Complete the table asap with clear and complete solutions.thank you! Bar MemberMagnitude , KNType of Bar ForceABВСBDDEDF150 KN36.00 m8.00 mC4.00 m6.00 m6.00 m180 KN35o250 KN

Answered: Image /qna-images/answer/993de6d1-3096-4586-80b7-0336c3e99312.jpg

Answered: Complete the table asap with clear and…

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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C 5.8 m 3.2 m 1 hi 12 3,5 m %3D h2 4.5 m h2 L203X152X25.4 Shape Steel A992 Gr. 50 Gage of Angles 75 mm g1 T 75 mm g2 Bolt h 28 mm Bolt Diameter, dbolt 50 mm A aspace 65 mm b,pace d. 90 mm Сэрасе 70 mm 12 dapace 90 mm espace 95 mm fapace I || |||
gs9a7d6fg 87bas5b6 t87sa6 78ysadf87y 78asdgfh87wawhtw
PROBLEM 01 REFER TO FIGURE 01, DETERMINE THE INTERNAL FORCES ACTING ON THE FOLLOWING MEMBERS: (STATE COMPRESSION OR TENSION) o FG o FC GH CH
The pin-jointed space truss shown in Fig. P.15.14 is pinned to supports 0, 4, 5 and 9 and is loaded by a force P in the x direction and a force 3P in the negative y direction at the point 7. Find the rotation of the member 27 about the z axis due to this loading. All members have the same cross-sectional area, A, and Young's modulus, E. (Hint. Calculate the deflections in the x direction of joints 2 and 7.) Ans. 382P19AE.
Situation 6: AC3x35 has a length of 1.40 m. Fy-248 MPa. The properties are given below. Properties of C - Channel with L-1.40 meters. A-703 mm^2 d = 76 mm bf - 34.8 mm tf-6.93 mm kdes - 17.5 mm tw - 3.35 mm rx - 30.5 mm ry - 10.0 mm Cw-0.0741 x 10^9 mm 6 ro 39.90 mm H-0.646 J-9.41 x 10^3 mm 4 1. Which of the following most nearly gives the axial capacity (Pu) considering Flexural Buckling if support conditions are both are pinned? 56 KN O 48 KN O 77 KN O 65 KN
1) L ACP 4 = 4i37'is' - 1546'4.8" L ACP 25:51' 14" 2) Longh 6708.4 Sin i5'46'5' AP = Sin 25'32'40" AP = 4227.5s m 3 Lengh BP = /B0 - 0-a AP Sin n I Sin o 138 41'15" 422755 Sin 138'41'ı5" Sin I5'46' 5" /02 70. 27
Зсcos'x - 1 To evaluate 1- sin?x dx, the integrand can be transformed to A J(3tan?x - csc²x) dx ® J3cos x (1- sec²x) dx © (3 cos x - sec?x) dx O S(3cos?x + csc?x) dx D (Зс
W200x19 L25x25x4.8 Determine Ix, Iy , Rx, Sx, Sy
78 56856t57tb 8g7t8gub jntguyi8bh jngb nitg78ib nuhjtg78ib nt78igb n
(plus sights) BS (+) Station HI BMA 6.87' 1031.19' TP1 4.98' TP2 7.43' TP3 0.98' TP4 4.57' TP5 8.20' трь 3.65' TBMX 8.52' TP7 2.22' TP8 7.04' TP9 4.74' TPID 3.68' ΤΡΙ 5.90' BMA Σ = (minus sights) (minus sights) FS (-) IFS (-) 5.32' 6.27' 2.84' 8.34' 3.76 6.52' 5.97' 5.89¹ 4.63' 2.39' 6.00⁰ 8.27 2.60' Check: Misclosure: Permissible Misclosure: Adjustment: Elevation 1024.32 102398 Published Information for BM 303- elevation = 1024.32' (NAVD88) Fieldnote information: BS on BM 303 = 6.87' FS on TP1= 5.32' BS on TP1 = 4.98' FS on TP2 = 6.27' BS on TP2= 7.43' FS on TP3= 2.84' BS on TP3= 0.98' FS on TP4 = 8.34' BS on TP4 = 4.57' FS on TP5= 3.76' BS on TP5= 8.20' FS on TP6= 6.52' BS on Tp6= 3.65' FS on TBM R22 = 5.97' BS on TBM R22 = 8.52' FS on TP7 5.89' BS on TP7= 2.22' FS on TP8= 4.63' BS on TP8 7.04' FS on TP9 = 2.39' BS on TP9 = 4.74' FS on TP10= 6.00' BS on TP10 = 3.68' FS on TP11 = 8.27' BS on TP11= 5.90' FS on BM 303 = 2.60'
Given that the velocity af article chamges with time such that (v16-t), find: A-Draw a B0btain the distance soveridin the fivst(user? sTheraceleration whencil tob sec. velocity-fime gre fom tMhe fivetc4 sec)?
DETERMINE THE maGniTUOE an o DIRE C1O OF RES ULTANT Of TWO FORCES 5kN 650 300

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