8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement σ m ≤ σ all . For the selected design, determine ( a ) the actual value of σ m in the beam, ( b ) the maximum value of the principal stress σ max at the junction of a flange and the web. 8.9 Loading of Prob. 5.73 and selected W530 × 92 shape. Fig. P5.73

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Answer 1

Textbook Question

Chapter 8.2, Problem 9P

8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement σ_m ≤ σ_all. For the selected design, determine (a) the actual value of σ_m in the beam, (b) the maximum value of the principal stress σ_max at the junction of a flange and the web.

8.9 Loading of Prob. 5.73 and selected W530 × 92 shape.

Fig. P5.73

Chapter 8.2, Problem 9P, 8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem

(a)

Expert Solution

To determine

The actual value of $σm$ in the beam.

Answer to Problem 9P

The actual value of $σm$ in the beam is $137.5 MPa_$ .

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $W530×92$ .

Calculation:

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

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 1

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

$∑MA=0−70(3)−(5×11)×112−70(8)+Dy(11)=0−210−302.5−560+11Dy=0Dy=97.5 kN$

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

$∑Fy=0Ay−70−70−(5×11)+Dy=0Ay−140−55+97.5=0Ay=97.5 kN$

Shear force:

Show the calculation of shear force as follows;

$VA=97.5 kN$

$VB−VA=−5×3VBLeft=−15+VA=−15+97.5=82.5 kN$

$VBRight=82.5−70=12.5 kN$

$VC−VB=−5×5−70VCLeft=−95+VB=−95+82.5=−12.5 kN$

$VCRight=−12.5−70=−82.5 kN$

$VD−VC=−5×3−70VD=−15−70+VC=−85−12.5=−97.5 kN$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 2

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$12.5x=12.55−x62.5−12.5x=12.5x25x=62.5x=2.5 m$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$MA=0$

$MB−MA=(82.5×3)+12×(97.5−82.5)×3MB=247.5+22.5+MA=270+0=270 kN-m$

$Mmax−MB=12×12.5×2.5Mmax=15.625+MB=15.625+270=285.625 kN-m$

$MC−Mmax=−12×12.5×2.5MC=−15.625+Mmax=−15.625+285.625=270 kN-m$

$MD=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 3

Refer to the Figure 3;

The maximum bending moment in the beam is $|Mmax|=285.625 kN-m$ .

Write the section a property for a $W530×92$ rolled steel section as table 2.

Dimension	Unit( $mm$ )
d	533 mm
$bf$	209 mm
$tf$	15.6 mm
$tw$	$10.2 mm$
I	$554×106 mm4$
$S$	$2080×103 mm3$

Here, d is depth of the section, $bf$ is breadth of the flange, $tf$ is thickness of the flange, $Ix$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=12d$ (1)

Substitute $533 mm$ for d in Equation (1).

$c=12×533=266.5 mm$

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=|M|maxS$ (2)

Here, $|M|max$ is maximum bending moment and $S$ is the section modulus.

Substitute $285.625 kN-m$ for $|M|max$ and $2080×103 mm3$ for $S$ in Equation (2).

$σm=285.625 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =285.625×1032.08×10−3=137,319,711.5 Pa(1 Mpa106 Pa)$

$σm=137.5 Pa$

Thus, the actual value of $σm$ in the beam is $137.5 MPa_$ .

(b)

Expert Solution

To determine

The maximum value of principal stress $σmax$ at the junction of a flange and the web.

Answer to Problem 9P

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Explanation of Solution

Calculation:

Find the value $yb$ as follows:

$yb=c−tf$ (3)

Here, c is the centroid and $tf$ is the thickness of flange.

Substitute $266.5 mm$ for c and $15.6 mm$ for $tf$ in Equation (3).

$yb=266.5−15.6=250.9 mm$

Find the area of flange $(Af)$ of section using the relation:

$Af=bftf$ (4)

Here, $bf$ is the width of the flange and $tf$ is the thickness of the flange.

Substitute $209 mm$ For $bf$ and $15.6 mm$ for $tf$ in Equation (4).

$Af=209×15.6=3,260.4 mm2$

Find the centroid of flange $y¯$ using the relation:

$y¯=12(c+yb)$ (5)

Substitute $266.5 mm$ for c and $250.9 mm$ for $yb$ in Equation (5).

$y¯=12(266.5 +250.9)=258.7mm$

Find the first moment about neutral axis $(Q)$ as follows:

$(Q)=Afy¯$ (6)

Here, $Af$ is the area of flange and $y¯$ is the centroid.

Substitute $3,260.4 mm2$ for $Af$ and $258.7mm$ for $y¯$ in Equation (6).

$(Q)=3,260.4 ×258.7=843.47×103 mm3$

At mid span the value of $V=0$ and $τb=0$ .

Find the maximum value of principal stress $σmax$ as follows:

$σb=ybcσm$ (7)

Here, actual value of normal stress $yb$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $137.5MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (7).

$σb=250.9266.5 ×137.5=129.45 MPa$

At section B and C.

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=MS$ (8)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $270 kN-m$ for $M$ and $2080×103 mm3$ for $S$ in Equation (8).

$σm=270 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =270×1032.08×10−3=129,807,692.3 Pa(1 Mpa106 Pa)$

$σm=129.808 Pa$

Find the value of $σb$ as follows:

$σb=ybcσm$ (9)

Substitute $129.808 MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (9).

$σb=250.9266.5 ×129.808=0.9414×129.808=122.20 MPa$

Find the shear stress at b $(τb)$ as follows:

$τb=VQIt$ (10)

Substitute $Afy¯$ for Q in Equation (10).

$τb=VAfy¯Itw$ (11)

Substitute $82.5 kN$ for V, $3,260.4 mm2$ for $Af$ , $258.7mm$ for $y¯$ , $554×10−6 m4$ for I, and $10.2 mm$ for $tw$ in Equation (11).

