Solve Prob. 5.85, assuming that the cross section of the beam is inverted, with the flange of the beam resting on the supports at B and C.
5.85 Determine the largest permissible distributed load w for the beam shown, knowing that the allowable normal stress is +80 MPa in tension and –130 MPa in compression.
Fig. P5.85
The largest permissible distributed load w.
Answer to Problem 86P
The largest permissible distributed load (w) is
Explanation of Solution
Given information:
The allowable normal stress of the material in tension is
The allowable normal stress of the material in compression is
Calculation:
Show the free-body diagram of the beam as in Figure 1.
Determine the vertical reaction at point C by taking moment about point B.
Determine the vertical reaction at point B by resolving the vertical component of forces.
Shear force:
Show the calculation of shear force as follows;
Show the calculated shear force values as in Table 1.
Location (x) m | Shear force (V) kN |
A | 0 |
B (Left) | –0.2w |
B (Right) | 0.25w |
C (Left) | –0.25w |
C (Right) | 0.2w |
D | 0 |
Plot the shear force diagram as in Figure 2.
Location of the maximum bending moment:
The maximum bending moment occurs where the shear force changes sign.
Refer to Figure 2;
Use the similar triangle concept.
The maximum bending moment occurs at a distance of 0.45 m from left end of the beam.
Bending moment:
Show the calculation of the bending moment as follows;
Show the calculated bending moment values as in Table 2.
Location (x) m | Bending moment (M) kN-m |
A | 0 |
B | –0.02w |
Max BM | 0.01125w |
C | –0.02w |
D | 0 |
Plot the bending moment diagram as in Figure 3.
Show the free-body diagram of the T-section as in Figure 4.
Determine the centroid in y-axis
Here, the area of the section 1 is
Refer to Figure 4;
Substitute
Determine the moment of inertia (I) using the equation.
Here, the depth of the section 1 is
Substitute 60 mm for
Refer to Figure 4;
Tension at Points B and C:
Refer to Figure 3;
Determine the distributed load (w) using the relation.
Substitute –0.02w for
Compression at Points B and C:
Refer to Figure 3;
Determine the distributed load (w) using the relation.
Substitute –0.02w for
Tension at Points E:
Refer to Figure 3;
Determine the distributed load (w) using the relation.
Substitute 0.01125w for
Compression at Points E:
Refer to Figure 3;
Determine the distributed load (w) using the relation.
Substitute 0.01125w for
Refer to the calculated distribution loads; the smallest value controls the design.
Therefore, the largest permissible distributed load (w) is
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Chapter 5 Solutions
EBK MECHANICS OF MATERIALS
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