MITx 2.091x

Can anyone help me on this question:

HW9_3: Design against buckling in a SD truss

We are considering again the truss in the figure, from homework problem HW4_2. The truss is loaded at joint C as indicated, and we had determined the axial forces in each of the bars using the method of joints.

From the solution to HW4_2, we had obtained:

N(AB)=2W
N(BC)=W
N(BD)=W
N(BE)=−2√W
N(CD)=−2√W
N(ED)=−W

Now you can take the magnitude of the applied load to be W=2 kN and the length L=4 m . Assume that the bars of the truss have a circular cross section and are made of steel, with a Young's modulus E=210 GPa and a failure strength (in tension and compression)
σ(f)=250 MPa

HW9_3_1
(24 points possible)

If you want a safety factor, SF=4 against both failure of the material, due to stresses that exceed its
failure strength, and collapse of the truss due to buckling of bars under compressive loading, obtain the numeric values, in mm, of the minimum radius of each of the bars of the truss.

r(AB) = not answered (in mm)
r(BC)= not answered (in mm)
r(BD) = not answered (in mm)
r(BE) = not answered (in mm)
r(CD) = not answered (in mm)
r(ED) = not answered (in mm)

If you tried to copy and paste the drawing or sketch of this truss here, it did not work because copy and paste is not allowed here.

From what you said, moment of inertia of the cross section must be increased or length of section reduced assuming you are doing Euler column bucking. As the question implies be careful going up on the diameter to avoid buckling and getting into simple excessive stress failure.
I am not about to get into MITx questions though. When the 8:01x students discovered this site we were swamped and the questions were clearly exam questions. As a graduate and former faculty member at MIT I was very unhappy with the performance.

Its fine. Damon

Here is the image though.

courses.edx/c4x/MITx/2.01x/asset/images_HW9_3_1.png

just add (dot)org after "courses.edx" in order to view the pictures of the truss.

Ltb

To determine the minimum radius of each bar in the truss, we need to consider both the failure of the material due to stress exceeding its failure strength and the collapse of the truss due to buckling of bars under compressive loading. The safety factor is given as SF = 4.

To calculate the minimum radius, we can use the Euler's buckling formula which is given as:

P_critical = (π^2 * E * I) / (L^2)

Where P_critical is the critical compressive load, E is the Young's modulus, I is the moment of inertia of the cross-section, and L is the length of the bar.

For a circular cross-section, the moment of inertia is given as:

I = (π * r^4) / 4

Where r is the radius of the cross-section.

Now, let's calculate the minimum radius for each bar:

1. For bar AB:
Using the given axial force N(AB) = 2W and the applied load W = 2 kN, we know that N(AB) = 4 kN.
We can calculate the critical compressive load as:
P_critical = N(AB) / SF = 4 kN / 4 = 1 kN
Substituting the values into the Euler's buckling formula:
1 kN = (π^2 * 210 GPa * (π * r^4) / 4) / (4 m)^2
Simplifying the equation, we can solve for r.

2. Similarly, we can repeat the above steps for bars BC, BD, BE, CD, and ED to calculate their minimum radii.

It is important to note that without the given values for the axial forces and load, it is not possible to provide the exact numeric values of the minimum radii for each bar. Make sure to refer to the solution or the previous problem to obtain the values of N(AB), N(BC), N(BD), N(BE), N(CD), and N(ED).