This chapter discusses the analysis and design of beams, which are structural members that support loads applied at different points. Beams can be subjected to concentrated loads or distributed loads. Beams are classified based on their support conditions, with statically determinate beams having three unknowns and statically indeterminate beams having more than three unknowns. Shear and bending moment diagrams are constructed to determine the internal shear and moment forces in the beam resulting from the applied loads. The positive and negative directions of shear and bending moment are defined.

1. Chapter 5The 1st chapter in 2nd term

2. Introduction :In this chapter we care for the analysis and the design of beams ,structural members supporting loads applied at different points ,In most cases ,the loads are perpendicular to the axis of the beam ,when loads aren’t at right angles they also produce axial forces .

3. The loading at beams consists of Concentrated loads :P1,P2 expressed in newtons ,pounds.

4. Distributedloads(w) :Expressed in (N/m) or (b/ft)The load from (A) to (B) is said to be “uniformly distributed”

5. Classification of beamsBeams are classified due to the way they are supported .The distance “L” shown called the span .Beams a,b,c are said to be statically determinate beamsBeams d,e,f are said to be statically intermediate beams that conclude more than three unknowns .

7. Other beams are connected by hinges to form a simple continuous structure .It would be noted that the reaction at the supports involves four unknowns .

8. “Shear & Bending Moment Diagram”If the force is uniform distributed as Fo for a length of span L ,so the resultant = FoL acts in the medium of length .We notice that Fo is uniformly distributed force .The total force value is FoL

9. If the force isn’t uniformly distributed (w=ax) ,as shown .So ,wmax= aL , wmin = 0the total resultant force =area of the force triangle = 0.5L wmax = L (aL) =0.5a L2The resultant force =0.5a L2Point of effect (see in fig.)

10. The positive & negative Shear “S”If we cut section through abeam :Positive shear : look  (from left to right) Negative shear : look  (from right to left)

11. The positive & negative Moment “M”Physically : we consider the +ve moment ,if the effect of external forces acting on the beam tend to bend it as shown .

12. Mathematically : we consider the +ve moment as shown

13. So we can conclude the +ve (shear & moment) We can also conclude the –ve (shear & moment)