.4)

(4.6)(4.6)

(4.7)(4.7)

(5.1)(5.1)

qshearZ

d q0 K Vz $ QEzEIyy

1419.807187Therefore, the shear flow at the right end of the horizontal piece = qStart d q

shearYCq

shearZ = 920.5995799

Note that his is physically impossible, since this is a free surface, but the approximations in our simple theory + the presence of the vertical piece cause this to be non-zero.

We could just integrate down the vertical member to point B to get the increment and then add "qStart" to the increment. However, it turns out that it is instructive to look at the contributions of Vy and Vz seperately.Let's continue the integration for QEy and QEz.QEy d QEy C E 2 $ 2.5 $ 5.0$.5

3.187500000 108

QEz d QEz C E 2 $ .25Kzc $ 5.0 $ .50.

q0 d 0 :

qshearY

d q0 K Vy $ QEyEIzz

K1212.361332q0 d 0 :

qshearZ

d q0 K Vz $ QEzEIyy

0.q

Bd q

shearYCq

shearZ:

print `The total shear flow at B = ,̀ qB

;

The total shear flow at B =, K1212.361332

Quick calculation of shear centerMust set Vz =0 for this calculation to work out correctly (set it at the top of the file)

Let's sum moments about y=0 and z= middle of vertical member. This is chosen so that we do not have to consider the moment due to the shear flow in the vertical member. The result is sbar d 'sbar':shearFlowMoment d int Kq12 $ 5.25, sbar = 0 ..5.0 $ 2

K29687.21138I cheated a little bit. I knew that the shearflow in the upper and lower horizontal pieces were the same, so I did it for the upper part and multiplied by 2. You should confirm this.The applied force Vy must be applied at a location (i.e. through a particular z-coordinate) such that it creates this moment.In terms of the original coordinate system, the moment due to Vy is -(zsc - .25)*Vy. Hence, we can calculate the z-coordinate as follows: ans d solve K zscK .25 $ Vy = shearFlowMoment, zsc ;

3.218721138The distance from the median line is (in the -z direction)abs K.25 C ans

2.968721138

if Vz s0 then print `Need to re-do this calculation with Vz set to zero` ; fi:

Need to re-do this calculation with Vz set to zero

4c_example_4.7_pathOptionsNew.mw p. 1 of 9

• •

• • 4c_example_4.7_pathOptionsNew.mw

This procedure makes an additional approximation optional... since the wall is assumed to be thin, the variation of the coordinate through the wall thickness is ignored. (negligible). This simplifies automating the calculation.

For example, we can make the following approximation:

If you define thinWallOption = `veryThin`, this approximation will be used.

I also shifted the original coordinate system. The original y now goes through the middle of the vertical piece.

update this sketch... use variables rather than numbers.


(1)(1)

restart : currentdir ; with linalg : with plots : #Digits d 15: #thinWallOption d veryThin;

"C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear"

Define the geometry and loads Vy d 1.0e4; Vz d K1.0e4; #Vyd0: #It is interesting to see what each load contributes to the shear flow#Vz d 0: numPoints d 4 :numSeg d 3 :

YZd 5.25, 4.75 , 5.25, 0 , K5.25, 0 , K5.25, 4.75 : evalm YZ ;conn d 1, 2 , 2, 3 , 3, 4 : evalm conn ; t d .5, .5, .5 :Ed 1.00 107, 3.00 107, 1.00 107 ;

10000.K10000.


(1.1)(1.1)

(2.2)(2.2)

5.25 4.755.25 0K5.25 0K5.25 4.75

1 22 33 4

1.000000000 107, 3.000000000 107, 1.000000000 107

Determine the location of the weighted centroidWe only need to calculate the z centroid, since symmetry tells us that yc = 0.0.L d vector numSeg :yc d 0 : # for this configweightedAread 0 :ZweightedFirstMomentd 0 :for i from 1 to numSeg do printlevel d 0 :p1 d conn i, 1 : p2 d conn i, 2 : y1 d YZ p1, 1 : z1 d YZ p1, 2 : y2 d YZ p2, 1 : z2 d YZ p2, 2 : L i d sqrt y2Ky1 2 C z2Kz1 2 ;print `Coordinates for segment & length= ,̀ y1, z1, y2, z2, L i ;

y d y1 $ 1 Ks

L iCy2$

sL i

:

z d z1 $ 1 Ks

L iCz2$

sL i

:

weightedAread weightedArea C int E i $ t i , s = 0 ..L i ;ZweightedFirstMoment d ZweightedFirstMoment C int z$E i $ t i , s = 0 ..L i ; od;

zc d ZweightedFirstMomentweightedArea

;

print `weightedArea = ,̀ weightedArea ; print `ZweightedFirstMoment = `, ZweightedFirstMoment ; print `zc = ,̀ zc ; #zc C.25;

Coordinates for segment & length= , 5.25, 4.75, 5.25, 0, 4.750000000Coordinates for segment & length= , 5.25, 0, K5.25, 0, 10.50000000

Coordinates for segment & length= , K5.25, 0, K5.25, 4.75, 4.7500000000.5503048780

weightedArea = , 2.050000000 108

ZweightedFirstMoment = , 1.12812500 108

zc = , 0.5503048780

Shift the coordinates to the weighted centroidal axesfor i from 1 to numPoints do YZ i, 1 d YZ i, 1 Kyc : YZ i, 2 d YZ i, 2 K zc : od: evalm YZ ;


(3.1.1)(3.1.1)

(2.1.1)(2.1.1)

