I.Solve the given matrix problems below:13121111-2A =2-9-3B =33C=34D =24-3-131754-150 14523533-1F =47E =85-3364-2427a. det(A) and det(A+C)Tb. [(A-C)*g]•1 if g = -3

Answered: Image /qna-images/answer/243a29bb-826a-4c8a-a192-2afb70c897b3.jpg

Answered: Solve the given matrix problems below:

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Must include explanation , Don't Just copy ,I will rate accordingly. Q: A square ABCD has sides equal of 200 mm (Figure Q2 (a)). Forces of 150 N each act along AB and CD and 250 N acting along CB and AD. Evaluate the moment of the couple, which will keep the system in equilibrium.
A closed cylindrical vessel of diameter 15 cm and length 100 cm contains water up to a height of 80 cm. The vessel is rotated at speed of 500 rpm about its vertical axis. Find the height in cm of the paraboloid formed. a. 56.06 b. 20.4 c. 891.7 d. 645.23
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8.108 If the intensity of the line loading is w = [(40x − x2)/40] N/mm, where x is measured in mm, use integration to find the resultant. (Ans: R = 0, CR = 9000 N · m CCW)
Situation 5. The column is used to support the floor which exerts a force of 3000 lb on the top of the column. The effect of soil pressure along its side is distributed as shown. 10. Replace this loading by an equivalent resultant force in lb. a. 3253.9 b. 4260.0 c. 3003.3 d. 3918.0 11. Specify where it acts along the column, measured from its base A in ft. a. 3.86 b. 1.62 c. 5.14 d. 1.4
The opposite figure composed of a triangular part ( base =b =10.2 m, height h=5.1 m } , ρ1=3.3 kg/m2) and a semicircular part of diameter a=10.2 m and density ρ2=2 kg/m2. mass of the plate is 1- mass of the plate is: (A.167.55 - B.556.25 - C.204.41 - D.589.76 -E.438.97) 2- XG of the plate is: (A.22.07 - B.15.41 - C.2.09 - D.23.97 - E.9.51) 3- . YG of the plate is: (A.5.85 - B.8.89 - C.3.62 - D.4.56 - E.7.72 ) 4- IXX semicircular part is: (A. 23166.32 - B. 23697.66 - C. 21306.64 - D. 2656.69 - E. 17587.28 ) 5- IXX for total shape is : (A. 3028.77 - B. 10055.53 - C. 7935.39 - D. 3695.10 - E. 10661.29 ) 6- . IYY for total shape is (A.40836.78 - B.28515.34 - C.17602.06 - D.3872.45 -E.44357.19 )
The opposite figure has a triangular part of density ρ1=8.5 kg/m2, and a quadrant of ρ2=17 kg/m2. for a=8.8 m and b=6.9 m, 1- the mass of the total shape is: (A. 286.00 - B. 893.74 - C. 779.08 - D. 1590.85 - E. 428.99) 2- XG= (A. 12.24 - B. 36.33 - C. 25.29 - D. 33.32 - E. 10.04 ) 3- YG= (A. 3.63 - B. 1.43 - C. 2.75 - D. 2.14 - E. 4.45 )
The cylinder having a diameter D ft weighs W lb and is L ft long. Determine the reaction (lb.) at A neglecting friction. Determine the reaction at B (lb.) neglecting friction. Weight density of water is 62.4 lb/ft3 http://Sp.gr Gs = 1.6 D = 4 ft W = 8171 lb L = 8 ft 5. Reaction at A (lb.) is ____________. 6. Reaction at B (lb.) is ____________. Direction is upwards or downwards?
P value is 7.1 kN
Vn = 6*36*1/(4*1.1) = 49.09 am I wrong? I believe in the calculation in the solution the coefficient m was not taken into account. And Pn is not 54 kips either, with the given formula it comes up as 6*36*0.5/(4*1.25) = 21.6 and in which case the design strength would be 21.6 kips. Please let me know if there's something I'm missing from the formulas.

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