Tech III HW 4 Beam Design & Multiframe Homework Solutions

UNIVERSITY of MARYLAND

School of Architecture, Planning and Preservation

ARCH 412Technology III

Prof. Deborah Oakley

Beam Design (i.e., Selection) Problems

For all problems, consider loads to act only on the major axis of the section and neglect the

member self weight. All members are simply supported.

1. A wood floor is supported by members of Southern Pine No. 2 spaced at 24 o.c. and

spanning 14. The total floor load is 40 psf. Select the size required based on flexural stress

and check shear stress.

2. Select an A36 steel beam to carry a total uniform load of 2500 lbs/ft spanning 30. Size for

flexural stress and check shear stress.

3. Select an A572 steel girder to carry two concentrated loads of 40 kips each located at

the 1/3 points for a span of 36 feet. Size for flexural stress and check shear stress.

Multiframe Problems

For all problems, provide screen captures from the Multiframe plot window.

4. For the beam below, what are the bending and shear stresses at the point where the 3.2

k/ft load stops? Where is the point of maximum moment? Verify this with manual

calculations by cutting a section at this location. Select an A36 steel beam to carry this

force for flexural and shear stresses.

5. What is the optimal length of the overhanging portion of a propped cantilever beam

with a main span of 23 feet carrying a uniform load of 2 k/ft? Express this as a ratio of the

supported end (by length). What makes this an optimal length?

6. Consider a two span continuous beam of 30 length on each span and compare this

with creating two simply supported spans by releasing their ends. Create two load cases,

the first with uniform load of 3 k/ft on each span, the second with 3 k/ft load on the left

span and 1.5 k/ft on the right span. Which case produces the worst condition? Select

A36 steel members for each case.

7. Compare the flexural and shear stresses and deflections of a W12x14 member that spans

14 and carries a uniform load of 1 k/ft with a square steel bar 2.1 x 2.1 of the same

span and loading. What are the stresses and deflections for each? How far can the bar

shape safely span? (Note: If you dont have that section in Custom 2 shapes, create it

by going to Edit Sections Add Standard Section and enter the length and width

for a flat bar type shape.)

8. Go to the Kibel Gallery. What is the name of the multi-use hall located in Katowice,

Poland? What is the name of its unique structural principal?

27ft

12ft 15 ft

3.2 k/ft

ARCH 412 Technology III Beam Selection & Multiframe Homework

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Problem #4, Multiframe Plot. Note that to make an analysis of the loading, you must first

select a trial beam. Since the moment and shear are independent of the beam

selected (if neglecting the beam self weight), then any selection will be fine. Actual

beam sizing will then proceed from the resultant moments and shears.

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Point of zero

shear and Mmax

10.76


#5) Propped Cantilever (ie., Overhanging Beam), Optimal Length

Look at this one in terms of the beam deflections. For a 23 long simple span beam with a load of

2 k/ft, a maximum moment of 132 k-ft (neglecting beam weight) requires an A36 beam size of

W16x45. This beam has a maximum center deflection of 0.74.

Following are three different trials using this beam section. The first with an 6 cantilever end, a

second with a 12 cantilever end, and the last with a 10 cantilever end:

With a 6 Cantilever End:

Note the deflection of the free end is upwards at 0.42 and the maximum center span

deflection is 0.62, which is slightly less than the 0.74 for a simple span condition.

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23 6


With a 12 Cantilever End:

Now the deflection of the free end is downwards at 0.64 and the main span deflection is

reduced to 0.27

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23 12


Going back and forth we can zero in on a length that produces zero deflection on the free end,

while also reducing the amount of deflection on the main span:

With a cantilever end of 10, we see the following:

With a cantilever end of 10, the deflection is virtually zero, and the main span deflection is still

reduced to 0.41 versus 0.62 for the first case.

We may conclude a general rule of thumb by expressing this as a % of the total beam

length. 10/33 = 0.30. Or, to put it another way, if we keep the cantilever span at no

more than 1/3 of the overall length, well optimize the behavior of the overall beam.

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23 10


#6) Multiframe flexural stress results for a two span beam, 30 long each span. Controlling case is

with 3k/ft load on each span. Maximum moment occurs at center support is 337.5 K-ft. This is a

negative moment resulting in top tension on the beam. A W2x84 is selected for this condition.

Load Case 1: 3 k/ft on each span (flexural stress). Blue line represents deflected shape. Max.

deflection is 0.33:

Load Case 2: 1.5 k/ft on left span & 3 k/ft on right span (flexural stress). Notice that while the

positive moment (top compression) on the left span and negative moment at the center

support (top tension) have both decreased, the positive moment on the right span has

increased. The load on the left side had been acting to counterbalance the load on the right

span in load case 1. Deflection on the right span has also increased to 0.44:

The same span condition and same member size (W24x84), but with the ends free to rotate

(end releases in Multiframe). Note that now there is no negative moment at the center

support, which is always the case where a member has a pinned connection. Maximum positive

moment (top compression) yields a bending stress of 20.66 ksi, which is the same as the negative

bending stress at the center support for Load Case 1 of the continuous beam above. The

maximum deflection, however has increased to 0.8

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#7) Multiframe results for a 14 long W12x14 steel member with 1k/ft uniform load:

Maximum stress is 19.7 ksi

Maximum Deflection is 0.34 or (14ftx12in/ft)/0.34in = 494.1 or l/494 of span

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Results for same span with a 2.1 x2.1 square bar

Note that the maximum stress has now skyrocketed to 190.5

The deflection is an impossibly large 18.4 or (14ftx12in/ft)/18.4in = 9.13 or l/9 of span!

Each of these members has a cross sectional area of roughly 4.2in2 The moment of inertia for the W12x14, though, is 88.6 in4 versus the 2.1 square bar with an

Ix = (2.14)/12 = 1.62 in4. The difference in Ix is 88.6 / 1.62 = 54.7, so thats how much stiffer the

W12x14 is! If we multiply the deflection of the W12x14 by this amount, we get 0.34in x 54.7 = 18.6,

which corresponds closely (allowing for rounding errors) to the deflection of the 2.1 square bar.

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Tech III HW 4 Beam Design & Multiframe Homework Solutions