1. Note: for part a), you can do this without a calculator; it might be easier to work out the problem symbolically, and then substitute the numbers at the end of the problem. C L 2L L The axial rod is fixed to the two walls, as shown above. Portion AB is composed of one material (EAB = E), while portion BD is composed of a different material (EBD = 2E). Portion AB has cross sectional area AAB = A, while portion BD has cross sectional area ABD = 2A. EAB = E = 15×106 psi AAB = A = 1 in² EBD = 2E = 30×10° psi ABD = 2A = 2 in² LAB = L = 10 in. LBC = 2L = 20 in. LcD = L = 10 in. An external force, P = 100 kips, is applied at point C as shown below. a) Determine the reaction forces at wall A and D, Ra and Rp. Specify the direction, (→) or (). b) Deflection at point C. Specify the direction, (→) or (+). (Ans: Ra = 100/7 kips →, Rp = 600/7 kips →, &= 1.429×10² in +)

1. Note: for part a), you can do this without a calculator; it might be easier to work out the problem symbolically, and then substitute the numbers at the end of the problem. C L 2L L The axial rod is fixed to the two walls, as shown above. Portion AB is composed of one material (EAB = E), while portion BD is composed of a different material (EBD = 2E). Portion AB has cross sectional area AAB = A, while portion BD has cross sectional area ABD = 2A. EAB = E = 15×106 psi AAB = A = 1 in² EBD = 2E = 30×10° psi ABD = 2A = 2 in² LAB = L = 10 in. LBC = 2L = 20 in. LcD = L = 10 in. An external force, P = 100 kips, is applied at point C as shown below. a) Determine the reaction forces at wall A and D, Ra and Rp. Specify the direction, (→) or (). b) Deflection at point C. Specify the direction, (→) or (+). (Ans: Ra = 100/7 kips →, Rp = 600/7 kips →, &= 1.429×10² in +)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freedom under the influence of that force. Here, three degrees of freedom are from the translational motion while the other three are from the rotation motion. However, sometimes the bodies are not allowed to undergo any motion and are fixed, under such conditions the body is said to be constrained.

Question

1. Note: for part a), you can do this without a calculator; it might be easier to work out the
problem symbolically, and then substitute the numbers at the end of the problem.
В
2L
The axial rod is fixed to the two walls, as shown above. Portion AB is composed of one material
(EAB = E), while portion BD is composed of a different material (EBD = 2E). Portion AB has
cross sectional area AAB = A, while portion BD has cross sectional area ABD = 2A.
EAB = E = 15×106 psi
AAB = A = 1 in²
EBD = 2E = 30×10° psi
ABD = 2A = 2 in²
LAB = L = 10 in.
LBC = 2L = 20 in.
LcD = L = 10 in.
An external force, P = 100 kips, is applied at point C as shown below.
a) Determine the reaction forces at wall A and D, Ra and Rp. Specify the direction, (→) or (+).
b) Deflection at point C. Specify the direction, (→) or (+).
(Ans: RA = 100/7 kips →, Rp = 600/7 kips →, & = 1.429x10² in –)

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