$τb=82.5 kN(103 N1 kN)×3,260.4 mm2(1 m2106 mm2)×258.7mm(1 m103 mm)554×10−6 ×10.2 mm(1 m103 mm)=82.5×103×3.26×10−3×0.2587554×10−6 ×0.0102 =12,312,834.4 Pa(1 MPa106 Pa)=12.3143 MPa$

Find the maximum shearing stress (R) using the relation:

$R=(σb2)2+τb2$ (12)

Here, $σb$ is normal bearing stress and $τm$ is the shearing stress.

Substitute $122.20 MPa$ for $σb$ and $12.3143 MPa$ for $τb$ in Equation (12).

$R=(122.20 2)2+(12.3143)2=(3,733.2+151.641)=62.32 MPa$

Determine the maximum value of the principle stress using the relation:

$σmax=σb2+R$ (13)

Here, R is the maximum shearing stress and $σb$ is normal bearing stress.

Substitute $122.20 MPa$ for $σb$ and $62.32 MPa$ for R in Equation (13).

$σmax=122.20 MPa2+62.32=61.1+62.32=123.4 MPa$

Based on results,

Select the maximum value of principal stress $σmax$ .

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

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Answer 2

Textbook Question

Chapter 8.2, Problem 9P

8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement σ_m ≤ σ_all. For the selected design, determine (a) the actual value of σ_m in the beam, (b) the maximum value of the principal stress σ_max at the junction of a flange and the web.

8.9 Loading of Prob. 5.73 and selected W530 × 92 shape.

Fig. P5.73

Chapter 8.2, Problem 9P, 8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem

(a)

Expert Solution

To determine

The actual value of $σm$ in the beam.

Answer to Problem 9P

The actual value of $σm$ in the beam is $137.5 MPa_$ .

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $W530×92$ .

Calculation:

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

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 1

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

$∑MA=0−70(3)−(5×11)×112−70(8)+Dy(11)=0−210−302.5−560+11Dy=0Dy=97.5 kN$

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

$∑Fy=0Ay−70−70−(5×11)+Dy=0Ay−140−55+97.5=0Ay=97.5 kN$

Shear force:

Show the calculation of shear force as follows;

$VA=97.5 kN$

$VB−VA=−5×3VBLeft=−15+VA=−15+97.5=82.5 kN$

$VBRight=82.5−70=12.5 kN$

$VC−VB=−5×5−70VCLeft=−95+VB=−95+82.5=−12.5 kN$

$VCRight=−12.5−70=−82.5 kN$

$VD−VC=−5×3−70VD=−15−70+VC=−85−12.5=−97.5 kN$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 2

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$12.5x=12.55−x62.5−12.5x=12.5x25x=62.5x=2.5 m$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$MA=0$

$MB−MA=(82.5×3)+12×(97.5−82.5)×3MB=247.5+22.5+MA=270+0=270 kN-m$

$Mmax−MB=12×12.5×2.5Mmax=15.625+MB=15.625+270=285.625 kN-m$

$MC−Mmax=−12×12.5×2.5MC=−15.625+Mmax=−15.625+285.625=270 kN-m$

$MD=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 3

Refer to the Figure 3;

The maximum bending moment in the beam is $|Mmax|=285.625 kN-m$ .

Write the section a property for a $W530×92$ rolled steel section as table 2.

Dimension	Unit( $mm$ )
d	533 mm
$bf$	209 mm
$tf$	15.6 mm
$tw$	$10.2 mm$
I	$554×106 mm4$
$S$	$2080×103 mm3$

Here, d is depth of the section, $bf$ is breadth of the flange, $tf$ is thickness of the flange, $Ix$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=12d$ (1)

Substitute $533 mm$ for d in Equation (1).

$c=12×533=266.5 mm$

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=|M|maxS$ (2)

Here, $|M|max$ is maximum bending moment and $S$ is the section modulus.

Substitute $285.625 kN-m$ for $|M|max$ and $2080×103 mm3$ for $S$ in Equation (2).

$σm=285.625 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =285.625×1032.08×10−3=137,319,711.5 Pa(1 Mpa106 Pa)$

$σm=137.5 Pa$

Thus, the actual value of $σm$ in the beam is $137.5 MPa_$ .

(b)

Expert Solution

To determine

The maximum value of principal stress $σmax$ at the junction of a flange and the web.

Answer to Problem 9P

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Explanation of Solution

Calculation:

Find the value $yb$ as follows:

$yb=c−tf$ (3)

Here, c is the centroid and $tf$ is the thickness of flange.

Substitute $266.5 mm$ for c and $15.6 mm$ for $tf$ in Equation (3).

$yb=266.5−15.6=250.9 mm$

Find the area of flange $(Af)$ of section using the relation:

$Af=bftf$ (4)

Here, $bf$ is the width of the flange and $tf$ is the thickness of the flange.

Substitute $209 mm$ For $bf$ and $15.6 mm$ for $tf$ in Equation (4).

$Af=209×15.6=3,260.4 mm2$

Find the centroid of flange $y¯$ using the relation:

$y¯=12(c+yb)$ (5)

Substitute $266.5 mm$ for c and $250.9 mm$ for $yb$ in Equation (5).

$y¯=12(266.5 +250.9)=258.7mm$

Find the first moment about neutral axis $(Q)$ as follows:

$(Q)=Afy¯$ (6)

Here, $Af$ is the area of flange and $y¯$ is the centroid.

Substitute $3,260.4 mm2$ for $Af$ and $258.7mm$ for $y¯$ in Equation (6).

$(Q)=3,260.4 ×258.7=843.47×103 mm3$

At mid span the value of $V=0$ and $τb=0$ .

Find the maximum value of principal stress $σmax$ as follows:

$σb=ybcσm$ (7)

Here, actual value of normal stress $yb$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $137.5MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (7).

$σb=250.9266.5 ×137.5=129.45 MPa$

At section B and C.

