Breaking a Plain Concrete Beam

Outline

1. Introduction
   
2. Shear Strength
   
3. Loads that Shear the Beam
   
4. Bending Strength
   
5. Loads that Break the Beam in Bending
   
6. Summary Loads
   
7. Safe Loads
   
8. Conclusion
   
9. References
   

1. Introduction

In our earlier lessons (here and here) we developed diagrams and expressions for the load on our plain concrete `Make and Break' beam. Actually, the expressions are suitable for a reinforced beam, also, and beams of other dimensions, except that we would need to tweak the self-weight a bit. And the methods could be used generally for beams in general, both in the lab, and in practice (actual structures).

From those lessons we came away with expressions for internal loads (shear and moment) in terms of some uniformly distributed applied load, and a concentrated applied load at mid-span. We came up with those expressions in terms of `exact' conditions for testing in the lab, and design, or future conditions, that contain factors of safety for uncertainty. And those expressions are shown below.

At the `exact' (lab) `level' ...

V = V _{due to sw} + V _{due to uniform load} + V _{due to conc load P}

V _max = 155 lb + 2.58 ω _LL + P / 2 .

M = M sw_{M sw} + M _{uniform load} + M _P

M _max = 237 lb-ft + 4.5 ω _LL ft ² + 1.5 P .

At the `design' level (factored for uncertainty in future events) ...

V _{u, max} = 199 lb + 4.13 ω _LL ft + 0.8 P

M _u = 284 lb-ft + 7.2 ω _LL ft² + 2.4 P ft.

Now let's solve for the actual numbers ... the uniform live load, ω _LL, on the beam to break it, and the concentrated load, P, at mid-span, to exactly break it. We will do this by determining the strength of the beam, and setting the strength equal to the load (load expressions above). We don't necessarily know whether the beam will break in shear, or bending, so we will check both. And, we will also calculate `safe' loads (ω _LL or P), say, if we want to use this beam for some future use. (But, note, in real life we won't ever use plain concrete in a beam ... but it makes for illustration.)

2. Shear Strength

We will now calculate the loads to shear the beam. We will look at two cases: 1) the uniform load ω _LL, say made up of students standing on the beam, and 2) the concentrated load P. We will take these loads to not act concurrently. To do this we must first calculate the shear strength of our plain concrete beam.

From Chapter 22 of the ACI CodVn

V_n = (4/3) √f '_c b h

WherVn

V_n is the `nominal' (what I call `perfect world') shear strength,

f '_c is the specified concrete strength,

b is the beam width, and

h is the beam depth.

We do not yet know the concrete strength of our beam, but let's say (use) 3500 psi.

SVn

V_n = (4/3) √3500 psi (5 in.)(11 in.) Vn

V_n = 4338 lb.

3. Loads that Shear the Beam

Now we can set load equal to strength, and solve ...

For the uniform live load ...

Load = Strength and solve for ω _LL ...

V _max = 155 lb + 2.58 ω _LL = = 4338 lb ... (note that we let P = 0 and it went away).

... ω _LL = 1621 lb/ft.

Wow, this is a lot. If an average college student weighs 150 lb, then this corresponds to ...

1621 lb/ft divided by 150 lb/college student = 10.8 college students per foot of beam ... (pretty crowded).

Since we are loading 6 ft of beam, that would be W = ω L = 10.8 students / ft (6 ft) = 65 students ... (the entire class!).

For the concentrated load P ...

V _max = 155 lb + P / 2 = 4338 lb ... (where the uniform load term is not used)

Solving for P ... P = 8366 lb.

Note that 65 students are going to weigh ... W = 65 (150) = 9750 lb.

So, the load is about the same, for shear, whether concentrated or distributed.

If the beam doesn't break first,flexurexure, which we will now look at.

Flexurexure Strength

For Chapter 22 of the ACI Code ...

M_n = 5 √f '_c S ...

Where,

M_n is the `nominal' (perfect world) bendflexureexure) strength,

`5' is a `number' ... that goes along with the √f '_c to give us an extreme fiber rupture strength. The value 5 we could say is at the conservative end of a range of values from, say, 5 to 7.5. Once when we added fiber reinforcement it seemed we were getting numbers closer to 7.5. But, we'll use the `5'. `5' is `the Code' (and we always obey the Code).

√f '_c ... is our specified concrete strength, and

S ... is the section modulus, in this bhse bh² / 6.

Using 3500 psi for √f '_c and our dimensions of 5 in. and 11 in. ...

S = (11 in.)(5 in.)² / 6 = 45.8 in.³.

(Remember, we laid this thing like a plank, so the greater dimension is the width, not the depth, and as such, we get a pretty small S.)

So,

M_n = 5 √3500 psi 45.8 in.³

M_n = 13,548 lb-in.

