EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Textbook Question
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Chapter 6.6, Problem 68P

6.65 through 6.68 An extruded beam has the cross section shown. Determine (a) the location of the shear center O, (b) the distribution of the shearing stresses caused by the vertical shearing force V shown applied at O.

Chapter 6.6, Problem 68P, 6.65 through 6.68 An extruded beam has the cross section shown. Determine (a) the location of the

Fig. p6.67

(a)

Expert Solution
Check Mark
To determine

Find the location of the shear center O.

Answer to Problem 68P

The location of the shear center O is 10.22mm_.

Explanation of Solution

Calculation:

Calculate the moment of inertia as shown below.

I=bd312+A(yy¯)2

Here, b is the width of the section, d is the height of the section, A is the area of the beam, and (yy¯) is the centroid of the beam from the neutral axis.

Calculate the moment of inertia for whole section as shown below.

I=2×[112×30×63+30×6×452]+2×[112×30×43+30×4×152]+112×6×903=730,080+54,320+364,500=1.1489×106mm4

Calculate the forces acting along the member as shown below.

F=qdy (1)

Here, τ is the shear stress.

Sketch the cross section of flange as shown in Figure 1.

EBK MECHANICS OF MATERIALS, Chapter 6.6, Problem 68P , additional homework tip  1

Refer to Figure 1.

Calculate the first moment of area as shown below.

Q=Ay¯ (2)

Calculate the first moment of area for AB as shown below.

Q(x)=yts

Calculate the horizontal shear per unit length as shown below.

q=VQI (3)

Here, V is the vertical shear.

Substitute yts for Q in Equation (3).

q=V×ytsI=VytsI

Calculate the force FAB as shown below.

Substitute VytsI for q and apply the limits in Equation (1).

F=0bVytsIds=VytI(s22)0b=Vytb22I (4)

For flange AB and flange HJ:

Substitute 45mm for y, 6mm for t, 30mm for b, and 1.1489×106mm4 for I in Equation (4).

FAB=FHJ=V×45×6×3022×1.1489×106=0.10575V

For flange DE and flange FG:

Substitute 15mm for y, 4mm for t, 30mm for b, and 1.1489×106mm4 for I in Equation (4).

FDE=FFG=V×15×4×3022×1.1489×106=0.0235V

Sketch the shear flow as shown in Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 6.6, Problem 68P , additional homework tip  2

Refer to Figure 2.

Calculate the eccentricity as shown below.

Ve=45FAB+15FDE+15FFG+45FHJ

Substitute 0.10575V for FAB, 0.0235V for FDE, 0.0235V for FFG, and 0.10575V for FHJ.

Ve=45×0.10575V+15×0.0235V+15×0.0235V+45×0.10575Ve=10.22mm

Therefore, the location of the shear center O is 10.22mm_.

(b)

Expert Solution
Check Mark
To determine

Find the distribution of the shearing stresses caused by the vertical shearing force.

Answer to Problem 68P

The shearing stress at point B, E, G, and J is 0_.

The shearing stress at point A and H is 41.1MPa_.

The shearing stress at point just above D and just below F is 68.5MPa_.

The shearing stress at point just to the right of D and just to the right of F is 13.7MPa_.

The shearing stress at point just below D and just above F is 77.7MPa_.

The shearing stress at point K is 81.1MPa_.

Explanation of Solution

Given information:

The vertical shear is 35kN.

Calculation:

Refer to part (a).

The moment of inertia I=1.1489×106mm4.

Calculate the shear stress as shown below.

τ=VQIt (5)

At point B, E, G, and J:

Calculate the first moment of area as shown below.

Q=0

Hence, the shearing stress at point B, E, G, and J is 0_.

At point A and H:

Calculate the first moment of area as shown below.

Q=30×6×45=8,100mm3

The thickness of the section is 6mm.

Calculate the shear stress as shown below.

Substitute 35kN for V, 8,100mm3 for Q, 6mm for t, and 1.1489×106mm4 for I in Equation (5).

τ=35kN×1,000N1kN×8,100mm31.1489×106mm4×6mm=283.5×1066.8934=41.1N/mm2×1MPa1N/mm2=41.1MPa

Hence, the shearing stress at point A and H is 41.1MPa_.

At point just above D and just below F:

Calculate the first moment of area as shown below.

Q=30×6×45+6×30×30=8,100+5,400=13,500mm3

The thickness of the section is 6mm.

Calculate the shear stress as shown below.

Substitute 35kN for V, 13,500mm3 for Q, 6mm for t, and 1.1489×106mm4 for I in Equation (5).

τ=35kN×1,000N1kN×13,500mm31.1489×106mm4×6mm=472.5×1066.8934=68.5N/mm2×1MPa1N/mm2=68.5MPa

Hence, the shearing stress at point just above D and just below F is 68.5MPa_.

At point just to the right of D and just to the right of F:

Calculate the first moment of area as shown below.

Q=30×4×15=1,800mm3

The thickness of the section is 4mm.

Calculate the shear stress as shown below.

Substitute 35kN for V, 1,800mm3 for Q, 4mm for t, and 1.1489×106mm4 for I in Equation (5).

τ=35kN×1,000N1kN×1,800mm31.1489×106mm4×4mm=63×1064.5956=13.7N/mm2×1MPa1N/mm2=13.7MPa

Hence, the shearing stress at point just to the right of D and just to the right of F is 13.7MPa_.

At point just below D and just above F:

Calculate the first moment of area as shown below.

Q=30×6×45+6×30×30+30×4×15=8,100+5,400+1,800=15,300mm3

The thickness of the section is 6mm.

Calculate the shear stress as shown below.

Substitute 35kN for V, 15,300mm3 for Q, 6mm for t, and 1.1489×106mm4 for I in Equation (5).

τ=35kN×1,000N1kN×15,300mm31.1489×106mm4×6mm=535.5×1066.8934=77.7N/mm2×1MPa1N/mm2=77.7MPa

Hence, the shearing stress at point just below D and just above F is 77.7MPa_.

At point just below D and just above F:

Calculate the first moment of area as shown below.

Q=30×6×45+6×30×30+30×4×15=8,100+5,400+1,800=15,300mm3

The thickness of the section is 6mm.

Calculate the shear stress as shown below.

Substitute 35kN for V, 15,300mm3 for Q, 6mm for t, and 1.1489×106mm4 for I in Equation (5).

τ=35kN×1,000N1kN×15,300mm31.1489×106mm4×6mm=535.5×1066.8934=77.7N/mm2×1MPa1N/mm2=77.7MPa

Hence, the shearing stress at point just below D and just above F is 77.7MPa_.

At point K:

Calculate the first moment of area as shown below.

Q=30×6×45+6×30×30+30×4×15+6×15×7.5=8,100+5,400+1,800+675=15,975mm3

The thickness of the section is 6mm.

Calculate the shear stress as shown below.

Substitute 35kN for V, 15,975mm3 for Q, 6mm for t, and 1.1489×106mm4 for I in Equation (5).

τ=35kN×1,000N1kN×15,975mm31.1489×106mm4×6mm=559.125×1066.8934=81.1N/mm2×1MPa1N/mm2=81.1MPa

Therefore, the shearing stress at point K is 81.1MPa_.

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Chapter 6 Solutions

EBK MECHANICS OF MATERIALS

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