Question

Deflection due to Axial Loads X 4 kN/m 100 kN 100 m Figure: Linearly Elastic, Prismatic bar loaded with distributed and point axial loads. Given that the bar's cross sectional area, A = 0.0025 m² and Young's Modulus, E = 120 GPa, complete the following. Part 1: Find deflection, $\delta$, of the end of the bar (x= 100 m). [Select] Part 2: Find maximum deflection, $\delta_{max}$, occurring anywhere in the bar. [Select]

          Deflection due to Axial Loads
X
4 kN/m
100 kN
100 m
Figure: Linearly Elastic, Prismatic bar loaded with distributed and point axial loads.
Given that the bar's cross sectional area, A = 0.0025 m² and Young's Modulus, E = 120 GPa, complete the following.
Part 1: Find deflection, $\delta$, of the end of the bar (x= 100 m).
[Select]
Part 2: Find maximum deflection, $\delta_{max}$, occurring anywhere in the bar.
[Select]
        
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Deflection due to Axial Loads
X
4 kN/m
100 kN
100 m
Figure: Linearly Elastic, Prismatic bar loaded with distributed and point axial loads.
Given that the bar's cross sectional area, A = 0.0025 m² and Young's Modulus, E = 120 GPa, complete the following.
Part 1: Find deflection, δ, of the end of the bar (x= 100 m).
[Select]
Part 2: Find maximum deflection, δmax, occurring anywhere in the bar.
[Select]

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Answer choices for both parts are: 33.3mm 37.5mm 100mm 133mm Deflection due to Axial Loads 4 kN/m 100 kN 100 m Figure: Linearly Elastic, Prismatic bar loaded with distributed and point axial loads Given that the bar's cross-sectional area, A = 0.0025 m^2 and Young's Modulus, E = 120 GPa, complete the following: Part 1: Find deflection, δ, of the end of the bar (x = 100 m). [Select ] Part 2: Find maximum deflection, δ_max, occurring anywhere in the bar. [Select]
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problem-104-brass-rod-e-m0-gfa-with-cross-sectional-area-0f-250-mm-is-loaded-by-forces-p-i5-kn-pz-tokn-and-p-8-kn-segment-lengths-of-the-bar-arc-0-20-m-b-075-m-and-l2-m-the-change-in-length-41351

Brass rod (E = 200 GPa) with cross-sectional area of 250 mm² is loaded by forces P₁ = 5 kN, P₂ = 10 kN, and P₃ = 8 kN. Segment lengths of the bar are a = 2.0 m, b = 0.75 m, and c = 1.2 m. The change in length of the bar is approximately: (A) 0.9 mm (B) 1.6 mm (C) 2.1 mm (D) 3.4 mm. Solution 1.04: Draw the axial load diagram first. Start by drawing the axial load diagram from the free end (not the fixed end). Calculate the maximum force independently in each section. Calculate the expansion/contraction in each section independently as well. Make sure your units are consistent.

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problem-104-brass-rod-e-m0-gfa-with-cross-sectional-area-0f-250-mm-is-loaded-by-forces-p-i5-kn-pz-tokn-and-p-8-kn-segment-lengths-of-the-bar-arc-0-20-m-b-075-m-and-l2-m-the-change-in-length-41351

Brass rod (E = 200 GPa) with cross-sectional area of 250 mm² is loaded by forces P₁ = 5 kN, P₂ = 10 kN, and P₃ = 8 kN. Segment lengths of the bar are a = 2.0 m, b = 0.75 m, and c = 1.2 m. The change in length of the bar is approximately: (A) 0.9 mm (B) 1.6 mm (C) 2.1 mm (D) 3.4 mm. Solution 1.04: Draw the axial load diagram first. Start by drawing the axial load diagram from the free end (not the fixed end). Calculate the maximum force independently in each section. Calculate the expansion/contraction in each section independently as well. Make sure your units are consistent.

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problem-104-brass-rod-e-m0-gfa-with-cross-sectional-area-0f-250-mm-is-loaded-by-forces-p-i5-kn-pz-tokn-and-p-8-kn-segment-lengths-of-the-bar-arc-0-20-m-b-075-m-and-l2-m-the-change-in-length-41351

Brass rod (E = 200 GPa) with cross-sectional area of 250 mm² is loaded by forces P₁ = 5 kN, P₂ = 10 kN, and P₃ = 8 kN. Segment lengths of the bar are a = 2.0 m, b = 0.75 m, and c = 1.2 m. The change in length of the bar is approximately: (A) 0.9 mm (B) 1.6 mm (C) 2.1 mm (D) 3.4 mm. Solution 1.04: Draw the axial load diagram first. Start by drawing the axial load diagram from the free end (not the fixed end). Calculate the maximum force independently in each section. Calculate the expansion/contraction in each section independently as well. Make sure your units are consistent.

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Transcript

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00:02 So in this problem, we have a brass rod, and we are trying to find the change in the length of the brass rod.
00:13 And so this is the diagram given approximately, and so we have a series of three forces being applied on our rod at different locations.
00:22 So first let's write down our expression for, i guess, change in length, deformation.
00:30 So delta l is equal to force times the length of that, length of that particular segment where the force is being applied over the cross -sectional area times young's modulus.
00:48 And so another given in the problem is that the area is equal to 250 millimeters squared.
00:58 So now we're going to apply this formula for different segments within our brass.
01:04 Rod.
01:07 So we can say that the total deformation, delta l, will be equal to delta l -a -b, so delta l -a -b being the section over here, plus our delta l -b -c, plus our delta l -cd.
01:32 So c -d will be the segment over here.
01:36 So now what we can do is write down on an expanded form of our expression, so we can say delta l -a -b is f -1 times l -a -b over a -e plus f -2 times l -b -c over a -e, plus f -3, l -c -d over a -e.
02:13 So notice that all these forces are actually different.
02:17 We have an f -1.
02:18 F2 and f3...
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