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=MS$ (8)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $270 kN-m$ for $M$ and $2080×103 mm3$ for $S$ in Equation (8).

$σm=270 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =270×1032.08×10−3=129,807,692.3 Pa(1 Mpa106 Pa)$

$σm=129.808 Pa$

Find the value of $σb$ as follows:

$σb=ybcσm$ (9)

Substitute $129.808 MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (9).

$σb=250.9266.5 ×129.808=0.9414×129.808=122.20 MPa$

Find the shear stress at b $(τb)$ as follows:

$τb=VQIt$ (10)

Substitute $Afy¯$ for Q in Equation (10).

$τb=VAfy¯Itw$ (11)

Substitute $82.5 kN$ for V, $3,260.4 mm2$ for $Af$ , $258.7mm$ for $y¯$ , $554×10−6 m4$ for I, and $10.2 mm$ for $tw$ in Equation (11).

$τb=82.5 kN(103 N1 kN)×3,260.4 mm2(1 m2106 mm2)×258.7mm(1 m103 mm)554×10−6 ×10.2 mm(1 m103 mm)=82.5×103×3.26×10−3×0.2587554×10−6 ×0.0102 =12,312,834.4 Pa(1 MPa106 Pa)=12.3143 MPa$

Find the maximum shearing stress (R) using the relation:

$R=(σb2)2+τb2$ (12)

Here, $σb$ is normal bearing stress and $τm$ is the shearing stress.

Substitute $122.20 MPa$ for $σb$ and $12.3143 MPa$ for $τb$ in Equation (12).

$R=(122.20 2)2+(12.3143)2=(3,733.2+151.641)=62.32 MPa$

Determine the maximum value of the principle stress using the relation:

$σmax=σb2+R$ (13)

Here, R is the maximum shearing stress and $σb$ is normal bearing stress.

Substitute $122.20 MPa$ for $σb$ and $62.32 MPa$ for R in Equation (13).

$σmax=122.20 MPa2+62.32=61.1+62.32=123.4 MPa$

Based on results,

Select the maximum value of principal stress $σmax$ .

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

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Answer 3

Textbook Question

Chapter 8.2, Problem 9P

8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement σ_m ≤ σ_all. For the selected design, determine (a) the actual value of σ_m in the beam, (b) the maximum value of the principal stress σ_max at the junction of a flange and the web.

8.9 Loading of Prob. 5.73 and selected W530 × 92 shape.

Fig. P5.73

Chapter 8.2, Problem 9P, 8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem

(a)

Expert Solution

To determine

The actual value of $σm$ in the beam.

Answer to Problem 9P

The actual value of $σm$ in the beam is $137.5 MPa_$ .

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $W530×92$ .

Calculation:

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

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 1

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

$∑MA=0−70(3)−(5×11)×112−70(8)+Dy(11)=0−210−302.5−560+11Dy=0Dy=97.5 kN$

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

$∑Fy=0Ay−70−70−(5×11)+Dy=0Ay−140−55+97.5=0Ay=97.5 kN$

Shear force:

Show the calculation of shear force as follows;

$VA=97.5 kN$

$VB−VA=−5×3VBLeft=−15+VA=−15+97.5=82.5 kN$

$VBRight=82.5−70=12.5 kN$

$VC−VB=−5×5−70VCLeft=−95+VB=−95+82.5=−12.5 kN$

$VCRight=−12.5−70=−82.5 kN$

$VD−VC=−5×3−70VD=−15−70+VC=−85−12.5=−97.5 kN$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 2

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$12.5x=12.55−x62.5−12.5x=12.5x25x=62.5x=2.5 m$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$MA=0$

$MB−MA=(82.5×3)+12×(97.5−82.5)×3MB=247.5+22.5+MA=270+0=270 kN-m$

$Mmax−MB=12×12.5×2.5Mmax=15.625+MB=15.625+270=285.625 kN-m$

$MC−Mmax=−12×12.5×2.5MC=−15.625+Mmax=−15.625+285.625=270 kN-m$

$MD=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 3

Refer to the Figure 3;

The maximum bending moment in the beam is $|Mmax|=285.625 kN-m$ .

Write the section a property for a $W530×92$ rolled steel section as table 2.

Dimension	Unit( $mm$ )
d	533 mm
$bf$	209 mm
$tf$	15.6 mm
$tw$	$10.2 mm$
I	$554×106 mm4$
$S$	$2080×103 mm3$

Here, d is depth of the section, $bf$ is breadth of the flange, $tf$ is thickness of the flange, $Ix$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=12d$ (1)

Substitute $533 mm$ for d in Equation (1).

$c=12×533=266.5 mm$

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=|M|maxS$ (2)

Here, $|M|max$ is maximum bending moment and $S$ is the section modulus.

Substitute $285.625 kN-m$ for $|M|max$ and $2080×103 mm3$ for $S$ in Equation (2).

$σm=285.625 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =285.625×1032.08×10−3=137,319,711.5 Pa(1 Mpa106 Pa)$

$σm=137.5 Pa$

Thus, the actual value of $σm$ in the beam is $137.5 MPa_$ .

(b)

Expert Solution

To determine

The maximum value of principal stress $σmax$ at the junction of a flange and the web.

Answer to Problem 9P

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Explanation of Solution

Calculation:

Find the value $yb$ as follows:

$yb=c−tf$ (3)

Here, c is the centroid and $tf$ is the thickness of flange.

Substitute $266.5 mm$ for c and $15.6 mm$ for $tf$ in Equation (3).

$yb=266.5−15.6=250.9 mm$

Find the area of flange $(Af)$ of section using the relation:

$Af=bftf$ (4)

Here, $bf$ is the width of the flange and $tf$ is the thickness of the flange.

Substitute $209 mm$ For $bf$ and $15.6 mm$ for $tf$ in Equation (4).

