Equivalents G = 6M = 4L = 8 1. What must be the pull in a L(10) meters tape in order to remove the sagcorrection, given the following?a = M x10“ m?E = 2x10" N/m?Po = G kgWeight of tape, 11 NThe tape length is supported at both ends.

Sag correction is given byCS=w2l324P2Let P=pullCS=11 N210324P2Cs=5041.667P2

1. What must be the pull in a L(10) meters tape in order to remove correction, given the following? a = M x10“ m² Weight of tape, 11 N E= 2x10" N/m² Po = G kg The tape lenoth is supported at both ends

1. What must be the pull in a L(10) meters tape in order to remove correction, given the following? a = M x10“ m² Weight of tape, 11 N E= 2x10" N/m² Po = G kg The tape lenoth is supported at both ends

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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Question

Equivalents
G = 6
M = 4
L = 8

1. What must be the pull in a L(10) meters tape in order to remove the sag
correction, given the following?
a = M x10“ m?
E = 2x10" N/m?
Po = G kg
Weight of tape, 11 N
The tape length is supported at both ends.

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