Part 2: A girder shown below in figure 1 is subject to the following concurrent forces acting at angles indicated to the horizontal neutral axis. The girder is also supporting a mass of 5000kg 30kN 20kN GIRDER 45⁰ figure (1) 60⁰ 5000Kg Sni 30⁰ gravity 9.81 : 40kN
Part 2: A girder shown below in figure 1 is subject to the following concurrent forces acting at angles indicated to the horizontal neutral axis. The girder is also supporting a mass of 5000kg 30kN 20kN GIRDER 45⁰ figure (1) 60⁰ 5000Kg Sni 30⁰ gravity 9.81 : 40kN
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Please solve both sub-parts (i & ii) completely and accurate with correct calculations. Thank you!
![PASS
f
force
i. Using graphical techniques, i.e. use a polygon of forces with a suitable scale to determine the
resultant force created by the all forces present. ( magnitude and direction)
ii. Using analytical techniques i.e. use force resolution methods to determine the horizontal and
Vertical component of each force present, then combine them vectorily to determine resultant
tan(A+B) =
created by the applied forces. ( magnitude and direction)
18/100100 na
but ol
iii Analyse the vorint
has stoltgras bit of pilloyol sel bbu o amor Isaidgang sej
signs bas obintigens al bail or seferino) monogin seu mungib oma erit quiz ni
Formulae & Identities for Trigonometry:gital wolod wode rabe Aud
musu latnosinod srit af badetont
sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
cos(A+B) = cosAcosB - sinAsinB
cos(A-B) = cosAcosB + sinAsinB
sin2A = 2sinAcosA
cos2A = cos²A-sin²A
cos2A = 2cos²A-1
sin²A+ cos²A = 1
tan A + tan B
1-tan A tan B
R sin(0-a)= asin - bcos
and
tan2A =
and
MADE
2 tan A
1-tan² A
R sin(+a)= asin + bcos](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b80f2bf-5a47-4fbd-9670-e2e34003eb04%2F638d0f4d-7e13-41b0-908b-da233ab2ed16%2Fyhre9z_processed.jpeg&w=3840&q=75)
Transcribed Image Text:PASS
f
force
i. Using graphical techniques, i.e. use a polygon of forces with a suitable scale to determine the
resultant force created by the all forces present. ( magnitude and direction)
ii. Using analytical techniques i.e. use force resolution methods to determine the horizontal and
Vertical component of each force present, then combine them vectorily to determine resultant
tan(A+B) =
created by the applied forces. ( magnitude and direction)
18/100100 na
but ol
iii Analyse the vorint
has stoltgras bit of pilloyol sel bbu o amor Isaidgang sej
signs bas obintigens al bail or seferino) monogin seu mungib oma erit quiz ni
Formulae & Identities for Trigonometry:gital wolod wode rabe Aud
musu latnosinod srit af badetont
sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
cos(A+B) = cosAcosB - sinAsinB
cos(A-B) = cosAcosB + sinAsinB
sin2A = 2sinAcosA
cos2A = cos²A-sin²A
cos2A = 2cos²A-1
sin²A+ cos²A = 1
tan A + tan B
1-tan A tan B
R sin(0-a)= asin - bcos
and
tan2A =
and
MADE
2 tan A
1-tan² A
R sin(+a)= asin + bcos
![Part 2: A girder shown below in figure 1 is subject to the following concurrent forces acting at angles
indicated to the horizontal neutral axis. The girder is also supporting a mass of 5000kg
30kN
20kN
GIRDER
45⁰
figure (1)
60⁰
5000Kg
30⁰
gravity 9.81
:
40kN](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b80f2bf-5a47-4fbd-9670-e2e34003eb04%2F638d0f4d-7e13-41b0-908b-da233ab2ed16%2F8k4azhw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Part 2: A girder shown below in figure 1 is subject to the following concurrent forces acting at angles
indicated to the horizontal neutral axis. The girder is also supporting a mass of 5000kg
30kN
20kN
GIRDER
45⁰
figure (1)
60⁰
5000Kg
30⁰
gravity 9.81
:
40kN
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