Strength of Materials I EGCE201 กำ��ลั�งวั�สดุ 1

Instructor: ดุร.วัรรณส�ร� พั�นธ์�อุไร (อุ.ปู�)ห้�อุงทำ��ง�น: 6391 ภ�ควั�ช�วั�ศวักำรรมโยธ์�

E-mail: egwpr@mahidol.ac.th

โทำรศ�พัทำ�: 66(0) 2889-2138 ต่"อุ 6391

Design of beam for bending

Design of beam for bending

Steps in designing a beam for bending

Assume that E,G,all, and all for the material selected are known1.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 Smin .3.From available tables, select beams with S>Smin .Considerations:

-small weight per unit length

-small displacement (show in the latter chapter)

1.

2.

Design Example

1. Modeling distributed loads as equivalent concentrated force

2. Use the knowledge gained from lecture 4 (either by formingequations or by inspection) to form moment diagram.

1. the shear force is linear, the moment variesparabolically.2. the area under the shear diagram is – so is the correspondent moment

A = 0.5(9)(-2250) = -10,125

M = -10,125

Design Example (continued)

Mmax = 18,000 lb-ft =216,000 lb-in

Vmax = 3750 lb

3. From 2, we find

4.

Complete Beam Analysis Example

• For the beam loaded and supported as shown, determine the max tensile and compressive stresses and where they occur. Given

1. Determine the location of the centroid.

• The area of the cross section (A) = 18.75 in2

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.

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).

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-ftA2 = 0.5(3.5ft)(52.5-35 kip)+(3.5ft)(35 kip) = 153.13 kip-ftA3 = 0.5(7ft)(-35kip) = -122.5 kip-ft

The moment diagram is as shown.

A1+A2=

A1=

A1+A2+A3=0

RecallPositive moment, top beam is in compression and the bottom is under tension

6. Next, we compute stressesAt 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.

NextShear Stress in Beams

A cutting plane is passed through a beam at an arbitraryspanwise location, the internalreactions are required for Equilibrium are a bendingmoment and a shear force.

The moment and shear forceas shown are considered positive.

The shear and normal stressesacting on an element of area arerepresented as forces by multiplyingthem by the area (dA)

3 out of 6 equations of equilibrium involve the normal force xdAPure Bending

3 out of 6 equations of equilibrium involve the shearing force xydA, xzdA

1.2.

From 1.

Vertical shearing stresses exist in a transverse section of the beam if a shear force exists at that section

Shear stress on a horizontal plane

H

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. ofthat part of the section located above the line y=y1

HPQ

Ix

Shear Flow, q

H

x

PQ

I P VBut

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

qVQ

I

Example

• Determine the shear flow (q) of the following cross section

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

q Q 0 5 5 8.

• 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 (+,-).

Q A y

y

Transverse shear stresses in beam

• The horizontal shear flow at C

• A shear force exists on a horizontal plane passing C

qVQ

I

For a narrow rectangular beam

b < h/4

For an I- beam web

flange

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.

1. Compute the moment of inertia

2. Compute the nail spacing

3. Compute stress distribution

Displacements in beams

Next week