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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Chapter 8, Problem 65RP

(a) Knowing that σall = 24 ksi and σall = 14.5 ksi, select the most economical wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for τm, τm, and the principal stress σmax at the junction of a flange and the web of the selected beam.

Fig. P8.65

Chapter 8, Problem 65RP, (a) Knowing that all = 24 ksi and all = 14.5 ksi, select the most economical wide-flange shape that

(a)

Expert Solution
Check Mark
To determine

Select the most economical wide flange shape section.

Answer to Problem 65RP

The most economical wide flange shape section is W14×22_.

Explanation of Solution

Given information:

The allowable bending stress σall is 24ksi.

The shear stress τall is 14.5ksi.

Calculation:

Sketch the free body diagram of beam as shown in figure 1.

Mechanics of Materials, 7th Edition, Chapter 8, Problem 65RP , additional homework tip  1

Here, RA is reaction at point A and RD is reaction at point D.

Calculate the reaction for the given structure:

Taking moment about D,

MD=0

15RA+(10.5×9×2)+(3×12.5)=015RA+(189)+(37.5)=015RA+226.5=0RA=15.1kips

Summation of vertical force zero.

Fy=0

15.1+18=12.5+RDRD=15.4kips

Calculate the shear force diagram as follows:

Shear force at A is SFA=0.

Shear force at B,

SFB=15.12×9=15.118=2.9kips

Shear force at C,

SF@c=15.12×912.5=15.11812.5=15.4kips

Shear force at D,

SFD=15.1(2×9)12.5+15.4=15.11812.5+15.4=0

Maximum shear force occurs at the point where SF=0.

Consider a section xx at a distance x from A.

SF@xx=15.1(2×x)=02x=15.1x=7.55ft

Calculate the bending moment as follows:

Bending moment at point A,

BMA=0

Bending moment at point B,

BMB=15.1×9(2×9×92)=135.981=54.9kipsft

Bending moment at point C,

BMC=15.1×12(2×9(92+3))=181.2135=46.2kipsft

Bending moment at point D,

BMD=15.1×152×9(92+6)12.5×3=226.518937.5=0

Sketch the shear force and bending moment diagram as shown in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 8, Problem 65RP , additional homework tip  2

Refer to Figure 2.

The maximum bending moment is 57.003kipsft.

Find the value of Smin using the relation:

Smin=|M|maxσall (1)

Here, |M|max is maximum moment.

Substitute 57.003kipsft for |M|max and 24ksi for σall in Equation (1).

Smin=57.003×1224=28.502in3

Write the section properties shape as shown in Table 2.

ShapeS(in3)
W16×2638.4
W14×2229.0
W12×2633.4
W10×3032.4
W8×3531.2

Select the section W14×22 wide flange section.

The most economical wide flange shape section is W14×22_.

Write the section properties wide flange section as shown in Table 3.

ShapeW14x22
Area, A6.49in2
Depth, d13.74in
Web thickness, tw0.230 in.
Width bf5.000 in.
Thickness tf0.335 in.

Find the area of web (Aweb) using the relation:

Aweb=dtw (2)

Here, d is the depth of the section and tw is thickness of web.

Substitute 13.7in. for d and 0.230in. for tw in Equation (2).

Aweb=13.7×0.230=3.151in2

(b)

Expert Solution
Check Mark
To determine

The value to be expected for σm,τm and the principal stress σmax at the junction of a flange and the web of the selected beam.

Answer to Problem 65RP

The normal stress σm at point is 23.6ksi_

The shear stress τm at point is 4.89ksi_

The maximum principal stress is 22.4ksi_

Explanation of Solution

Calculation:

Point E:

Find the normal stress at point E using the relation:

σm=MES (3)

Here, ME moment at point E and S is the section property.

Substitute 57.003kipsft for ME and 29in3 for S in Equation (3).

σm=57.003kipsft(12kipsin1kipsft)29.0=57.003×1229.0=23.587ksi23.6ksi

Thus, the normal stress σm at point is 23.6ksi_

Find the value of C using the relation:

c=d2 (4)

Here, d is the depth of section.

Substitute 13.7in for d in Equation (4).

c=13.72=6.85in.

Find the value of yb using the relation:

yb=ctf (5)

Here, tf is the thickness of flange.

Substitute 6.85in. for c and 0.335in. for tf in Equation (5).

yb=6.850.335=6.515in.

Find the normal stress σb at point E using the relation:

σb=ybcσm (6)

Here, σm is the normal stress,

Substitute 6.515in. for yb, 6.85in. for c, and 23.587ksi for σm in Equation (6).

σb=6.5156.85×23.587=0.9510×23.587=22.43ksi

Find the shear stress τm at point E using the relation:

The point E is located at top. Since Q is zero.

Thus, the shear stress at point τm is zero.

Point C:

Find the normal stress at point E using the relation:

σm=McS (7)

Here, Mc moment at point C and S is the section property.

Substitute 46.2kipsft for MC and 29in3 for S in Equation (7).

σm=46.2kipsftkipsft(12kipsin1kipsft)29.0=46.2×1229.0=19.1172ksi

Find the shear stress point C using the relation:

τm=|V|Aweb (8)

Here, V is shear force and Aweb is the area of web.

Substitute 15.4kips for V and 3.151in2 for Aweb in Equation (8).

τm=15.43.151=4.8877ksi

Thus, the shear stress τm at point is 4.89ksi_

Find the σb point c using the relation:

σb=ybcσm (9)

Substitute 6.515in. for yb, 6.85in. for c, and 19.1172ksi for σm in Equation (9).

σb=6.5156.85×19.1172=0.9510×19.1172=18.1823ksi

Find the R using the relation:

R=(σb2)2+τm (10)

Here, σb is the normal stress at point c and τm is shearing stress at point C.

Substitute 18.1823ksi for σb and 4.8877ksi for τm in Equation (10).

R=(18.18232)2+4.8877=82.64+23.88=10.3216ksi

Find the maximum principal stress using the relation:

σmax=σb2+R (11)

Substitute 18.1823ksi for σb for 10.3216ksi for R in Equation (11).

σmax=18.18232+10.3216=9.090615+10.3216=19.41ksi

Compare the results,

Select the maximum value of stress for Point B is controls.

Thus, the maximum principal stress is 22.4ksi_.

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Chapter 8 Solutions

Mechanics of Materials, 7th Edition

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