5.25 4.1996951225.25 K0.5503048780K5.25 K0.5503048780K5.25 4.199695122

Calculate the weighted moments of inertia

EIzz and EyzEIzzd 0 :EIyy d 0 :for i from 1 to numSeg do printlevel d 0 :p1 d conn i, 1 : p2 d conn i, 2 : y1 d YZ p1, 1 : z1 d YZ p1, 2 : y2 d YZ p2, 1 : z2 d YZ p2, 2 : #print `z1,z2= ,̀ z1, z2 ;

y d y1 $ 1 Ks

L iCy2$

sL i

:

z d z1 $ 1 Ks

L iCz2$

sL i

:

#print `y,z,L=`, y, z, L ;EIzzd EIzz C int y2 $ E i $ t i , s = 0 ..L i ;EIyy d EIyy C int z2 $ E i $ t i , s = 0 ..L i ; #print ̀ contrib=`, int z2 $ E i $ t i , s = 0 ..L i ;

localEI d E i12

$L i $ t i 3 :

if thinWallOption = ̀ veryThin` then localEI d 0.0 : fi:

if y2 = y1 then EIzz dEIzz C localEI : fi: if z2 = z1 then EIyy d EIyy C localEI : fi:

od;print `EIzz = ,̀ EIzz ;print `EIyy = ,̀ EIyy ;

EIzz = , 2.757239584 109

EIyy = , 2.984395644 108

Calculate the shear flow in each segment


(4.1)(4.1)

QEy d vector numSeg :QEz d vector numSeg : sumY d 0 : sumZ d 0 :for i from 1 to numSeg do printlevel d 0 :p1 d conn i, 1 : p2 d conn i, 2 : y1 d YZ p1, 1 : z1 d YZ p1, 2 : y2 d YZ p2, 1 : z2 d YZ p2, 2 :

y d y1 $ 1 Ks

L iCy2$

sL i

:

z d z1 $ 1 Ks

L iCz2$

sL i

:

QEy i d int E i $ y $ t i , s = 0 ..s C sumY;QEz i d int E i $ z $ t i , s = 0 ..s C sumZ; sumY d eval QEy i , s = L i : sumZ d eval QEz i , s = L i : print `sumY, sumZ= ,̀ sumY, sumZ ; od;print `QEy for each segment = ,̀ evalm QEy ; print `QEz for each segment = ,̀ evalm QEz ;

sumY, sumZ=, 1.246875000 108, 4.333650915 107

sumY, sumZ=, 1.246875000 108, K4.333650913 107

sumY, sumZ=, 0., 0.02QEy for each segment = ,

2.6250000 107 s, 7.8750000 107 sK7.500000 106 s2 C1.246875000 108, K2.6250000 107 sC1.246875000 108

QEz for each segment = , 2.099847561 107 sK2.500000 106 s2, K8.254573170 106 sC4.333650915 107, K2.751524390 106 sC2.500000 106 s2 K4.333650913 107

Loop over segments

shearFlow d vector numSeg : p d vector numSeg : offset d 0 : segmentColor d red, green, blue :for i from 1 to numSeg do print `Results for segment ,̀ i ;

q0 d 0 : qshearY

d q0 K Vy $ QEy iEIzz

;

print `Shear flow due to Vy = ,̀ qshearY

;

q0 d 0 : qshearZ

d q0 K Vz $ QEz iEIyy

;

print `Shear flow due to Vz = `, qshearZ

; shearFlow i d q

shearYCq

shearZ;

print `Total shear flow in segment = ,̀ shearFlow i ;


p i d plot sCoffset, shearFlow i , s = 0 ..L i , color = segmentColor i , title = `Shear flow... look at sketch to see meaning of coordinate "s"` , thickness = 3 :

print `offset, length = `, offset, L i ;offset d offset C L i : od: display convert p, list ;

Results for segment , 1Shear flow due to Vy = , K95.20391392 s

Shear flow due to Vz = , 703.6089753 sK83.76905405 s2

Total shear flow in segment = , 608.4050614 sK83.76905405 s2offset, length = , 0, 4.750000000

Results for segment , 2Shear flow due to Vy = , K452.2185911K285.6117417 sC27.20111826 s2

Shear flow due to Vz = , 1452.103351K276.5911144 sTotal shear flow in segment = , 999.8847599K562.2028561 sC27.20111826 s2

offset, length = , 4.750000000, 10.50000000Results for segment , 3

Shear flow due to Vy = , K452.2185911C95.20391391 sShear flow due to Vz = , K1452.103350K92.19703814 sC83.76905405 s2

Total shear flow in segment = , K1904.321941C3.00687577 sC83.76905405 s2offset, length = , 15.25000000, 4.750000000


(4.1.1)(4.1.1)

(4.1.2)(4.1.2)

(4.1.3)(4.1.3)

5 10 15 20

K1500

K1000

K500

0

500

1000

Shear flow... look at sketch to see meaning of coordinate "s"

Maximum value in segments... assumes that there is a zero slope in curve for the segment,which is not always the case. If there is not, tghe extreme value occurs at an endpoint.for i from 1 to numSeg do sMax d solve diff shearFlow i , s = 0, s : print sMax, eval shearFlow i , s = sMax ; od:Check endpoint valuesfor i from 1 to numSeg do print eval shearFlow i , s = 0 , eval shearFlow i , s = L i ; od:

0., 999.884760999.8847599, K1904.321941K1904.321941, 0.000001

Shear flow at points A and Beval shearFlow 1 , s = 4.5

1041.499431eval shearFlow 2 , s = 5.25

K1201.949413

Shear center............ for this calculation we need to go to the top of the


(4.1.1.1)(4.1.1.1)

(4.1.1.4)(4.1.1.4)

