Shear and Moment Diagrams for a Simple Beam Part 2

1. Uniform Load - Revisited
   
2. Shear
   
3. Bending Moment
   
4. Factored Loads
   
5. Conclusion
   

1. Uniform Load - Revisited

Let's revisit the uniform load thing. In our previous example (here) the uniform load was limited to the self weight, and so it didn't really illustrate the `V @ d' thing as well as I would like to have. So, let's also investigate a uniform `laboratory Live load' applied to the beam. Visualize, perhaps, a line of students standing along the beam.

Now the load is applied to the top of the beam, so we can deduct the loads within distance `d' of the supports. Let's also assume that the students are lined up from support to support, and not out on the ends sticking past the supports.

2. Shear

Let our line load be ω _LL.

The total load, W _LL, is ω _LLl = ω _LL 6 ft.

The reactions are (both) ... R = W_LL /2 = ω _LL 6 ft / 2 = 3 ω _LL ft.

The maximum shear is ... V max = ω _LL l / 2 = 3 ω _LL ft ... same as the reaction.

For design we may use V @ d ...

V @ d = ω _LL l / 2 = ω _LL d = ω _LL (l / 2 - d) = ω _LL (3 ft - 5/12ft)

V @ d = 2.58 ω _LL ft.

So, if we have a uniform load and a concentrated load at mid-span, plus the self weight, we can express the total shear as,

V = ω _sw (l/2 - d/2) + ω _LL (l/2 - d) + P/2 ...

In our example, then,

V = 55.4 plf (6 ft/2 - (5/12 ft)/2) + 2.58 ω _LL + P / 2

V = 155 lb + 2.58 ω _LL + P / 2 .

3. Bending Moment ...

M _{max, LL} = ω _LLl ²/8 = ω _LL (6 ft)² / 8 = 4.5 ω _LL ft ²

M _{max, LL} = 4.5 ω _LL ft ².

The total moment in our example is, then ...

M _{max, Total} = M _{max, sw} + M _{max ω LL} + M _P

M _{max, Total} = 50 237 lb-ft + 4.5 ω _LL ft ² + 1.5 P .

4. Factored Loads

In terms of `design' for future use, the load factor on the uniform (student) Live load will be 1.6, so,

V_u = V_{u @ d/2 for sw} + V_{u @ d for ω LL} + V_{u for P} ,

which for our example is,

V _u = 199 lb + 1.6 (2.58 ω _LL) + 0.8 P

V _u = 199 lb + 4.13 ω _LL ft + 0.8 P .

And for Moment ...

M _u = 60284 lb-ft + 1.6 (4.5 ω _LL ft²) + 2.4 P ft

M _u = 60 284 lb-ft + 7.2 ω _LL ft² + 2.4 P ft.

5. Conclusion

Now we are ready to solve for the `number of students to break the beam', or the `concentrated load P to break the beam', or any combination thereof. And, we are ready to do it at the `lab' level (no load factors), and `design for some future use' (with the load factors).

6. References

Shear and Moment Diagrams for a Simple Beam, Associated Content (Previous Lesson).