EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 6.6, Problem 80P

(a)

To determine

The points where the shearing stress is maximum and the values of the stress.

(a)

Expert Solution
Check Mark

Answer to Problem 80P

The shearing stress is maximum at the points of y=25a and z=23a_.

The maximum shear stress along the vertical leg is 2720Pta_.

The maximum shear stress along the horizontal leg is 14Pta_.

The maximum shear stress at the corner of the leg  is 34Pta_.

Explanation of Solution

Calculation:

Refer to sample problem 6.6 in the text book.

Combined stress along the vertical leg τe=3P(ay)(a+5y)4ta3 (1)

Combined stress along the horizontal leg τf=3P(az)(a3z)4ta3 (2)

Modify Equation (1).

τe=3P4ta3(a2ay+5ay5y2)=3P4ta3(a2+4ay5y2)

Calculate the point along the vertical leg differentiate both sides of the equation with respect to y as shown below.

dτedy=ddy(3P4ta3(a2+4ay5y2))=3P4ta3(4a10y)

Consider the condition dτedy=0. Hence,

3P4ta3(4a10y)=04a10y=010y=4ay=25a

Hence, the shearing stress is maximum at the points of y=25a_.

Calculate the maximum shear stress along the vertical leg as shown below.

Substitute 25a for y in Equation (1).

τm=3P4ta3(a25a)(a+5×25a)=3P4ta3×35a×3a=2720Pta

Hence, the maximum shear stress along the vertical leg is 2720Pta_.

Modify Equation (2).

τf=3P4ta3(a2az3az+3z2)=3P4ta3(a24az+3z2)

Calculate the point along the horizontal leg differentiate both sides of the equation with respect to z as shown below.

dτfdz=ddz(3P4ta3(a24az+3z2))=3P4ta3(4a+6z)

Consider the condition dτfdy=0. Hence,

3P4ta3(4a+6z)=04a+6z=06z=4az=23a

Hence, the shearing stress is maximum at the points of z=23a_.

Calculate the maximum shear stress along the horizontal leg as shown below.

Substitute 23a for z in Equation (2).

τm=3P4ta3(a23a)(a3×23a)=3P4ta3(13a)(a)=14Pta

Hence, the maximum shear stress along the horizontal leg is 14Pta_.

The corner points of the horizontal and vertical legs are y=0and z=0.

Calculate the maximum shear stress at the corner point as shown below.

Substitute 0 for y in Equation (1).

τm=3P4ta3(a0)(a+5×0)=3P4ta3×a2=34Pta

Substitute 0 for z in Equation (2).

τm=3P4ta3(a0)(a3×0)=3P4ta3a2=34Pta

Therefore, the maximum shear stress at the corner of the leg  is 34Pta_.

(b)

To determine

Show that the points are located on the neutral axis for the loading P.

(b)

Expert Solution
Check Mark

Answer to Problem 80P

The points y and z are located on the neutral axis for the loading P.

Explanation of Solution

Calculation:

Sketch the cross section along the neutral axis as shown below.

EBK MECHANICS OF MATERIALS, Chapter 6.6, Problem 80P

Refer to Figure 1.

Calculate the moment of inertia as shown below.

Along y axis:

Iy=13ta3

Along z axis:

Iz=112ta3

Consider the angle θ=45°.

Calculate angle ϕ as shown below.

tanϕ=IzIytanθ

Substitute 13ta3 for Iy, 112ta3 for Iz, and 45° for θ.

tanϕ=112ta313ta3tan45°tanϕ=14ϕ=tan1(14)ϕ=14.036°

Calculate the angle of the neutral axis from the horizontal as shown below.

θϕ=45°14.036°=30.964°

Calculate the location of the centroid as shown below.

Along y axis:

y¯=Ay¯A=at×a22at=14a

Along z axis:

z¯=Az¯A=at×a22at=14a

Calculate the neutral axis intersects for vertical leg as shown below.

y=y¯+z¯tan36.964°

Substitute 14a for y¯ and 14a for z¯.

y=14a+14a×tan36.964°=0.4a

Calculate the neutral axis intersects for horizontal leg as shown below.

z=z¯+y¯tan(45°+ϕ)

Substitute 14a for y¯, 14a for z¯, and 14.036° for ϕ.

z=14a+14a×tan(45°+14.036°)=14a+14a×tan(59.036°)=0.67a

Therefore, the points y and z are located on the neutral axis for the loading P

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Chapter 6 Solutions

EBK MECHANICS OF MATERIALS

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