STATICS : METHOD OF SECTIONS - GET THE GIVEN, WHAT IS ASKED, AND THE ANSWER. - MAKE A PROPER FREE BODY DIAGRAM. (IT DOES HAVE THE GREATER POINTS IN THIS ACTIVITY) - YOUR SUB ANSWERS MUST BE IN 6 DECIMAL PLACES. - YOUR FINAL ANSWER MUST BE IN 4 DECIMAL PLACES. - TELL THE DIRECTION AND WHETHER IT IS TENSILE OR COMPRESSION. - WRITE YOUR SOLUTION PROPERLY (EASY TO UNDERSTAND). - I PROVIDED A SAMPLE PROBLEM SO THAT YOU WILL KNOW HOW TO SOLVE IT.

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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STATICS : METHOD OF SECTIONS

- GET THE GIVEN, WHAT IS ASKED, AND THE ANSWER.

- MAKE A PROPER FREE BODY DIAGRAM. (IT DOES HAVE THE GREATER POINTS IN THIS ACTIVITY)

- YOUR SUB ANSWERS MUST BE IN 6 DECIMAL PLACES.

- YOUR FINAL ANSWER MUST BE IN 4 DECIMAL PLACES.

- TELL THE DIRECTION AND WHETHER IT IS TENSILE OR COMPRESSION.

- WRITE YOUR SOLUTION PROPERLY (EASY TO UNDERSTAND).

- I PROVIDED A SAMPLE PROBLEM SO THAT YOU WILL KNOW HOW TO SOLVE IT.

THANK YOU IN ADVANCE! 

 

Solve the forces in PQ, FL, QG, and QM.

Hint: Identify all zero-force members and remove them from the structure. In the summation of moments, look for a point where member forces at the supports converge so that their moment arm is zero. From here you will be able to solve PQ. Solve FL using summation of moments from the other half of the section. Solve QG and QM using Method of Joints.

R
1.5
1.5
1.5
1.5
1.5
Q 1.5 kN
1.2 kN
M
K
S
N
4.
3
E
G
J
2.5
1.5
2 kN
ID
(A
3 kN
Note: All measurements are in meters
Transcribed Image Text:R 1.5 1.5 1.5 1.5 1.5 Q 1.5 kN 1.2 kN M K S N 4. 3 E G J 2.5 1.5 2 kN ID (A 3 kN Note: All measurements are in meters
1
3
4. Method of Sections
Sample Problem 6
Given
10m
10m
10m
10m
Required Feo. FsR, Fsu
20 kN
18 kN
65
IK
H
30°
G
20
W
M
12
18 kN
20
70° 1 I
L
35°
9: 59.0362°
E
15 kN
25 kN
15 kN
10
55°
A
27 kN
12m
12m
1
3
4. Method of Sections
X: 59.0362°
Sample Problem 6
Soluti on'
X'
* JONT C
Fco cos(90-r) t 15 cos (90.r-10):
FCD:-16.3349KN, T
FCE
15KN
90-8
90-K-10
FCD
Fee
16.3349 KN, C
FCA
1
2
3
4. Method of Sections
X: 59.0362°
Sample Problem 6
JUINT
x Ź Fx'
オ
FRU
1s 1 FRs cos (2x -90)
18KN
18 + FRS ws
FRS -204 kN T
20-4KN, C
FRS
206-90
FRT
FRS
1
2
3
4. Method of Sections
X : 59.0362°
Fsz: -20:4KN
Sample Problem 6
ST is a
zero force member
d* JONT S
FscQ
FSR
12
FSR COSX + Fsu
12
FSu
-20.4 COS590362 t F su
Fsu 13. 6624 kN, T
Twg T wg
Transcribed Image Text:1 3 4. Method of Sections Sample Problem 6 Given 10m 10m 10m 10m Required Feo. FsR, Fsu 20 kN 18 kN 65 IK H 30° G 20 W M 12 18 kN 20 70° 1 I L 35° 9: 59.0362° E 15 kN 25 kN 15 kN 10 55° A 27 kN 12m 12m 1 3 4. Method of Sections X: 59.0362° Sample Problem 6 Soluti on' X' * JONT C Fco cos(90-r) t 15 cos (90.r-10): FCD:-16.3349KN, T FCE 15KN 90-8 90-K-10 FCD Fee 16.3349 KN, C FCA 1 2 3 4. Method of Sections X: 59.0362° Sample Problem 6 JUINT x Ź Fx' オ FRU 1s 1 FRs cos (2x -90) 18KN 18 + FRS ws FRS -204 kN T 20-4KN, C FRS 206-90 FRT FRS 1 2 3 4. Method of Sections X : 59.0362° Fsz: -20:4KN Sample Problem 6 ST is a zero force member d* JONT S FscQ FSR 12 FSR COSX + Fsu 12 FSu -20.4 COS590362 t F su Fsu 13. 6624 kN, T Twg T wg
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