(4.1.1.3)(4.1.1.3)

(4.1.1.6)(4.1.1.6)

(4.1.1.2)(4.1.1.2)

worksheet and set Vz=0print(`The shear flows are `, evalm(shearFlow));

The shear flows are , 608.4050614 sK83.76905405 s2, 999.8847599K562.2028561 sC27.20111826 s2, K1904.321941C3.00687577 sC83.76905405 s2

Let's sum moments about y=0 and z= middle of vertical member. This is chosen so that we do not have to consider the moment due to the shear flow in the vertical member. The result is shearFlowMoment d int KshearFlow 1 $ 5.25, s = 0 ..4.75 C int KshearFlow 3 $ 5.25, s = 0 ..4.75

11277.20109The applied force Vy must be applied at a location (i.e. through a particular z-coordinate) suchthat it creates this moment.In terms of the original coordinate system, the moment due to Vy is -(z - .25)*Vy. Hence, we can calculate the z-coordinate as follows:z d 'z':zc d solve K zK t 2

2 $ Vy = shearFlowMoment, z ;

K0.8777201090The distance from the median line is (in the -z direction)

abs Kt 2

2 C zc

1.127720109ZC d K0.8781249960;

K0.8781249960

Calculate the rate of rotation If loads are applied as shown, the beam will twist. This will cause additional shear stresses. Along themiddle of the rectangular segments, the first approximation of the additional shear stress will be zero. However, away from the middle, there will definitely be additional shear stress. Here is a sketch(from notes by Carlos Felippa of U. Colorado) of a rectangular region that gives some feeling for what is going on. The length of each arrow corresponds to the intensity of the shear stress.

For a cross-section made up of thin rectangular regions, the contribution of each rectangular region tothe total torsional stiffness is quite simple. Each rectangular region behaves like it is detached from the rest of the cross-section, so we simply caculate the torsional stiffness of each rectangle and add it


> >

(5.3)(5.3)

(5.2)(5.2)

(5.1)(5.1)

(5.4)(5.4)

up. The tosional stiffness of a rectangle is approximately GJ = G b t33

Once we know this, we proceed in exactly the same way we did for a multicell cross-section, but in this case we have multiple solid rectangles rather than multiple cells.Mx = GJ * kx where kx = rate of rotation and GJ = effective torional stiffnessWe assume a Poisson's ratio of .3

G d E 12 1C .3

, E 22 1C .3

, E 32 1C .3

;

Mx1 d G 1 $ abs YZ 2, 2 KYZ 1, 2 $t 1 3

3 $ kx;

Mx2 d G 2 $ abs YZ 3, 1 KYZ 2, 1 $t 2 3

3$kx;

Mx3 d G 3 $ abs YZ 4, 2 KYZ 3, 2 $t 3 3

3$kx;

Mx d Mx1 C Mx2 CMx3;3.846153846 106, 1.153846154 107, 3.846153846 106

7.612179487 105 kx

5.048076927 106 kx

7.612179487 105 kx

6.570512824 106 kx

What is the applied torque?If the shear forces were applied through the shear center, there would be no twist.Let's calculate equivalent loads that have the shear forces acting through the shear center. When we shift the location of a force, the magnitude of the force does not change, but there is a change in the specified moment. Consider the z-direction force. When we shift it to act through y=0, equivalence demands that both the original and the new load system create the same moment. Let's sum moments about y=0.

moment1 d 'moment_1': moment2 d 'moment2 ':

moment1 d solve Vz $ 5.5 = Vz $ 0 C moment1, moment1 ; K55000.

The y-direction force is being shifted from z=0 to z=zc. Let's calculate moments about z =0. moment2 d solve Vy $0 = KVy $ZC C moment2, moment2 ; appliedTorque d moment1 C moment2;

K8781.249960K63781.24996

ror d solve Mx = appliedTorque, kx ;tipRotation d 120 $ ror ;

tipRotationDegrees d evalftipRotation $ 180.

Pi;

K0.009707195111K1.164863413K66.74175724

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\5a_rectangularBoxBeam_summary.docx p. 1 of 3

Box Beam Analysis This summarizes the results for one version of the single-cell box beam. These calculations make the additional assumption about ignoring the moment of inertia about the centroid of each section. Details are in the Maple worksheet rectangularBoxBeam.mw. w=10 h=10

In what follows, we define qa to be the shear flow at the coordinate location y,z= (-w,-h)

ror = Rate of rotation =

=> Substitute this into the expressions for the shear flow for each segment.

Plot of shear flow… path = 1-2-3-4-1 for the case Mx= -1.0e5 Vy = 1.0e4. The value of Mx is the torque about the center.

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\5a_rectangularBoxBeam_summary.docx p. 2 of 3 Let's calculate the net force in each segment due to the shear flow. The direction of the force is in the direction "s". Here are results for several cases.

Unless one has the right mix of shear and moment, there will be twist.

C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear\5a_rectangularBoxBeam_summary.docx p. 3 of 3 Finally, consider the case that the shear force acts through the shear center (which means the moment is just right to prevent twist) Let's calculate the net force in each segment due to the shear flow. The direction of the force is in the direction "s". Now we see that segment 1 is picking up a lot more load than segment 3, which is expected, since the modulus is 3 times as large.

Don’t forget, positive force is in the direction of “s” not y or z. What if we had multiple cells? What is different?

• For one cell you assign the starting value of the shear flow at some arbitrary point to equal an unknown quantity and then started your integrations. (Later, you solve for this unknown.)

• For multiple cells, you will have to repeat what you did for one cell, which means you will have “n” unknown starting values if there are n cells.