But we need it in the same dimensions as our load equation ... lb-ft, so,

M_n= 13,548 lb-in. (ft/12 in.) = 1129 lb-ft.

That isflexuralxural strength of the section.

5. Loads that Break the BeaFlexureexure

To determine the loads that break the beam, we will set the load (expressions) equal to the strength (number).

For the uniform live load ...

M _{max, Total} = 237 lb-ft + 4.5 ω _LL ft ² = 1129 lb-ft ...

Solving for ω _LL ... ω _LL = 198 lb/ft.

Or, ... 198 lb/ft divided by 150 lb/student = 1.3 students per foot ... or 1.3 students/ft x 6 ft = 7.9 ... 8 students.

For concentrated load P ...

M _{max, Total} = 237 lb-ft + 1.5 P = 1129 lb-ft ...

Solving for P ... P = 595 lb.

Wow, that is not very much. (We'll see in the lab that the thing will break before we `hardly get started'.)

6. Summary Loads

So, now let's look at what controls.

For the uniform live load ...

It will break in shear at ω _LL = 1621 lb/ft ...

It will breaflexureexure at ω _LL = 198 lb/ft ... which means it will breaflexturexture before it breaks in shear ...

So, the `controlling' uniform live load is 198 lbflexureexure controls).

For the concentrated load P ...

It will break in shear at P = 8366 lb ... but it actually will never get there, because it will breaflexureexure, first, at P = 595 Flexureexure controls.)

So, the controlling concentrated load P is 595flexureexure controls.)

We have not taken the loads to act concurrently (well, except for self weight). We could look at all kinds of combinations. Say, if the uniform load is 150 lb/ft, what concentrated load P, added to that, would break the beam? (And, would it break in shearflexureexure.)

7. Safe Loads

IF we were to use these beams in some future use (structure) the building codes would require that we apply factors of safety. The numbers calculated in the section above are theoretical `exact' (perfect world) numbers. In other words, we could only allow students that weight 150 lb each to be a part of our experiment, and we will count exactly how many there are. And we (supposedly) know the `exact' amount of load P. But in some future situation we may have students of varying weights, and the concentrated load P, say, coming down from a column, may be carrying varying weights. So, we take into account the uncertainty of these future loads with load factors. And, in fact, we already did (in the earlier lessons), and the factored load expressions are repeated below.

And, in some future design situation, there are also uncertainties in our strength values. There is uncertainty in construction (dimensions, etc.); there is uncertainty in concrete strength. If we were using reinforcement there would be some reasonable uncertainties in its exact location, and so on. So, the building codes will also require us to use factors of safety with our calculated (perfect world) strength values. For plain concrete we accommodate this with strength reduction factors (denoted φ) of 0.55 for both shearflexureexure.

So, for (reduced strength) ...

Shear .Vn φ V_n = 0.55 (4338 lb) = 2386 Flexure

exure ... φ M_n = 0.55 (1129 lb) = 621 lb-ft.

And, from earlier, our `factored' load expressions are ...

V _{u, max} = 199 lb + 4.13 ω _LL ft + 0.8 P, ... and ...

M _u = 284 lb-ft + 7.2 ω _LL ft² + 2.4 P ft.

Solving for the `safe' uniform load ... (setting factored load = factored strength)

... ω _{LL (safeplf} 47 plflexureflexure governs) ...

plf 47 plf (6 ft) = 281 lb. ... And 281 divided by 150 lb/student = about ... less than 2 students. (Yikes!)

Solving for the `safe' concentrated load ... (setting factored load = factored strength)

... P _safe = 140 lb.

(The beam may carry only ONE student, safely, standing at mid-span. Scary!)

That's the one reason we don't use plain concrete for spanning members ... it's not very strong.

But more importantly, when it breaks it breaks suddenly and catastrophically. (All `three' students end up on the floor below, along with the broken beam.)

8. Conclusion

In this lesson we calculated the shear and bending strengths of a plain concrete beam. Then we set those values equal to expressions for load, and determined the loads that would break the beam, theoretically, and the loads that could be used `safely'. We used a beam that was oriented more like a plank (`on its side', inefficientflexureflexure) and flexureflexure controlled. We never will shear the beam - it will always break open first. And the values we obtained are ...

To break the beam ...

... ω _LL = 198 lb/fflexureflexure governs)

... P = 595 lflexureflexure governs)

Safe loads ...

... ω _Lplf47 plflexureflexure governs)

... P = 140 lflexureflexure governs)

NOTE that these are not concurrent; they are `either-or'.

Also note that in this example the safe loads are about one-fourth of the `breaking' loads ... representing an overall factor of safety of about 4.

Remember, we will make a plain concrete beam for `lab' purposes only. Building codes do not permit us to make plain concrete beams for structures.

9. References

Building Code Requirements for Structural Concrete, ACI 318, American Concrete Institute, P.O. BoxFarmingtonmington hills, Michigan, 48333.