Good morning!

New words Beam 梁 Tension 拉伸 Compression 压 Shear 剪 Torsion 扭转 Bending 弯曲 Concave side 凹面 Convex side 凸面

Exersize measure the tendency of one part of a beam to be slided with respect to the other part (shear force) the tendency for one part to be rotated with respect to the other part (bending moment)

Review: In last week, we have learned columns made of : timber steel concrete

New member –beam Question: in considering the beam what are we concerned with? Answer: the effects of the forces

Five forms of deformation Tension Compression Shear Bending Torsion

Internal forces For example: the internal forces that result from bending deformations are compressive on the concave side and tensile on the convex side

SF&BM measure the tendency of one part of a beam to be slided with respect to the other part (SF effect),and the tendency for one part to be rotated with respect to the other part(BM effect) The question we need resolve is to calculate the SF&BM

Shear Force

Calculation The SF at any section A in a straight beam is the algebraic sum of all vertical forces lefts of that point SF A =∑V L A

Eg1: 3KN 5KN A B C D 2M 2M 1M 8KN

STEPS Cut the beam at point B,so we will get part of the shear force of the beam 8KN 3KN A B

If we cut the beam at point C,what will happen to the SF? TRYING… What about point D? TRYING…

SFD If we plot the value of the SF at all point along the axis of a beam, we’ll obtain the Shear Force Diagram 8KN 5KN

Principle for solving the SFD Calculate the shear forces at the following significant positions: –At the start and end of the beam –At every support –At every point load –At the start and end of every distributed load

Conclusion: We find that at a point load the SF changes instantaneously and the SF diagram shows a step.

Eg2: A B C D 2M 2M 1M

Steps: First calculate the reaction at support A. Then cut the beam at point D

Finally plot the SFD 15 A B C D 2M 2M 1M

Conclusion: When there is an UDL,we see the SF change by equal intervals,and the SFD will have a constant slope.

Have a Try 4KN 6KN 3KN A ① ② B ③

Bending Moment BM L =∑A L L The BM at any section A in a straight beam is the algebraic sum of areas of the SFD left of that point

Eg3 A B C D 2M 2M 1M 8KN 5KN A B C D

NOTE Locate the positions of zero SF, as these will also be positions of maximum BM.

The result is: 0 16KN 26KN

HAVE A TRY 4KN 6KN 3KN A ① ② B ③

CONSIDER All we learned today is the simply supported beam, then what about continuous beam? Let’s talk about it on next lesson!

Homework: Exersize1 a/b/c/f

Bye bye See you next class