$Af=209×15.6=3,260.4 mm2$

Find the centroid of flange $y¯$ using the relation:

$y¯=12(c+yb)$ (5)

Substitute $266.5 mm$ for c and $250.9 mm$ for $yb$ in Equation (5).

$y¯=12(266.5 +250.9)=258.7mm$

Find the first moment about neutral axis $(Q)$ as follows:

$(Q)=Afy¯$ (6)

Here, $Af$ is the area of flange and $y¯$ is the centroid.

Substitute $3,260.4 mm2$ for $Af$ and $258.7mm$ for $y¯$ in Equation (6).

$(Q)=3,260.4 ×258.7=843.47×103 mm3$

At mid span the value of $V=0$ and $τb=0$ .

Find the maximum value of principal stress $σmax$ as follows:

$σb=ybcσm$ (7)

Here, actual value of normal stress $yb$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $137.5MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (7).

$σb=250.9266.5 ×137.5=129.45 MPa$

At section B and C.

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=MS$ (8)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $270 kN-m$ for $M$ and $2080×103 mm3$ for $S$ in Equation (8).

$σm=270 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =270×1032.08×10−3=129,807,692.3 Pa(1 Mpa106 Pa)$

$σm=129.808 Pa$

Find the value of $σb$ as follows:

$σb=ybcσm$ (9)

Substitute $129.808 MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (9).

$σb=250.9266.5 ×129.808=0.9414×129.808=122.20 MPa$

Find the shear stress at b $(τb)$ as follows:

$τb=VQIt$ (10)

Substitute $Afy¯$ for Q in Equation (10).

$τb=VAfy¯Itw$ (11)

Substitute $82.5 kN$ for V, $3,260.4 mm2$ for $Af$ , $258.7mm$ for $y¯$ , $554×10−6 m4$ for I, and $10.2 mm$ for $tw$ in Equation (11).

$τb=82.5 kN(103 N1 kN)×3,260.4 mm2(1 m2106 mm2)×258.7mm(1 m103 mm)554×10−6 ×10.2 mm(1 m103 mm)=82.5×103×3.26×10−3×0.2587554×10−6 ×0.0102 =12,312,834.4 Pa(1 MPa106 Pa)=12.3143 MPa$

Find the maximum shearing stress (R) using the relation:

$R=(σb2)2+τb2$ (12)

Here, $σb$ is normal bearing stress and $τm$ is the shearing stress.

Substitute $122.20 MPa$ for $σb$ and $12.3143 MPa$ for $τb$ in Equation (12).

$R=(122.20 2)2+(12.3143)2=(3,733.2+151.641)=62.32 MPa$

Determine the maximum value of the principle stress using the relation:

$σmax=σb2+R$ (13)

Here, R is the maximum shearing stress and $σb$ is normal bearing stress.

Substitute $122.20 MPa$ for $σb$ and $62.32 MPa$ for R in Equation (13).

$σmax=122.20 MPa2+62.32=61.1+62.32=123.4 MPa$

Based on results,

Select the maximum value of principal stress $σmax$ .

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

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Answer 4

Textbook Question

Chapter 8.2, Problem 9P

8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem of Chap. 5 to support a given loading at a minimal cost while satisfying the requirement σ_m ≤ σ_all. For the selected design, determine (a) the actual value of σ_m in the beam, (b) the maximum value of the principal stress σ_max at the junction of a flange and the web.

8.9 Loading of Prob. 5.73 and selected W530 × 92 shape.

Fig. P5.73

Chapter 8.2, Problem 9P, 8.9 through 8.14 Each of the following problems refers to a rolled-steel shape selected in a problem

(a)

Expert Solution

To determine

The actual value of $σm$ in the beam.

Answer to Problem 9P

The actual value of $σm$ in the beam is $137.5 MPa_$ .

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $W530×92$ .

Calculation:

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

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 1

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

$∑MA=0−70(3)−(5×11)×112−70(8)+Dy(11)=0−210−302.5−560+11Dy=0Dy=97.5 kN$

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

$∑Fy=0Ay−70−70−(5×11)+Dy=0Ay−140−55+97.5=0Ay=97.5 kN$

Shear force:

Show the calculation of shear force as follows;

$VA=97.5 kN$

$VB−VA=−5×3VBLeft=−15+VA=−15+97.5=82.5 kN$

$VBRight=82.5−70=12.5 kN$

$VC−VB=−5×5−70VCLeft=−95+VB=−95+82.5=−12.5 kN$

$VCRight=−12.5−70=−82.5 kN$

$VD−VC=−5×3−70VD=−15−70+VC=−85−12.5=−97.5 kN$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 2

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$12.5x=12.55−x62.5−12.5x=12.5x25x=62.5x=2.5 m$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$MA=0$

$MB−MA=(82.5×3)+12×(97.5−82.5)×3MB=247.5+22.5+MA=270+0=270 kN-m$

$Mmax−MB=12×12.5×2.5Mmax=15.625+MB=15.625+270=285.625 kN-m$

$MC−Mmax=−12×12.5×2.5MC=−15.625+Mmax=−15.625+285.625=270 kN-m$

$MD=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 3

Refer to the Figure 3;

The maximum bending moment in the beam is $|Mmax|=285.625 kN-m$ .

Write the section a property for a $W530×92$ rolled steel section as table 2.

Dimension	Unit( $mm$ )
d	533 mm
$bf$	209 mm
$tf$	15.6 mm
$tw$	$10.2 mm$
I	$554×106 mm4$
$S$	$2080×103 mm3$

Here, d is depth of the section, $bf$ is breadth of the flange, $tf$ is thickness of the flange, $Ix$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=12d$ (1)

Substitute $533 mm$ for d in Equation (1).

$c=12×533=266.5 mm$

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=|M|maxS$ (2)

Here, $|M|max$ is maximum bending moment and $S$ is the section modulus.

Substitute $285.625 kN-m$ for $|M|max$ and $2080×103 mm3$ for $S$ in Equation (2).

