Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 5.5, Problem 145P

Two cover plates, each 7.5 mm thick, are welded to a W460 × 74 beam as shown. Knowing that σall = 150 MPa for both the beam and the plates, determine the required value of (a) the length of the plates, (b) the width of the plates.

Fig. P5.144 and P5.145

Chapter 5.5, Problem 145P, Two cover plates, each 7.5 mm thick, are welded to a W460  74 beam as shown. Knowing that all = 150

(a)

Expert Solution
Check Mark
To determine

Find the length of the plates.

Answer to Problem 145P

The length of the plates is 4.49m_.

Explanation of Solution

The length of the cover plate is l=5m.

The thickness of the cover plate is t=7.5mm.

The width of the cover plate is b=200mm.

The allowable normal stress in the beam is σall=150MPa.

Show the free-body diagram of the prismatic beam as in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 5.5, Problem 145P , additional homework tip  1

Determine the vertical reaction at point B by taking moment at point A.

MA=0(40×8)×4+By(8)=0By=160kN

Determine the vertical reaction at point A by resolving the vertical component of forces.

Fy=0Ay(40×8)+By=0Ay320+160=0Ay=160kN

Show the free-body diagram of the section AD as in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 5.5, Problem 145P , additional homework tip  2

Determine the moment at point D by taking moment about point D.

MD=0MD160(4l2)+40×(4l2)×(4l2)2=0MD160(4l2)+20(4l2)2=0MD=160(4l2)20(4l2)2

Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.

The section modulus of the W460×74 section is S=1460×103mm3.

Determine the moment at point D (MD) of the beam using the relation.

σall=MDS

Here, the allowable stress of the beam is σall.

Substitute 150 MPa for σall, 160(4l2)20(4l2)2 for MD, and 1460×103mm3 for S.

150MPa×106Pa1MPa=160(4l2)20(4l2)2kN-m×1000N1kN1460×103mm3×(1m1000mm)3160(4l2)20(4l2)2=219000

Find the length of the plate using trial and error method.

l=4.49m

Therefore, the length of the plates is 4.49m_.

(b)

Expert Solution
Check Mark
To determine

Find the width of the plates.

Answer to Problem 145P

The width of the plates is 211mm_.

Explanation of Solution

The length of the cover plate is l=5m.

The thickness of the cover plate is t=7.5mm.

The width of the cover plate is b=200mm.

The allowable normal stress in the beam is σall=150MPa.

Show the free-body diagram of the prismatic beam as in Figure 3.

Mechanics of Materials, 7th Edition, Chapter 5.5, Problem 145P , additional homework tip  3

Determine the vertical reaction at point B by taking moment at point A.

MA=0(40×8)×4+By(8)=0By=160kN

Determine the vertical reaction at point A by resolving the vertical component of forces.

Fy=0Ay(40×8)+By=0Ay320+160=0Ay=160kN

Show the free-body diagram of the section AC as in Figure 4.

Mechanics of Materials, 7th Edition, Chapter 5.5, Problem 145P , additional homework tip  4

Determine the moment at point C by taking moment about point C.

MC=0MC160(4)+(40×4)×42=0MC640+320=0MC=320kN-m

Show the cross section of the beam as in Figure 5.

Mechanics of Materials, 7th Edition, Chapter 5.5, Problem 145P , additional homework tip  5

Determine the section modulus (S) of the beam using the relation.

S=MCσall

Here, the allowable normal stress in the beam is σall.

Substitute 320kN-m for MC and 150 MPa for S.

S=320kN-m150MPa×103kN/m21MPa=2133.33×106m3×(1000mm1m)3=2133.33×103mm3

Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.

The depth of the W460×74 section is d=457mm.

The moment of inertia of the W460×74 section is Ibeam=333×106mm4.

Determine the moment of inertia (I) of the beam using the relation;

I=Ibeam+2Iplates=Ibeam+2(bt312+A(y¯)2) (1)

Here, the moment of inertia of the beam is Ibeam, the width of the plate is b, the thickness of the plate is t, the cross sectional area of the plate is A, and the distance between the neutral axis and the center of the plates is y¯.

Refer to the cross section of the beam;

A=(b×7.5)mm2;y¯=4572+7.52=232.5mm

Substitute 333×106mm4 for Ibeam, 7.5 mm for t, (b×7.5)mm2 for A, and 232.5 mm for y¯ in Equation (1).

I=333×106+2(b×7.5312+b×7.5×232.52)

Determine the distance from the neutral axis (c) to outer most fibre as follows;

c=4572+7.5=236mm

Determine the width of the plates (b) using the relation.

S=Ic

Substitute 2133.33×103mm3 for S, 333×106+2(b×7.5312+b×7.5×232.52) for I, and 236 mm for c.

2133.33×103=333×106+2(b×7.5312+b×7.5×232.52)236mm333×106+2(b×7.5312+b×7.5×232.52)=503465880b=211mm

Therefore, the width of the plates is 211mm_.

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