• Calculate the moment due to the shear flow. You cannot use the simple formula involving the enclosed area, since the shear flow is not constant.

• Now you are basically back where you were for just torsion of multi-cell beams: you have “n” unknowns, “n-1” compatibility constraints since the twist rate is the same for all the cells, and one equation that equates the moment due to the shear flow to the total moment.

5b_rectangularBoxBeam_mod.mw p. 1 of 10

(1)(1)

• •

• •

(2)(2)

• •

• • rectangularBoxBeam.mw

This procedure makes an additional approximation... since the wall is assumed to be thin, the variation of the coordinate through the wall thickness is ignored. (i.e. assumed to be negligible). Thissimplifies automating the calculation.

To keep things from getting too messy, we will restrict ourselves to shear force in the y-direction andtorque onlyMake sure the box is symmetric about y=0, since this worksheet only handles lack of symmetryabout z=0. I suspect that if you make this rectangle too elongated or too asymmetrical, the approximate solutionis not very good. For example, if we set the modulus of two adjacent sides to a very low value, I do not expect you to recover the solution for a "right angle" cross-section.

restart : currentdir ; with linalg : with plots : #Digits d 15:"C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear"

Note that the dimensions are defined per the path along the middle of the walls.

Define the geometry and loadsIt is assumed that the shear force is acting through the shear center. At this point I do not know where that is! (Actually, for a homogeneous rectangular cross-section, I do, but in general I would not.) Before I finish, I will calculate the shear center. # Vy d 1.0e4: Mx d 50000: numPoints d 4 :numSeg d 4 : wd 10 : h d 10 :

YZd Kw,Kh , w,Kh , w, h , Kw, h : evalm YZ ;conn d 1, 2 , 2, 3 , 3, 4 , 1, 4 : evalm conn ; t d .5, .5, .5, .5 :


(2.2)(2.2)

(1.1)(1.1)

Ed 3.00 107, 1.00 107, 1.00 107, 1.0e7 : nu d .3 : #Ed 1.00 107, 1.00 107, 1.00 107, 1.0e7 : nu d .3 :

Gd vector 4 : for i from 1 to 4 do

G i d E i2$ 1 Cnu

;

od:K10 K10

10 K1010 10

K10 10

1 22 33 41 4

Determine the location of the weighted centroidWe only need to calculate the z centroid, since symmetry tells us that yc = 0.0.L d vector numSeg :yc d 0 : # for this configweightedAread 0 :ZweightedFirstMomentd 0 :YweightedFirstMomentd 0 :

for i from 1 to numSeg do printlevel d 0 :p1 d conn i, 1 : p2 d conn i, 2 : y1 d YZ p1, 1 : z1 d YZ p1, 2 : y2 d YZ p2, 1 : z2 d YZ p2, 2 : L i d sqrt y2Ky1 2 C z2Kz1 2 ;#print `Coordinates for segment & length= ,̀ y1, z1, y2, z2, L i ;

y d y1 $ 1 Ks

L iCy2$

sL i

:

z d z1 $ 1 Ks

L iCz2$

sL i

:

weightedAread weightedArea C int E i $ t i , s = 0 ..L i ;ZweightedFirstMoment d ZweightedFirstMoment C int z$E i $ t i , s = 0 ..L i ;YweightedFirstMoment d YweightedFirstMoment C int y$E i $ t i , s = 0 ..L i ;

od;

zc d ZweightedFirstMomentweightedArea

:

yc d YweightedFirstMomentweightedArea

:

#print `weightedArea = ,̀ weightedArea ; #print `ZweightedFirstMoment = `, ZweightedFirstMoment ; print `yc, zc = ,̀ yc, zc ;

yc, zc = , 0., K3.333333333


(2.1.1)(2.1.1)

(3.1.1)(3.1.1)

Shift the coordinates to the weighted centroidal axesfor i from 1 to numPoints do YZ i, 1 d YZ i, 1 Kyc : YZ i, 2 d YZ i, 2 K zc : od: evalm YZ ;

K10. K6.66666666710. K6.66666666710. 13.33333333K10. 13.33333333

Calculate the weighted moments of inertia

EIzz and EyzEIzzd 0 :EIyy d 0 :for i from 1 to numSeg do printlevel d 0 :p1 d conn i, 1 : p2 d conn i, 2 : y1 d YZ p1, 1 : z1 d YZ p1, 2 : y2 d YZ p2, 1 : z2 d YZ p2, 2 : #print `z1,z2= ,̀ z1, z2 ;

y d y1 $ 1 Ks

L iCy2$

sL i

:

z d z1 $ 1 Ks

L iCz2$

sL i

:

print `y,z,L=`, y, z, L i ;EIzzd EIzz C int y2 $ E i $ t i , s = 0 ..L i ;EIyy d EIyy C int z2 $ E i $ t i , s = 0 ..L i ; #print ̀ contrib=`, int z2 $ E i $ t i , s = 0 ..L i ;

od;print `EIzz = ,̀ EIzz ;print `EIyy = ,̀ EIyy ;

y,z,L=, K10.C1.000000000 s, K6.666666667, 20y,z,L=, 10., K6.666666667C0.9999999999 s, 20y,z,L=, 10.K1.000000000 s, 13.33333333, 20

y,z,L=, K10., K6.666666667C0.9999999999 s, 20EIzz = , 3.333333333 1010

EIyy = , 3.999999999 1010

Calculate the shear flow in each segmentWe have a little complication. We do not know the shear flow anywhere. Hence, we will simply start somewhere and use a variable (whose value is unknown) to identify the shear flow at that location. Once we calculate all of the shear flows, we can impose equilibrium. For example, if the imposed loads cause Mx and Vy to exist in the beam, then if we integrate to obtain the effects of the the shearflows, we had better recover those values of Mx and Vy.


