EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 8220100257063
Author: BEER
Publisher: YUZU
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Chapter 1.2, Problem 6P

Two brass rods AB and BC, each of uniform diameter, will be brazed together at B to form a nonuniform rod of total length 100 m that will be suspended from a support at A as shown. Knowing that the density of brass is 8470 kg/m3, determine (a) the length of rod AB for which the maximum normal stress in ABC is minimum, (b) the corresponding value of the maximum normal stress.

Fig. P1.6

Chapter 1.2, Problem 6P, Two brass rods AB and BC, each of uniform diameter, will be brazed together at B to form a

(a)

Expert Solution
Check Mark
To determine

The length of brass rod AB such that the maximum normal stress in ABC is minimum.

Answer to Problem 6P

The length of brass rod AB such that the maximum normal stress in ABC is minimum as 35.7mm_.

Explanation of Solution

Given information:

The total length of the rod is 100m.

The length of rod AB is a.

The length of rod BC is (100a).

The density of the brass is 8,470kg/m3.

The diameter of the rod AB is 15mm.

The diameter of the rod BC is 10mm.

Calculation:

Calculate the area (AAB) of rod AB using the relation:

AAB=π4dAB2 (1)

Here, dAB is diameter of rod AB.

Substitute 15mm for AB in Equation (1).

AAB=π4×(15)2=π4×225=176.715mm2×(1m2106mm2)=176.715×106m2

Calculate the area (ABC) of rod BC using the relation:

ABC=π4dBC2 (2)

Here, dBC is diameter of rod BC.

Substitute 10mm for BC in Equation (2).

ABC=π4×(10)2=π4×100=78.54mm2×(1m2106mm2)=78.54×106m2

Calculate the weights (WAB) of rod AB using the formula:

WAB=ρgAABlAB (3)

Here, ρ is density of brass rod, g is acceleration due to gravity, and lAB is length of rod AB.

Substitute 8,470kg/m3 for ρ, 9.81m/s2 for g, 176.715×106m2 for AAB, and a for lAB in Equation (3).

WAB=8,470×9.81×176.715×106×a=14.683a

Calculate the weights (WBC) of rod BC using the formula:

WBC=ρgABClBC (4)

Here, lBC is length of rod BC.

Substitute 8,470kg/m3 for ρ, 9.81m/s2 for g, 78.54×106m2 for ABC, and (100a) for lBC in Equation (4).

WBC=8,470×9.81×78.54×106×(100a)=6.5259×(100a)=652.596.5259a

Find the force (PA) acting at point A using the relation:

PA=WAB+WBC (5)

Here, WAB is a weight of rod AB and WBC is a weight of rod BC.

Substitute 14.683a for WAB and (652.596.5259a) for WBC in Equation (5).

PA=14.683a+652.596.5259a=8.157a+652.59=652.59+8.157a

Find the normal stress acting at point A using the formula:

σA=PAAAB (6)

Here, σA is normal stress acting at point A and PA is the force acting at point A.

Substitute (652.59+8.157a) for PA and 176.715×106m2 for AAB in Equation (6).

σA=652.59+8.157a176.715×106m2=(3.6930×106)+(46.160×103a)

Find the normal stress acting at point B using the formula:

σB=PBABC (7)

Here, σB is normal stress acting at point B and PB is the force acting at point B.

Consider PB is equal to WBC. So, the value of PB is (652.596.5259a).

Substitute (652.596.5259a) for PB and 78.54×106m2 for ABC in Equation (7).

σB=652.596.5259a78.54×106m2=8.3090×10683.090×103a

Find the length of brass rod AB for the maximum stress in ABC as minimum:

σA=σB (8)

Here, σA is normal stress acting at point A and σB is normal stress acting at point B.

Substitute (8.3090×10683.090×103a) for σB and [(3.6930×106)+(46.160×103a)] for σA in Equation (8).

3.6930×106+46.160×103a=8.3090×10683.090×103a129,250a=4,616,000a=35.71m

Thus, the length of brass rod AB is 35.7mm_.

(b)

Expert Solution
Check Mark
To determine

The value of maximum normal stress of brass rods.

Answer to Problem 6P

The maximum normal stress of brass rods is 5.34MPa_.

Explanation of Solution

Calculation:

Find the maximum normal stress of brass rod using the relation:

σA=3.6930×106+46.160×103a

Substitute 35.71m for a.

σA=3.6930×106+46.160×103(35.71)=3.6930×106+1,648,373.6=5.34×106Pa(1MPa106Pa)=5.34MPa

σB=8.3090×10683.090×103a.

Substitute 35.71m for a in.

σB=8.3090×10683.090×103(35.71)=8.3090×1062,967,143.9=5.34×106Pa(1MPa106Pa)=5.34MPa

Thus, the maximum normal stress of brass rods is 5.34MPa_.

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Chapter 1 Solutions

EBK MECHANICS OF MATERIALS

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