indicated to the horizontal neutral axis. The girder is also supporting a mass of 5000kg 30kN 20kN GIRDER 45⁰ 60⁰ 5000K 30⁰ 40kN
indicated to the horizontal neutral axis. The girder is also supporting a mass of 5000kg 30kN 20kN GIRDER 45⁰ 60⁰ 5000K 30⁰ 40kN
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Please solve sub-part- ii completely and accurate with correct calculations. Thank you!

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

Transcribed Image Text:force
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
created by the applied forces. ( magnitude and direction)
buit of oupind
tan(A+B) =
iii Analvee the variation
signs bas obutilonu al bait of sel
Formulae & Identities for Trigonometry: argital wolod wode pobu A
xe leven latositod sdt at bede topl
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(-a)= asin - bcos
and
tan2A =
somongin sau murga emsa sill goized it
and
2 tan A
1-tan² A
R sin(+a)= asin + bcos
IVIDA
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