(4.1)(4.1)

(4.1.1)(4.1.1)

In what follows, we define qa to be the shear flow at the coordinate location y,z= (-w,-h)

QEy d vector numSeg : q0 d qa :shearFlow d vector numSeg : q d vector 4 :for i from 1 to numSeg do #printlevel d 0:p1 d conn i, 1 : p2 d conn i, 2 :y1 d YZ p1, 1 : z1 d YZ p1, 2 :y2 d YZ p2, 1 : z2 d YZ p2, 2 :

y d y1 $ 1 Ks

L iCy2$

sL i

:

z d z1 $ 1 Ks

L iCz2$

sL i

:

QEy i d int E i $ y $ t i , s = 0 ..s ;

q i d q0 K Vy $ QEy iEIzz

;

q0 d eval q i , s = L i ; print `Segment ,̀ i, ̀ : q = ,̀ q i ; od;

qSave d copy q :

Segment, 1, : q = , qa K3.000000000 10-11 Vy K1.50000000 108 sC7.500000 106 s2

Segment, 2, : q = , qa K0.001500000000 Vy sSegment, 3, : q = , qa K0.03000000000 VyK3.000000000 10-11 Vy 5.0000000 107 sK2.500000 106 s2

Segment, 4, : q = , qa K0.03000000000 VyC0.001500000000 Vy s

Solve for z-coordinate of shear center (see procedure used in next section--- more straightforward)

rotation rate

ror d 12$4$h$w

$ intq 1

t 1 $G 1, s = 0 ..2$w C int

q 2t 2 $G 2

, s = 0 ..2$h C intq 3

t 3 $G 3, s = 0 ..2$w

C intq 4

t 4 $G 4, s = 0 ..2$h ;

K7.800000000 10-10 VyC4.333333334 10-8 qatorqueAboutCenter d int h $ q 1 , s = 0 ..2$w C int w $ q 2 , s = 0 ..2$h C int h $ q 3 , s = 0 ..2$w C int w

$q 4 , s = 0 ..2$h ;qaAns d solve Mx = torqueAboutCenter, qa : print `qa = ,̀ qa ;q d map eval, q, qa = qaAns :


(4.2.2)(4.2.2)

(4.1.3)(4.1.3)

(4.1.2)(4.1.2)

(4.2.1)(4.2.1)

(4.2.3)(4.2.3)

print `The shear flows are ,̀ evalm q ;eval ror, qa = qaAns ;

K10. VyC800. qaqa = , qa

The shear flows are , 0.01250000000 VyC0.001250000000 MxK3.000000000 10-11 Vy K1.50000000 108 sC7.500000 106 s2 , 0.01250000000 VyC0.001250000000 MxK0.001500000000 Vy s, K0.01750000000 VyC0.001250000000 MxK3.000000000 10-11 Vy 5.0000000 107 sK2.500000 106 s2 , K0.01750000000 VyC0.001250000000 MxC0.001500000000 Vy s

K2.383333332 10-10 VyC5.416666668 10-11 MxIf I want the rate of rotation to be zero, this equation tells me that I will need to apply both a moment and a shear force. Let's offset the shear force from the weighted centroid. Apply it at a distance "e" from the weighted centroid (inthe z-direction).The moment about the weighted centroid moment d solve eval ror, qa = qaAns = 0, Mx ; solve KVy$e = moment, e ;

4.399999996 VyK4.399999996

Solve for z-coordinate of shear center... calculated relative ot the original coordinate system

rotation rate

q d copy qSave :

ror d 12$4$h$w

$ intq 1

t 1 $G 1, s = 0 ..2$w C int

q 2t 2 $G 2

, s = 0 ..2$h C intq 3

t 3 $G 3, s = 0 ..2$w

C intq 4

t 4 $G 4, s = 0 ..2$h ;

K7.800000000 10-10 VyC4.333333334 10-8 qaqaAns d solve ror = 0, qa : print `qa = ,̀ qaAns ;q d map eval, q, qa = qaAns :print `The shear flows are ,̀ evalm q ;eval ror, qa = qaAns ; torqueAboutCenter d int h $ q 1 , s = 0 ..2$w C int w $ q 2 , s = 0 ..2$h C int h $ q 3 , s = 0 ..2$w C int w

$q 4 , s = 0 ..2$h : print `This shear flow gives a moment about the center = `, torqueAboutCenter ;

qa = , 0.01800000000 Vy

The shear flows are , 0.01800000000 VyK3.000000000 10-11 Vy K1.50000000 108 sC7.500000 106 s2 , 0.01800000000 VyK0.001500000000 Vy s, K0.01200000000 VyK3.000000000 10-11 Vy 5.0000000 107 sK2.500000 106 s2 , K0.01200000000 VyC0.001500000000 Vy s

1. 10-19 VyThis shear flow gives a moment about the center = , 4.400000000 Vy

This tells me that in order to impose a zero rotation, I will need to impose a moment as indicated.zsc d solve KVy $ zsc = torqueAboutCenter, zsc : print `z-coordinate of shear center = ,̀ zsc ;

z-coordinate of shear center = , K4.400000000


(4.3.1)(4.3.1)

> >

Plot for specified Mx and VyNote that the moment is the "total moment". If we impose Vy alone, there would be rotation.

q d copy qSave : torqueAboutCenter d int h $ q 1 , s = 0 ..2$w C int w $ q 2 , s = 0 ..2$h C int h $ q 3 , s = 0 ..2$w