$σm=285.625 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =285.625×1032.08×10−3=137,319,711.5 Pa(1 Mpa106 Pa)$

$σm=137.5 Pa$

Thus, the actual value of $σm$ in the beam is $137.5 MPa_$ .

(b)

Expert Solution

To determine

The maximum value of principal stress $σmax$ at the junction of a flange and the web.

Answer to Problem 9P

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Explanation of Solution

Calculation:

Find the value $yb$ as follows:

$yb=c−tf$ (3)

Here, c is the centroid and $tf$ is the thickness of flange.

Substitute $266.5 mm$ for c and $15.6 mm$ for $tf$ in Equation (3).

$yb=266.5−15.6=250.9 mm$

Find the area of flange $(Af)$ of section using the relation:

$Af=bftf$ (4)

Here, $bf$ is the width of the flange and $tf$ is the thickness of the flange.

Substitute $209 mm$ For $bf$ and $15.6 mm$ for $tf$ in Equation (4).

$Af=209×15.6=3,260.4 mm2$

Find the centroid of flange $y¯$ using the relation:

$y¯=12(c+yb)$ (5)

Substitute $266.5 mm$ for c and $250.9 mm$ for $yb$ in Equation (5).

$y¯=12(266.5 +250.9)=258.7mm$

Find the first moment about neutral axis $(Q)$ as follows:

$(Q)=Afy¯$ (6)

Here, $Af$ is the area of flange and $y¯$ is the centroid.

Substitute $3,260.4 mm2$ for $Af$ and $258.7mm$ for $y¯$ in Equation (6).

$(Q)=3,260.4 ×258.7=843.47×103 mm3$

At mid span the value of $V=0$ and $τb=0$ .

Find the maximum value of principal stress $σmax$ as follows:

$σb=ybcσm$ (7)

Here, actual value of normal stress $yb$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $137.5MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (7).

$σb=250.9266.5 ×137.5=129.45 MPa$

At section B and C.

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=MS$ (8)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $270 kN-m$ for $M$ and $2080×103 mm3$ for $S$ in Equation (8).

$σm=270 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =270×1032.08×10−3=129,807,692.3 Pa(1 Mpa106 Pa)$

$σm=129.808 Pa$

Find the value of $σb$ as follows:

$σb=ybcσm$ (9)

Substitute $129.808 MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (9).

$σb=250.9266.5 ×129.808=0.9414×129.808=122.20 MPa$

Find the shear stress at b $(τb)$ as follows:

$τb=VQIt$ (10)

Substitute $Afy¯$ for Q in Equation (10).

$τb=VAfy¯Itw$ (11)

Substitute $82.5 kN$ for V, $3,260.4 mm2$ for $Af$ , $258.7mm$ for $y¯$ , $554×10−6 m4$ for I, and $10.2 mm$ for $tw$ in Equation (11).

$τb=82.5 kN(103 N1 kN)×3,260.4 mm2(1 m2106 mm2)×258.7mm(1 m103 mm)554×10−6 ×10.2 mm(1 m103 mm)=82.5×103×3.26×10−3×0.2587554×10−6 ×0.0102 =12,312,834.4 Pa(1 MPa106 Pa)=12.3143 MPa$

Find the maximum shearing stress (R) using the relation:

$R=(σb2)2+τb2$ (12)

Here, $σb$ is normal bearing stress and $τm$ is the shearing stress.

Substitute $122.20 MPa$ for $σb$ and $12.3143 MPa$ for $τb$ in Equation (12).

$R=(122.20 2)2+(12.3143)2=(3,733.2+151.641)=62.32 MPa$

Determine the maximum value of the principle stress using the relation:

$σmax=σb2+R$ (13)

Here, R is the maximum shearing stress and $σb$ is normal bearing stress.

Substitute $122.20 MPa$ for $σb$ and $62.32 MPa$ for R in Equation (13).

$σmax=122.20 MPa2+62.32=61.1+62.32=123.4 MPa$

Based on results,

Select the maximum value of principal stress $σmax$ .

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The actual value of $σm$ in the beam.

Answer to Problem 9P

The actual value of $σm$ in the beam is $137.5 MPa_$ .

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $W530×92$ .

Calculation:

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

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 1

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

$∑MA=0−70(3)−(5×11)×112−70(8)+Dy(11)=0−210−302.5−560+11Dy=0Dy=97.5 kN$

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

$∑Fy=0Ay−70−70−(5×11)+Dy=0Ay−140−55+97.5=0Ay=97.5 kN$

Shear force:

Show the calculation of shear force as follows;

$VA=97.5 kN$

$VB−VA=−5×3VBLeft=−15+VA=−15+97.5=82.5 kN$

$VBRight=82.5−70=12.5 kN$

$VC−VB=−5×5−70VCLeft=−95+VB=−95+82.5=−12.5 kN$

$VCRight=−12.5−70=−82.5 kN$

$VD−VC=−5×3−70VD=−15−70+VC=−85−12.5=−97.5 kN$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 2

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$12.5x=12.55−x62.5−12.5x=12.5x25x=62.5x=2.5 m$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$MA=0$

$MB−MA=(82.5×3)+12×(97.5−82.5)×3MB=247.5+22.5+MA=270+0=270 kN-m$

$Mmax−MB=12×12.5×2.5Mmax=15.625+MB=15.625+270=285.625 kN-m$

$MC−Mmax=−12×12.5×2.5MC=−15.625+Mmax=−15.625+285.625=270 kN-m$

$MD=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 3

Refer to the Figure 3;

The maximum bending moment in the beam is $|Mmax|=285.625 kN-m$ .

Write the section a property for a $W530×92$ rolled steel section as table 2.

Dimension	Unit( $mm$ )
d	533 mm
$bf$	209 mm
$tf$	15.6 mm
$tw$	$10.2 mm$
I	$554×106 mm4$
$S$	$2080×103 mm3$

Here, d is depth of the section, $bf$ is breadth of the flange, $tf$ is thickness of the flange, $Ix$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=12d$ (1)

Substitute $533 mm$ for d in Equation (1).

