A simply supported composite timber beam is loaded as shown in Figure 2(a). The beam cross-section is shown in Figure 2(b). Plot the mid-span flexural stress and strain distribution diagrams of the beam.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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A simply supported composite timber beam is loaded as shown in Figure 2(a). The beam cross-section is shown in Figure 2(b). Plot the mid-span flexural stress and strain distribution diagrams of the beam.

Note:
Timber Grade 1 properties
ETI = 15 GPa
• Tyield, tension - T1 = 10 MPa
FIGURE 1
FIGURE 2
yield, compression - TI = 5.0 MPa
Timber
Grade 1
200 200
mm, min
100 mm
Timber Grade 2 properties
• E12=30 GPa
•
yield, tension-T2 = 7.5 MPa
● O
400 kN load
(a) Longitudinal column section
40 kN/m
/= 3.0 m
(a) Longitudinal beam section
200
Inm
yield, compression-T2= 3.0 MPa
200
inm
100 mm-
200
mm
400 KN Load
100 mm
(b) Plan view of column
200
mm
100 mm
100 mm
200
inm
200 mm
Timber
Grade 1
Timber
Grade 2
(b) Beam cross-section
Transcribed Image Text:Note: Timber Grade 1 properties ETI = 15 GPa • Tyield, tension - T1 = 10 MPa FIGURE 1 FIGURE 2 yield, compression - TI = 5.0 MPa Timber Grade 1 200 200 mm, min 100 mm Timber Grade 2 properties • E12=30 GPa • yield, tension-T2 = 7.5 MPa ● O 400 kN load (a) Longitudinal column section 40 kN/m /= 3.0 m (a) Longitudinal beam section 200 Inm yield, compression-T2= 3.0 MPa 200 inm 100 mm- 200 mm 400 KN Load 100 mm (b) Plan view of column 200 mm 100 mm 100 mm 200 inm 200 mm Timber Grade 1 Timber Grade 2 (b) Beam cross-section
Expert Solution
Step 1

Answer 

We know that flexural stress and strain have linear diagram. Using bending equation we can determine the stress and strains in given composite timber beam

Let us first convert composite section into timber grade 1 section. The equivalent timber grade 1 section is obtained by multiplying the dimesions of timber grade 2 parallel to N.A by modular ratio

Modular ratio = Elastic Modulus of strong material/Elastic Modulus of weak material

Modular ratio = 30/15

Modular ratio = 2

Civil Engineering homework question answer, step 1, image 1

Now we know that bending stress is given by ftop = MI*yt and fbottom = MI*yb

M = Moment at midspan

M = 40*328  (midspan moment  due to udl in simply supported beam =w*l28 )

M = 45 Kn-m

M = 45 * 106 N-mm

I = Moment of inertia of equivalent timber 1 section 

Calculation of 'I' is shown in steps below

steps

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