C int w$q 4 , s = 0 ..2$h :qaAns d solve torqueAboutCenter = Mx, qa ;q d map eval, q, qa = qaAns ;

qaAns := 0.01250000000 VyC0.001250000000 Mxq := 0.01250000000 VyC0.001250000000 MxK3.000000000 10-11 Vy K1.50000000 108 s

C7.500000 106 s2 , 0.01250000000 VyC0.001250000000 MxK0.001500000000 Vy s, K0.01750000000 VyC0.001250000000 MxK3.000000000 10-11 Vy 5.0000000 107 sK2.500000 106 s2 , K0.01750000000 VyC0.001250000000 MxC0.001500000000 Vy s

offset d 0 :for i from 1 to numSeg do p i d plot sCoffset, eval q i , Mx = 1.0e5, Vy = 1.0e4 , s = 0 ..L i , title

= `Shear flow... look at sketch to see meaning of coordinate "s"` , thickness = 3 :

#print `offset, length = `, offset, L i ;offset d offset C L i : od: display convert p, list ;

10 20 30 40 50 60 70 80

K100

0

100

200

300

400



(4.3.2)(4.3.2)offset d 0 :for i from 1 to numSeg do p i d plot sCoffset, eval q i , Mx = K1.0e5, Vy = 1.0e4 , s = 0 ..L i , title

= `Shear flow... look at sketch to see meaning of coordinate "s"` , thickness = 3 :

#print `offset, length = `, offset, L i ;offset d offset C L i : od: display convert p, list ;

10 20 30 40 50 60 70 80

K300

K200

K100

0

100

200


Let's calculate the net force in each segment due to the shear flow. The direction of the force is in the direction "s".L d 2$w, 2$h, 2$w, 2$h : print ` Mx= K1.0e5, Vy= 1.0e4 ` ; for i from 1 to 4 do force d int q i , s = 0 ..L i : print `Force for segment ,̀ i, eval force, Mx = K1.0e5, Vy = 1.0e4 ; od: print ; print ` Mx= 0, Vy= 1.0e4 ` ; for i from 1 to 4 do force d int q i , s = 0 ..L i : print `Force for segment ,̀ i, eval force, Mx = 0, Vy = 1.0e4 ; od:


> >

(4.4.4)(4.4.4)

> >

(4.4.3)(4.4.3)

(4.4.2)(4.4.2)

> >

> >

(4.3.3)(4.3.3)

> >

(4.4.1)(4.4.1)

print ; print ` Mx= -1.0e5, Vy= 0.0 ` ; for i from 1 to 4 do force d int q i , s = 0 ..L i : print `Force for segment ,̀ i, eval force, Mx =K1.0e5, Vy = 0.0 ; od:

Mx= -1.0e5, Vy= 1.0e4 Force for segment , 1, 3000.000000

Force for segment , 2, K3000.000000Force for segment , 3, K7000.000000Force for segment , 4, K3000.000000

Mx= 0, Vy= 1.0e4 Force for segment , 1, 5500.000000

Force for segment , 2, K500.0000000Force for segment , 3, K4500.000000Force for segment , 4, K500.0000000

Mx= -1.0e5, Vy= 0.0 Force for segment , 1, K2500.000000Force for segment , 2, K2500.000000Force for segment , 3, K2500.000000Force for segment , 4, K2500.000000

More thoughtsq d copy qSave : torqueAboutCenter d int h $ q 1 , s = 0 ..2$w C int w $ q 2 , s = 0 ..2$h C int h $ q 3 , s = 0 ..2$w

C int w$q 4 , s = 0 ..2$h ;netFyd int q 1 , s = 0 ..2$w K int q 3 , s = 0 ..2$w ;

torqueAboutCenter := K10. VyC800. qanetFy := 1.000000000 Vy

If we apply just a shear force that acts through the center, then the torque about the center is equal to zero. Let's consider that.

qaAns d solve torqueAboutCenter = 0, qa ;q d map eval, q, qa = qaAns ;

qaAns := 0.01250000000 Vyq := 0.01250000000 VyK3.000000000 10-11 Vy K1.50000000 108 sC7.500000 106 s2 , 0.01250000000 Vy

K0.001500000000 Vy s, K0.01750000000 VyK3.000000000 10-11 Vy 5.0000000 107 sK2.500000 106 s2 , K0.01750000000 VyC0.001500000000 Vy s

torqueAboutCenter d int h $ q 1 , s = 0 ..2$w C int w $ q 2 , s = 0 ..2$h C int h $ q 3 , s = 0 ..2$w C int w$q 4 , s = 0 ..2$h ;

netFyd int q 1 , s = 0 ..2$w K int q 3 , s = 0 ..2$w ;torqueAboutCenter := 0.netFy := 1.000000000 Vy

ror d 12$4$h$w

$ intq 1

t 1 $G 1, s = 0 ..2$w C int

q 2t 2 $G 2

, s = 0 ..2$h C intq 3

t 3 $G 3, s = 0

..2$w C intq 4

t 4 $G 4, s = 0 ..2$h :

print `The rate of rotation = `, ror ; The rate of rotation = , K2.383333334 10-10 Vy


(4.5.3)(4.5.3)

(4.5.1)(4.5.1)

(4.5.2)(4.5.2)

Even more thoughtsWhat are the forces in each segment if the shear force acts through the shear center (there will be just the right amount of moment to have the rate of rotation = 0).q d copy qSave :

ror d 12$4$h$w

$ intq 1

t 1 $G 1, s = 0 ..2$w C int

q 2t 2 $G 2

, s = 0 ..2$h C intq 3

t 3 $G 3, s = 0 ..2$w

C intq 4

t 4 $G 4, s = 0 ..2$h ;