$c=12×533=266.5 mm$

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=|M|maxS$ (2)

Here, $|M|max$ is maximum bending moment and $S$ is the section modulus.

Substitute $285.625 kN-m$ for $|M|max$ and $2080×103 mm3$ for $S$ in Equation (2).

$σm=285.625 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =285.625×1032.08×10−3=137,319,711.5 Pa(1 Mpa106 Pa)$

$σm=137.5 Pa$

Thus, the actual value of $σm$ in the beam is $137.5 MPa_$ .

(b)

Expert Solution

To determine

The maximum value of principal stress $σmax$ at the junction of a flange and the web.

Answer to Problem 9P

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Explanation of Solution

Calculation:

Find the value $yb$ as follows:

$yb=c−tf$ (3)

Here, c is the centroid and $tf$ is the thickness of flange.

Substitute $266.5 mm$ for c and $15.6 mm$ for $tf$ in Equation (3).

$yb=266.5−15.6=250.9 mm$

Find the area of flange $(Af)$ of section using the relation:

$Af=bftf$ (4)

Here, $bf$ is the width of the flange and $tf$ is the thickness of the flange.

Substitute $209 mm$ For $bf$ and $15.6 mm$ for $tf$ in Equation (4).

$Af=209×15.6=3,260.4 mm2$

Find the centroid of flange $y¯$ using the relation:

$y¯=12(c+yb)$ (5)

Substitute $266.5 mm$ for c and $250.9 mm$ for $yb$ in Equation (5).

$y¯=12(266.5 +250.9)=258.7mm$

Find the first moment about neutral axis $(Q)$ as follows:

$(Q)=Afy¯$ (6)

Here, $Af$ is the area of flange and $y¯$ is the centroid.

Substitute $3,260.4 mm2$ for $Af$ and $258.7mm$ for $y¯$ in Equation (6).

$(Q)=3,260.4 ×258.7=843.47×103 mm3$

At mid span the value of $V=0$ and $τb=0$ .

Find the maximum value of principal stress $σmax$ as follows:

$σb=ybcσm$ (7)

Here, actual value of normal stress $yb$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $137.5MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (7).

$σb=250.9266.5 ×137.5=129.45 MPa$

At section B and C.

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=MS$ (8)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $270 kN-m$ for $M$ and $2080×103 mm3$ for $S$ in Equation (8).

$σm=270 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =270×1032.08×10−3=129,807,692.3 Pa(1 Mpa106 Pa)$

$σm=129.808 Pa$

Find the value of $σb$ as follows:

$σb=ybcσm$ (9)

Substitute $129.808 MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (9).

$σb=250.9266.5 ×129.808=0.9414×129.808=122.20 MPa$

Find the shear stress at b $(τb)$ as follows:

$τb=VQIt$ (10)

Substitute $Afy¯$ for Q in Equation (10).

$τb=VAfy¯Itw$ (11)

Substitute $82.5 kN$ for V, $3,260.4 mm2$ for $Af$ , $258.7mm$ for $y¯$ , $554×10−6 m4$ for I, and $10.2 mm$ for $tw$ in Equation (11).

$τb=82.5 kN(103 N1 kN)×3,260.4 mm2(1 m2106 mm2)×258.7mm(1 m103 mm)554×10−6 ×10.2 mm(1 m103 mm)=82.5×103×3.26×10−3×0.2587554×10−6 ×0.0102 =12,312,834.4 Pa(1 MPa106 Pa)=12.3143 MPa$

Find the maximum shearing stress (R) using the relation:

$R=(σb2)2+τb2$ (12)

Here, $σb$ is normal bearing stress and $τm$ is the shearing stress.

Substitute $122.20 MPa$ for $σb$ and $12.3143 MPa$ for $τb$ in Equation (12).

$R=(122.20 2)2+(12.3143)2=(3,733.2+151.641)=62.32 MPa$

Determine the maximum value of the principle stress using the relation:

$σmax=σb2+R$ (13)

Here, R is the maximum shearing stress and $σb$ is normal bearing stress.

Substitute $122.20 MPa$ for $σb$ and $62.32 MPa$ for R in Equation (13).

$σmax=122.20 MPa2+62.32=61.1+62.32=123.4 MPa$

Based on results,

Select the maximum value of principal stress $σmax$ .

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Answer 8

(a)

Expert Solution

To determine

The actual value of $σm$ in the beam.

Answer to Problem 9P

The actual value of $σm$ in the beam is $137.5 MPa_$ .

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $W530×92$ .

Calculation:

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

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 1

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

$∑MA=0−70(3)−(5×11)×112−70(8)+Dy(11)=0−210−302.5−560+11Dy=0Dy=97.5 kN$

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

$∑Fy=0Ay−70−70−(5×11)+Dy=0Ay−140−55+97.5=0Ay=97.5 kN$

Shear force:

Show the calculation of shear force as follows;

$VA=97.5 kN$

$VB−VA=−5×3VBLeft=−15+VA=−15+97.5=82.5 kN$

$VBRight=82.5−70=12.5 kN$

$VC−VB=−5×5−70VCLeft=−95+VB=−95+82.5=−12.5 kN$

$VCRight=−12.5−70=−82.5 kN$

$VD−VC=−5×3−70VD=−15−70+VC=−85−12.5=−97.5 kN$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 2

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$12.5x=12.55−x62.5−12.5x=12.5x25x=62.5x=2.5 m$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$MA=0$

$MB−MA=(82.5×3)+12×(97.5−82.5)×3MB=247.5+22.5+MA=270+0=270 kN-m$

$Mmax−MB=12×12.5×2.5Mmax=15.625+MB=15.625+270=285.625 kN-m$

$MC−Mmax=−12×12.5×2.5MC=−15.625+Mmax=−15.625+285.625=270 kN-m$

$MD=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 3

Refer to the Figure 3;

The maximum bending moment in the beam is $|Mmax|=285.625 kN-m$ .

