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3 16 C H A P T E R 6 B E N D I N G

6

EXAMPLE 6.17

A composite beam is made of wood and reinforced with a steelstrap located on its bottom side. It has the cross-sectional areashown in Fig. 638 a . If the beam is subjected to a bending moment of

determine the normal stress at points B and C. Takeand E st = 200 GPa.E w = 12 GPa

M = 2 kN #m,

(a)

150 mm

20 mm

C

B150 mm

M 2 kN m_

y

9 mm

(b)

150 mm

20 mm

C

B

N

A

150 mm

The moment of inertia about the neutral axis is therefore

= 9.358110- 62 m4 +

c1

12

10.009 m

210.150 m

23 +

10.009 m

210.150 m

210.095 m - 0.03638 m

22

d

I NA = c112 10.150 m210.02 m23 + 10.150 m210.02 m210.03638 m - 0.01 m22d

y = yA A

=[0.01 m]10.02 m210.150 m2 + [0.095 m]10.009 m210.150 m20.02 m10.150 m2 + 0.009 m10.150 m2 = 0.03638 m

SOLUTION

Section Properties. Although the choice is arbitrary, here we willtransform the section into one made entirely of steel.Since steel has agreater stiffness than wood the width of the wood mustbe reduced to an equivalent width for steel.Hence n must be less thanone. For this to be the case, so that

The transformed section is shown in Fig. 638 b .The location of the centroid (neutral axis ), calculated from a

reference axis located at the bottom of the section, is

bst = nb w =12 GPa200 GPa

1150 mm2 = 9 mmn = E w>E st,1

E st 7 E w

2,

Fig. 638

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6.7 R EINFORCED C ONCRETE B EAMS 3 17

6Normal Stress. Applying the flexure formula, the normal stress at

and C is

A ns.

The normal-stress distribution on the transformed (all steel ) section isshown in Fig. 638 c.

The normal stress in the wood at B in Fig. 638a , is determined fromEq.621; that is,

A ns.

Using these concepts, show that the normal stress in the steel andthe wood at the point where they are in contact is and

respectively. The normal-stress distribution in the

actual beam is shown in Fig. 638 d .

s w = 0.210 MPa,s st = 3.50 MPa

s B = n s B =12 GPa200 GPa

128.56 MPa2 = 1.71 MPa

s C =2(103) N #m10.03638 m29.358110- 62 m4 = 7.78 MPa

s B =2(103) N #m10.170 m - 0.03638 m2

9.358110- 62 m4 = 28.6 MPaB

7.78 MPa

28.6 MPa

3.50 MPa

B

C

(c)

M 2 kN m

7.78 MPa

C

3.50 MPa0.210 MPa

1.71 MPa

B

(d)

M 2 kN m

Fig. 638 (cont.)