K7.800000000 10-10 VyC4.333333334 10-8 qaqaAns d solve ror = 0, qa : print `qa = ,̀ qaAns ;q d map eval, q, qa = qaAns :print `The shear flows are ,̀ evalm q ;eval ror, qa = qaAns ; tmp1 d int h $ q 1 , s = 0 ..2$w : tmp2 d int w $ q 2 , s = 0 ..2$h ; tmp3 d int h $ q 3 , s = 0 ..2$w : tmp4 d int w$q 4 , s = 0 ..2$h :

# torqueAboutCenter d int h $ q 1 , s = 0 ..2$w C int w $ q 2 , s = 0 ..2$h C int h $ q 3 , s = 0 ..2$w C int w$q 4 , s = 0 ..2$h :

torqueAboutCenter d tmp1 C tmp2 C tmp3 C tmp4 :

print `This shear flow gives a moment about the center = `, torqueAboutCenter ; print `Contributions of each segment` print tmp1 ; print tmp2 ; print tmp3 ; print tmp4 ;

qa = , 0.01800000000 Vy

The shear flows are , 0.01800000000 VyK3.000000000 10-11 Vy K1.50000000 108 sC7.500000 106 s2 , 0.01800000000 VyK0.001500000000 Vy s, K0.01200000000 VyK3.000000000 10-11 Vy 5.0000000 107 sK2.500000 106 s2 , K0.01200000000 VyC0.001500000000 Vy s

1. 10-19 Vy0.6000000000 Vy

This shear flow gives a moment about the center = , 4.400000000 VyContributions of each segment

6.600000000 Vy2

0.6000000000 VyK3.400000000 Vy0.6000000000 Vy

for i from 1 to 4 do force d int q i , s = 0 ..L i : print `Force for segment ,̀ i, eval force, Vy = 1.0e4 ; od:

Force for segment , 1, 6600.000000Force for segment , 2, 600.0000000

Force for segment , 3, K3400.000000Force for segment , 4, 600.0000000


(4.5.4)(4.5.4)> >

> >

evalm qSave ;qa K3.000000000 10-11 Vy K1.50000000 108 sC7.500000 106 s2 , qa K0.001500000000 Vy s, qa

K0.03000000000 VyK3.000000000 10-11 Vy 5.0000000 107 sK2.500000 106 s2 , qaK0.03000000000 VyC0.001500000000 Vy s

10_rightAngle_4.17_broken.mw p. 1 of 6

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• • • •

• •

• •

• •

• •

• •

> >

• •

• •

problem_4.17.mw-----calculations are wrong--- this is more complicated than it seems

Need a better sketch! Had to do some guessing to figure out where the original origin is located.

A&H solution is wrong

Summary of what we need to do and calculateweighted centroid (since it is homogeneous, the centroid is the same)shift coordinate systemsection properties EIyy, EIzz, EIyz relative to new coordinate systemstatically equivalent loads that act through the weighted centroidMy(x), Mz(x) using FBD or by integrating the equilibrium equations in terms of stress resultantscurvatures diff(v,x,x) and -diff(w,x,x) using the structural constitutive equations (will need toinvert a matrix, since the constitutive matrix is not diagonal)axial strainaxial stressmaximum axial stressdisplacements v0(100) and w0(100) Note: This beam is also going to twist, but based on our simplifed theory, that will not affect the displacements v and w on the x-axis. Also, note that the weighted centroid is not located on the beam!


4. 4.

6. 6.

3. 3.

5. 5.

7. 7.

1. 1.

2. 2.

Summary with numbers (there are errors!)The following has the numbers taken straight from the solutions manual. The steps are correct, but there are errors. You will rework this problem for HW and use the correct numbers.

weighted centroid (since it is homogeneous, the centroid is the same)zbar = -1.18421 ybar = 1.18421shift coordinate systemYou might wish to draw a sketch with the shifted coordinate system. The alternative is to remember to shift coordinates whenever they are used in subsequent calculations.section properties EIyy, EIzz, EIyz relative to new coordinate systemIyy = Izz = 11.25 Iyz = 6.66statically equivalent loads that act through the weighted centroid

My(x), Mz(x) using FBD or by integrating the equilibrium equations in terms of stress resultants

curvatures diff(v,x,x) and -diff(w,x,x) using the structural constitutive equations (will need to invert a matrix, since the constitutive matrix is not diagonal)

axial strain


(4.2)(4.2)

9. 9.

(4.1)(4.1)

(4.3)(4.3)

> >

> >

8. 8.

(1)(1)

10. 10.

axial stressmultiply strain with modulus (no thermal load)maximum axial stressLook for where the axial strain is largest. => Look at how the curvatures vary. In the absence of thermal loads, you know the strains are largest on the outside of the beam. You might wantto plot the variation of strain with x.

displacements v0(100) and w0(100) Note: This beam is also going to twist, but based on our simplifed theory, that will not affect the displacements v and w on the x-axis. Also, note that the weighted centroid is not located on the beam!

You know the curvatures. Integrate them to obtain expressions for v0(x) and w0(x). These expressions will have integration constants. Use the kinematic BC's to determine the integration constants. Now you have the functions v0(x) and w0(x).

