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Question 1: A three-bar structural assemblage with both ends fixed, subject to forces F2 and F3 at the locations along its length, is given in Figure 1. Young's modulus for 3 Elements are E=200 GPa. The lengths ($l$) and cross-sectional areas ($A$) for the three bar elements are shown in the Figure. Use the finite element method and hand calculation, to determine: (1) The displacements at node No. 2 and node No. 3. (2) The reaction forces at node No. 1 and node No. 4. (3) The approximated normal stresses within the three elements. $l_1 = 20 \text{ mm}$ $A_1 = 4 \text{ cm}^2$ $l_2 = 30 \text{ mm}$ $A_2 = 6 \text{ cm}^2$ $l_3 = 40 \text{ mm}$ $A_3 = 8 \text{ cm}^2$ F?=2 kN F?=2 kN Figure 1: A three-bar structure subject to two forces

          Question 1: A three-bar structural assemblage with both ends fixed, subject to forces F2 and F3 at the locations along its length, is given in Figure 1.
Young's modulus for 3 Elements are E=200 GPa. The lengths ($l$) and cross-sectional areas ($A$) for the three bar elements are shown in the Figure.
Use the finite element method and hand calculation, to determine:
(1) The displacements at node No. 2 and node No. 3.
(2) The reaction forces at node No. 1 and node No. 4.
(3) The approximated normal stresses within the three elements.
$l_1 = 20 \text{ mm}$ 
$A_1 = 4 \text{ cm}^2$
$l_2 = 30 \text{ mm}$
$A_2 = 6 \text{ cm}^2$
$l_3 = 40 \text{ mm}$
$A_3 = 8 \text{ cm}^2$
F?=2 kN
F?=2 kN
Figure 1: A three-bar structure subject to two forces

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Question 1: A three-bar structural assemblage with both ends fixed, subject to forces F2 and F3 at the locations along its length, is given in Figure 1.
Young's modulus for 3 Elements are E=200 GPa. The lengths (l) and cross-sectional areas (A) for the three bar elements are shown in the Figure.
Use the finite element method and hand calculation, to determine:
(1) The displacements at node No. 2 and node No. 3.
(2) The reaction forces at node No. 1 and node No. 4.
(3) The approximated normal stresses within the three elements.
l1 = 20 mm
A1 = 4 cm^2
l2 = 30 mm
A2 = 6 cm^2
l3 = 40 mm
A3 = 8 cm^2
F?=2 kN
F?=2 kN
Figure 1: A three-bar structure subject to two forces

Added by Brian J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Supreeta N

10/12/2023

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For Element 1: Length (l) = 40mm = 0.04m Cross-sectional area (A) = 8cm^2 = 0.0008m^2 Young's modulus (E) = 200 GPa = 200x10^9 Pa [k1] = (200x10^9 x 0.0008) / 0.04 * [[1, -1], [-1, 1]] [k1] = 4x10^6 * [[1, -1], [-1, 1]] [k1] = [[4x10^6, -4x10^6], [-4x10^6, Show more…

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Please use the following equation (Elemental Matrix) to start your calculation. You do not have to derive this equation from the start in using the "Direct Formulation Technique" or "Minimal Total Potential Energy Technique". [k]=(AE)/(L)[[1,-1],[-1,1]] Question 1: A three-bar structural assemblage with both ends fixed, subject to forces F_(2) and F_(3) at the locations along its length, is given in Figure 1. Young's modulus for 3 Elements are E=200GPa. The lengths (l) and cross-sectional areas (A) for the three bar elements are shown in the Figure. Use the finite element method and hand calculation, to determine: (1) The displacements at node No. 2 and node No. 3. (2) The reaction forces at node No. 1 and node No. 4. (3) The approximated normal stresses within the three elements. Figure 1: A three-bar structure subject to two forces Question 1: A three-bar structural assemblage with both ends fixed,subject to forces F and F3 at the locations along its length,is given in Figure l. Young's modulus for 3 Elements are E=200 GPa.The lengths and cross-sectional areas A for the three bar elements are shown in the Figure. Use the finite element method and hand calculation.to determine (1 The displacements at node No.2 and node No.3. (2 The reaction forces at node No.l and node No.4 (3) The approximated normal stresses within the three elements l=40mm A=8cm2 1=30mm A=6 cm2 l=20mm A=4cm2 1 2 3 F=2kN F3=2kN Figure 1: A three-bar structure subject to two forces

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