     

y c

    max      

y

    To determine the normal stress in a beam subjected to bending we need to know the moment acting at that point; the distance from the neutral axis; and the moment of inertia of the beam.



x

 

E

  

y

         

My I

  

Flexure Formula

 max     

Mc I

   Normal stresses in a beam of linearly elastic material: (a) side view of beam showing distribution of normal stresses, and (b) cross section of beam showing the

z

axis as the neutral axis of the cross section

Positive curvature results from positive applied moments. For positive curvature compression occurs at the top surface. The beams shown have non symmetric cross-sections and the neutral axis is therefore not at the center.

Taking moments about the neutral z-axis (for each small slice the force is stress x area and the distance from the z-axis is y)

dM

  

x ydA M M



E

 

A

  

A



x ydA y

2

dA



EI

 

x



E



x

 

E y

 Remember

I neutral

 

y

2

dA

S



bh

2 6

Doubly symmetric cross-sectional shapes

If a beam is doubly symmetric (i.e. symmetric in the z and y directions, the neutral axis will be at the center of the beam.

  

Mc I

 

M S

 max 

M S

Where S=section modulus and is a geometric factor. For beam design we can calculate the required section modulus and then select.

S

 

d

3 32

FIG. 5-13

Example 5-2.

Wire bent around a drum

M

  

EI

   O

For the beam shown determine the maximum tensile and compressive stresses in the beam.

1) Calculate the maximum +ve and –ve moments Need to draw shear and moment diagrams M pos =2.025 kN.m M neg =-3.6 kN.m

2) Find neutral axis

c

2   

y i A i A i

 2   40 2   12  12   80 80    74  12   276 276  12  61 .

52

mm

3) Find moment of inertia about the neutral axis

I z



I

1 

I

2 

I

3

I

1  1 12

bh

3 

A

1

d

1 2  276  12 3 12   276  12   12 .

48  2  555600

mm

4

I

2 

I

3  12  80 3 12   80  12   21 .

52  2  956600

mm

4

I z

 2 , 469 , 000

mm

4

3) Calculate the maximum +ve and –ve stresses there are four combinations M c 2 pos c 1 ; M pos c 2 ; M neg c 1 ; M neg  1  

M pos c

1

I

  2 .

025 133600   15 .

2

MPa

 2  3   

M I I pos c

2 

M neg c

1  2 .

025 40100   3 .

6 133600  50 .

5

MPa

Tensile  26 .

9

MPa

 4  

M neg c

2

I

  3 .

6 40100   89 .

8

MPa

Compressive

A simple beam of span length 21 ft must support a uniform load q=2000lb/ft as shown. Neglecting the weight of the beam select a structural beam of wide flange shape to support the loads.

1) Determine M max Draw shear diagram Calculate M 2) Calculate required section modulus Calculate or look up max stress 3) Select a beam W12x50 S=64.7

X 1 =9430 ft M max =88,920 lb-ft S min =M max /  allow =59.3in

3 Design of Beams

A cantilever beam AB of length L is being designed to support a concentrated load at the free end . The cross-section is rectangular. Calculate the height as a function of distance so that the beam is fully stressed. I.e. at every point



=



allow

 max 

M S M



Px



allow



M S



Px bh x

2 / 6  6

Px bh x

2

S



bh x

2 6

h x

 6

Px b



allow h b

 6

PL b



allow h x



h b x L