Now let's do our own calculation

restart : currentdir ; with linalg : Digits d 15 : # Vy d 1.0e4; Vz d K1.0e4;

"C:\W\whit\Classes\304_2012_ver_3\_Notes\5_Torsion_and_transverseShear"

Determine the location of the weighted centroidWe only need to calculate the z centroid, since symmetry tells us that yc = 0.0.E := 1.00 107, 1.00 107 ;

1.000000000 107, 1.000000000 107

Ad 4.75 $ .5, 4.75 $ .5 ; A d 2.5, 2.5 ;

2.375, 2.3752.5, 2.5

If we do they way in A&HySubCentroid d K2.25, 0 ; zSubCentroidd 0. , 2.25

K2.25, 00., 2.25


(4.4)(4.4)

(5.1.3)(5.1.3)

(5.1.4)(5.1.4)

(4.5)(4.5)

(5.1.1)(5.1.1)

(5.1.2)(5.1.2)

#ySubCentroid d K2.5 C .25, 2.5K 4.752

;

#zSubCentroid d K2.5 C 4.752

, 2.5K .25

>i = 1

2

Ei A

i zSubCentroid

i;

5.625000000000 107

zcd>i = 1

2

Ei A

i zSubCentroid

i

sum E i $A i , i = 1 ..2;

print `zc = ,̀ zc ; zSubCentroid d evalm zSubCentroidKzc ;

1.12500000000000zc = , 1.12500000000000

K1.12500000000000 1.12500000000000

Calculate the weighted moments of inertia---------- change this to use Maple procedure

EIzzbase d vector 3, 4.75, .5, 4.75

4.75 0.5 4.75height d vector 3, .5, 10.5 , .5

0.5 10.5 0.5Calculate the weighted moments of inertia relative to the centroid of each regionEIzz0 d vector 3 : for i from 1 to 3 do

EIzz0 i d 112


od: print `EIzz0 =`, EIzz0

Error, invalid subscript selector

EIzz0 =, 4.94791666666667 105 4.82343750000000 108 EIzz03Now use the parallel axis theorem to calculate the weighted moment of inertia for the section#EIzz := sum( EIzz0[i]+ ySubCentroid[i]^2 * E[i]*A[i], i = 1..3);EIzz d 0 :for i from 1 to 3 do EIzz d EIzz C EIzz0 i CySubCentroid i ^2 * E i * A i ; od:Error, invalid subscript selectorprint `EIzz = ,̀ EIzz ;

EIzz = , 6.09401041666667 108


(6.2)(6.2)

(6.1)(6.1)

(5.2.3)(5.2.3)

(5.2.4)(5.2.4)

(5.2.1)(5.2.1)

(5.2.2)(5.2.2)

EIyyReverse the roles of base and heightheight d vector 3, 4.75, .5, 4.75

4.75 0.5 4.75base d vector 3, .5, 10.5 , .5

0.5 10.5 0.5

Calculate the weighted moments of inertia relative to the centroid of each regionEIyy0 d vector 3 : for i from 1 to 3 do

EIyy0 i d 112


od: print `EIyy0 =`, EIyy0

Error, invalid subscript selector

EIyy0 =, 4.46549479166667 107 1.09375000000000 106 EIyy03Now use the parallel axis theorem to calculate the weighted moment of inertia for the sectionEIyy d 0 :for i from 1 to 3 do EIyy d EIyy C EIyy0 i CzSubCentroid i ^2 * E i * A i ; od:

print `EIyy= `, EIyy ;EIyy= , 1.09029947916667 108

EIyz

Calculate the shear flow top horizontal pieceI will use the integral formula rather than the summation.

t d .5 : y d 5.25 :QEy d int E 1 $ y $ t , s = 0 ..s ;QEz d int E 1 $ 5 KsKzc $ t, s = 0 ..s ;

1.9375000 107 sK2.500000 106 s2Due to the shear Vy, we obtainq0 d 0 :

qshearY

d q0 K Vy $ QEyEIzz

K0.0430750822614418 Vy sDue to the shear Vz, we obtainq0 d 0 :

qshearZ

d q0 K Vz $ QEzEIyy


(7.3)(7.3)

(7.5)(7.5)

(6.5)(6.5)

(6.3)(6.3)

(7.2)(7.2)

(7.4)(7.4)

(7.1)(7.1)

(7.6)(7.6)

(6.4)(6.4)

(7.7)(7.7)

K9.17179196273957 10-9 Vz 1.9375000 107 sK2.500000 106 s2Therefore, the shear flow at "A" = shearFlowd q

shearYCq

shearZ

K0.0430750822614418 Vy sK9.17179196273957 10-9 Vz 1.9375000 107 sK2.500000 106 s2

At point "A", s= 4.5. eval shearFlow, s = 4.5

K0.193837870176488 VyK0.335343643637666 Vz

Calculate the shear flow in vertical memberI will use the integral formula rather than the summation.

Start by calculating the values at the beginning of the vertical integration.QEy0d eval QEy, s = 4.75

1.2468750000 108

QEz0 d eval QEz, s = 4.753.56250000000 107

Let's continue the integration for QEy and QEz.QEy d QEy0 C int E 2 $ 5.25Ks $t, s = 0 ..s ;QEy

B d eval QEy, s = 5.25 ;

1.2468750000 108 C2.6250000 107 sK2.500000 106 s2

1.935937500000 108

QEz d QEz0 C int E 2 $ .25Kzc $t, s = 0 ..s ;QEz

B d eval QEz, s = 5.25 ;

3.56250000000 107 K4.375000 106 s1.26562500000 107

At point B q0 d 0 :

qshearY

d q0 K Vy $ QEy

B

EIzzK0.317678731678133 Vy

q0 d 0 :

qshearZ

d q0 K Vz $ QEz

B

EIyyK0.116080492028423 Vz

qBd q

shearYCq

shearZ:

print `The total shear flow at B = ,̀ qB

;

The total shear flow at B =, K0.317678731678133 VyK0.116080492028423 Vz

chapter 5 torsion and transverse shear of thin-walled...

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