1. Strength of Materials I EGCE201 กำลังวัสดุ 1 Instructor: ดร. วรรณสิริ พันธ์อุไร ( อ. ปู ) ห้องทำงาน : 6391 ภาควิชาวิศวกรรมโยธา E-mail: egwpr@mahidol.ac.thegwpr@mahidol.ac.th โทรศัพท์ : 66(0) 2889-2138 ต่อ 6391

2. Design of beam for bending

5. Steps in designing a beam for bending Assume that E,G,  all, and  all for the material selected are known 1.Construct the diagrams corresponding to the specified loading conditions for the beam and define |V| max and |M| max. 2.Assume that the design of the beam is controlled by the normal stress at +,-c in the section and determine S min. 3.From available tables, select beams with S>S min. Considerations: -small weight per unit length -small displacement (show in the latter chapter) 1. 2.

6. Design Example 1. Modeling distributed loads as equivalent concentrated force 2. Use the knowledge gained from lecture 4 (either by forming equations or by inspection) to form moment diagram. 1. the shear force is linear, the moment varies parabolically. 2. the area under the shear diagram is – so is the correspondent moment A = 0.5(9)(-2250) = -10,125 M = -10,125

7. Design Example (continued) M max = 18,000 lb-ft =216,000 lb-in V max = 3750 lb 3. From 2, we find 4.

8. Complete Beam Analysis Example For the beam loaded and supported as shown, determine the max tensile and compressive stresses and where they occur. Given

9. 1. Determine the location of the centroid. The area of the cross section (A) = 18.75 in 2 2. Determine the area moment of inertia w.r.t. the centroidal axis (as shown in figure on the lower left) and use transfer formula to calculate I of the section.

10. 3. Begin the analysis by defining reactions at A and B using FBD, one writes the equations of equilibrium. 4. Next, the shear diagram can be constructed (see previous example or lecture 4 for more details).

11. 5. Now the moment diagram can be constructed. The moment is the area under the shear force diagram and the three areas are A1 = 0.5(3.5ft)(-17.5kip) = -30.63 kip-ft A2 = 0.5(3.5ft)(52.5-35 kip)+(3.5ft)(35 kip) = 153.13 kip-ft A3 = 0.5(7ft)(-35kip) = -122.5 kip-ft The moment diagram is as shown. A1+A2= A1= A1+A2+A3=0 Recall Positive moment, top beam is in compression and the bottom is under tension

12. 6. Next, we compute stresses At x=3.5 ft, moment is negative so the top is in tension while the bottom is under compression. 7.At x=7 ft, moment is positive so the top is under compression while the bottom is in tension.

13. Next Shear Stress in Beams

14. A cutting plane is passed through a beam at an arbitrary spanwise location, the internal reactions are required for Equilibrium are a bending moment and a shear force. The moment and shear force as shown are considered positive. The shear and normal stresses acting on an element of area are represented as forces by multiplying them by the area (dA)

15. 3 out of 6 equations of equilibrium involve the normal force  x dA Pure Bending 3 out of 6 equations of equilibrium involve the shearing force  xy dA,  xz dA 1. 2. From 1. Vertical shearing stresses exist in a transverse section of the beam if a shear force exists at that section

16. Shear stress on a horizontal plane H

17. Horizontal shear derivation Observing that x is constant over the cross section, the expression for H is written as Q is the first area moment w.r.t. to the N.A. of that part of the section located above the line y=y1

18. Shear Flow, q But Along a horizontal plane a distance y1 above the NA, the horizontal shear per unit length of beam The ratio of H/x is termed the shear flow and is denoted by “q” where

19. Example Determine the shear flow (q) of the following cross section

20. Solution The cross section is broken into 3 sections and the second area moment of inertia Given V = 8000 lb, one compute The shear flow can now be expressed as

21. Compute Q Q is the first area moment w.r.t the N.A. A is the area of the cross section above the plane for which q is being determined. is the location of that area w.r.t the N.A (+,-).

22. Transverse shear stresses in beam The horizontal shear flow at C A shear force exists on a horizontal plane passing C

23. For a narrow rectangular beam b < h/4

24. For an I- beam web flange

25. Example A shear force acts on a cross section. The cross section shown is made by nailing planks together. 1. Use shear flow to define the required nail spacing if each nail supports 700 lb shear before failure. 2. Compute shear stress at various locations in the cross section.

26. 1. Compute the moment of inertia

27. 2. Compute the nail spacing

28. 3. Compute stress distribution

29. Displacements in beams Next week