ISL TL = 499.9666667 --20)382) A horizontal distance of (500 m) must be lay out with a steel tape, length of the steel tape olunder standard conditions is (30 m). Field conditions indicate that standard conditions apply except the measured temperature is 50 °C and applied tension 120'N, Determine the correct -b)L E --50)3 The standard temperature20 °C. *21* Coefficient of thermal expansion of steel tape = 11.6x10-6 °C Po Standard tension = 50N 35 الله ميشه له horizontal distance to be laid out, if W L Mass of the tape = 0.05 kg/m P² 5)Modulus of elasticity (E) = 210*106 N/m2 120) Cross-sectional area of the tape = 0.02 cm² 1.05 +X5 +1.15 = 28 X5= 3-3- X5 = 1.1 1.1-2.4 = -1.3 5675-1.3= 55.4 -
ISL TL = 499.9666667 --20)382) A horizontal distance of (500 m) must be lay out with a steel tape, length of the steel tape olunder standard conditions is (30 m). Field conditions indicate that standard conditions apply except the measured temperature is 50 °C and applied tension 120'N, Determine the correct -b)L E --50)3 The standard temperature20 °C. *21* Coefficient of thermal expansion of steel tape = 11.6x10-6 °C Po Standard tension = 50N 35 الله ميشه له horizontal distance to be laid out, if W L Mass of the tape = 0.05 kg/m P² 5)Modulus of elasticity (E) = 210*106 N/m2 120) Cross-sectional area of the tape = 0.02 cm² 1.05 +X5 +1.15 = 28 X5= 3-3- X5 = 1.1 1.1-2.4 = -1.3 5675-1.3= 55.4 -
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question

Transcribed Image Text:Answer the
do
(0.0026
L'= L te=) 30 + (-0.0026)
s)? (L+= 29.9974
and Explain types of errors in me
nents,
1 (L'/L)ME TL = ( 29.9974/30) *500
H
TL =
499.9566657
382) Arizontal distance of (500 m) must be lay out with a steel tape, length of the steel tape
der standard conditions is (30 m). Field conditions indicate that standard conditions apply
except the measured temperature is 50 °C and applied tension 120'N, Determine the correct
T= (T-Ts) Lx
T= (50-2032) A
=T= 0.0lounder
P= (P-P₂)L
AE
P-(120-50)3 The standard temperature 20 °C.
0.02*21*Coefficient of thermal expansion of steel tape = 11.6x10-6 °C
70.005)
Standard tension = 50N ∙W
C-W² Mass of the tape = 0.05 kg/m
2p²
-(0.05 jan
Total = 0. olo44 +0.005-0.
Total (€) =
mans.
له میشه راه horizontal distance to be laid out, if
Modulus of elasticity (E) = 210*106 N/m2
24 (120) Cross-sectional area of the tape = 0.02 cm²
0.022
022
1.05 +X5 +1·15=3₁3
X5=3-3-22
X5=11
1.1-2.4 -1.3
567 5-1.3= 55.4
fall
1₁4.5-2.28 = -0183
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