Write the section a property for a $W530×92$ rolled steel section as table 2.

Dimension	Unit( $mm$ )
d	533 mm
$bf$	209 mm
$tf$	15.6 mm
$tw$	$10.2 mm$
I	$554×106 mm4$
$S$	$2080×103 mm3$

Here, d is depth of the section, $bf$ is breadth of the flange, $tf$ is thickness of the flange, $Ix$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=12d$ (1)

Substitute $533 mm$ for d in Equation (1).

$c=12×533=266.5 mm$

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=|M|maxS$ (2)

Here, $|M|max$ is maximum bending moment and $S$ is the section modulus.

Substitute $285.625 kN-m$ for $|M|max$ and $2080×103 mm3$ for $S$ in Equation (2).

$σm=285.625 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =285.625×1032.08×10−3=137,319,711.5 Pa(1 Mpa106 Pa)$

$σm=137.5 Pa$

Thus, the actual value of $σm$ in the beam is $137.5 MPa_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The actual value of $σm$ in the beam.

Answer 13

Answer to Problem 9P

The actual value of $σm$ in the beam is $137.5 MPa_$ .

Answer 14

Explanation of Solution

Given information:

Refer to problem 5.73 in chapter 5 in the textbook.

The shape of the rolled steel section is $W530×92$ .

Calculation:

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

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 1

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

$∑MA=0−70(3)−(5×11)×112−70(8)+Dy(11)=0−210−302.5−560+11Dy=0Dy=97.5 kN$

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

$∑Fy=0Ay−70−70−(5×11)+Dy=0Ay−140−55+97.5=0Ay=97.5 kN$

Shear force:

Show the calculation of shear force as follows;

$VA=97.5 kN$

$VB−VA=−5×3VBLeft=−15+VA=−15+97.5=82.5 kN$

$VBRight=82.5−70=12.5 kN$

$VC−VB=−5×5−70VCLeft=−95+VB=−95+82.5=−12.5 kN$

$VCRight=−12.5−70=−82.5 kN$

$VD−VC=−5×3−70VD=−15−70+VC=−85−12.5=−97.5 kN$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) kN
A	97.5
B (Left)	82.5
B (Right)	12.5
C (Left)	–12.5
C (Right)	–82.5
D	–97.5

Plot the shear force diagram as in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 2

Location of the maximum bending moment:

The maximum bending moment occurs where the shear force changes sign.

Refer to Figure 2;

Use the method of similar triangle.

$12.5x=12.55−x62.5−12.5x=12.5x25x=62.5x=2.5 m$

The maximum bending moment occurs at a distance 5.5 m from the left end of the beam.

Bending moment:

Show the calculation of the bending moment as follows;

$MA=0$

$MB−MA=(82.5×3)+12×(97.5−82.5)×3MB=247.5+22.5+MA=270+0=270 kN-m$

$Mmax−MB=12×12.5×2.5Mmax=15.625+MB=15.625+270=285.625 kN-m$

$MC−Mmax=−12×12.5×2.5MC=−15.625+Mmax=−15.625+285.625=270 kN-m$

$MD=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) kN-m
A	0
B	270
Max BM	285.625
C	270
D	0

Plot the bending moment diagram as in Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 8.2, Problem 9P , additional homework tip 3

Refer to the Figure 3;

The maximum bending moment in the beam is $|Mmax|=285.625 kN-m$ .

Write the section a property for a $W530×92$ rolled steel section as table 2.

Dimension	Unit( $mm$ )
d	533 mm
$bf$	209 mm
$tf$	15.6 mm
$tw$	$10.2 mm$
I	$554×106 mm4$
$S$	$2080×103 mm3$

Here, d is depth of the section, $bf$ is breadth of the flange, $tf$ is thickness of the flange, $Ix$ is moment of inertia about x-axis, and $S$ is section modulus.

Find the value of C using the relation:

$c=12d$ (1)

Substitute $533 mm$ for d in Equation (1).

$c=12×533=266.5 mm$

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=|M|maxS$ (2)

Here, $|M|max$ is maximum bending moment and $S$ is the section modulus.

Substitute $285.625 kN-m$ for $|M|max$ and $2080×103 mm3$ for $S$ in Equation (2).

$σm=285.625 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =285.625×1032.08×10−3=137,319,711.5 Pa(1 Mpa106 Pa)$

$σm=137.5 Pa$

Thus, the actual value of $σm$ in the beam is $137.5 MPa_$ .

Answer 15

(b)

Expert Solution

To determine

The maximum value of principal stress $σmax$ at the junction of a flange and the web.

Answer to Problem 9P

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Explanation of Solution

Calculation:

Find the value $yb$ as follows:

$yb=c−tf$ (3)

Here, c is the centroid and $tf$ is the thickness of flange.

Substitute $266.5 mm$ for c and $15.6 mm$ for $tf$ in Equation (3).

$yb=266.5−15.6=250.9 mm$

Find the area of flange $(Af)$ of section using the relation:

$Af=bftf$ (4)

Here, $bf$ is the width of the flange and $tf$ is the thickness of the flange.

Substitute $209 mm$ For $bf$ and $15.6 mm$ for $tf$ in Equation (4).

$Af=209×15.6=3,260.4 mm2$

Find the centroid of flange $y¯$ using the relation:

$y¯=12(c+yb)$ (5)

Substitute $266.5 mm$ for c and $250.9 mm$ for $yb$ in Equation (5).

$y¯=12(266.5 +250.9)=258.7mm$

Find the first moment about neutral axis $(Q)$ as follows:

$(Q)=Afy¯$ (6)

Here, $Af$ is the area of flange and $y¯$ is the centroid.

Substitute $3,260.4 mm2$ for $Af$ and $258.7mm$ for $y¯$ in Equation (6).

$(Q)=3,260.4 ×258.7=843.47×103 mm3$

At mid span the value of $V=0$ and $τb=0$ .

Find the maximum value of principal stress $σmax$ as follows:

$σb=ybcσm$ (7)

Here, actual value of normal stress $yb$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $137.5MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (7).

$σb=250.9266.5 ×137.5=129.45 MPa$

At section B and C.

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=MS$ (8)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $270 kN-m$ for $M$ and $2080×103 mm3$ for $S$ in Equation (8).

$σm=270 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =270×1032.08×10−3=129,807,692.3 Pa(1 Mpa106 Pa)$

$σm=129.808 Pa$

Find the value of $σb$ as follows:

$σb=ybcσm$ (9)

Substitute $129.808 MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (9).

$σb=250.9266.5 ×129.808=0.9414×129.808=122.20 MPa$

Find the shear stress at b $(τb)$ as follows:

$τb=VQIt$ (10)

Substitute $Afy¯$ for Q in Equation (10).

$τb=VAfy¯Itw$ (11)

Substitute $82.5 kN$ for V, $3,260.4 mm2$ for $Af$ , $258.7mm$ for $y¯$ , $554×10−6 m4$ for I, and $10.2 mm$ for $tw$ in Equation (11).

$τb=82.5 kN(103 N1 kN)×3,260.4 mm2(1 m2106 mm2)×258.7mm(1 m103 mm)554×10−6 ×10.2 mm(1 m103 mm)=82.5×103×3.26×10−3×0.2587554×10−6 ×0.0102 =12,312,834.4 Pa(1 MPa106 Pa)=12.3143 MPa$

Find the maximum shearing stress (R) using the relation:

$R=(σb2)2+τb2$ (12)

Here, $σb$ is normal bearing stress and $τm$ is the shearing stress.

Substitute $122.20 MPa$ for $σb$ and $12.3143 MPa$ for $τb$ in Equation (12).

$R=(122.20 2)2+(12.3143)2=(3,733.2+151.641)=62.32 MPa$

Determine the maximum value of the principle stress using the relation:

$σmax=σb2+R$ (13)

Here, R is the maximum shearing stress and $σb$ is normal bearing stress.

Substitute $122.20 MPa$ for $σb$ and $62.32 MPa$ for R in Equation (13).

$σmax=122.20 MPa2+62.32=61.1+62.32=123.4 MPa$

Based on results,

Select the maximum value of principal stress $σmax$ .

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The maximum value of principal stress $σmax$ at the junction of a flange and the web.

Answer 20

Answer to Problem 9P

The maximum value of principal stress $σmax$ at the junction of a flange and the web is $129.5 MPa_$ .

Answer 21

Explanation of Solution

Calculation:

Find the value $yb$ as follows:

$yb=c−tf$ (3)

Here, c is the centroid and $tf$ is the thickness of flange.

Substitute $266.5 mm$ for c and $15.6 mm$ for $tf$ in Equation (3).

$yb=266.5−15.6=250.9 mm$

Find the area of flange $(Af)$ of section using the relation:

$Af=bftf$ (4)

Here, $bf$ is the width of the flange and $tf$ is the thickness of the flange.

Substitute $209 mm$ For $bf$ and $15.6 mm$ for $tf$ in Equation (4).

$Af=209×15.6=3,260.4 mm2$

Find the centroid of flange $y¯$ using the relation:

$y¯=12(c+yb)$ (5)

Substitute $266.5 mm$ for c and $250.9 mm$ for $yb$ in Equation (5).

$y¯=12(266.5 +250.9)=258.7mm$

Find the first moment about neutral axis $(Q)$ as follows:

$(Q)=Afy¯$ (6)

Here, $Af$ is the area of flange and $y¯$ is the centroid.

Substitute $3,260.4 mm2$ for $Af$ and $258.7mm$ for $y¯$ in Equation (6).

$(Q)=3,260.4 ×258.7=843.47×103 mm3$

At mid span the value of $V=0$ and $τb=0$ .

Find the maximum value of principal stress $σmax$ as follows:

$σb=ybcσm$ (7)

Here, actual value of normal stress $yb$ is distance between centroid of the section to the centre of flange and c is the centroid.

Substitute $137.5MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (7).

$σb=250.9266.5 ×137.5=129.45 MPa$

At section B and C.

Find the maximum value of normal stress $(σm)$ in the beam using the relation:

$σm=MS$ (8)

Here, $M$ is bending moment at B and $S$ is the section property.

Substitute $270 kN-m$ for $M$ and $2080×103 mm3$ for $S$ in Equation (8).

$σm=270 kN-m(103 N-mkN-m)2080×103 mm3(10−9 m31 mm3) =270×1032.08×10−3=129,807,692.3 Pa(1 Mpa106 Pa)$

$σm=129.808 Pa$

Find the value of $σb$ as follows:

$σb=ybcσm$ (9)

Substitute $129.808 MPa$ for $σm$ , $250.9 mm$ for $yb$ , and $266.5 mm$ for $c$ in Equation (9).

$σb=250.9266.5 ×129.808=0.9414×129.808=122.20 MPa$

Find the shear stress at b $(τb)$ as follows:

$τb=VQIt$ (10)

Substitute $Afy¯$ for Q in Equation (10).

$τb=VAfy¯Itw$ (11)

Substitute $82.5 kN$ for V, $3,260.4 mm2$ for $Af$ , $258.7mm$ for $y¯$ , $554×10−6 m4$ for I, and $10.2 mm$ for $tw$ in Equation (11).

$τb=82.5 kN(103 N1 kN)×3,260.4 mm2(1 m2106 mm2)×258.7mm(1 m103 mm)554×10−6 ×10.2 mm(1 m103 mm)=82.5×103×3.26×10−3×0.2587554×10−6 ×0.0102 =12,312,834.4 Pa(1 MPa106 Pa)=12.3143 MPa$