CagayanStateUniversit y carig campus tel. No. (078) 304 0818 college of engineering
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Module 1, Unit 1 Topic 1 Homework, Unit 1 Topic 1 Homework, Refresher Problems Module 4-5, Volumes-of-Solids-with-Known-Cross-Sections
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- L’= 29.9838 m. (Correct distance beween the ends of the tape)
| Cs=w2L3/24P2 (1) full tape length
Since W=wL or W2=w2L2, and substituting this to (1), we obtain: Cs= W2L/24p2 (2) length at every support Where: Cs= correction due to sag or the difference between the tape reading and the horizontal distance between supports (m) w= weight of tape per unit length (kg/m) W= total weight of tape between supports (kg) L= interval between supports or the unsupported length of tape (m) P= tension or pull applied on the tape (kg) Enrichment Activities: A subtense bar 2.00 meters long is set up near the middle of two lines A-B. Using a theodolite set up at point A, the angle subtended reads 00025’30”. At point B, the angle reads 000 23’15”. Find the distance A-B. Given: 𝜙A = 000 25’30” 𝜙B = 00023’15” DA =unknown DB = unknown DA-B = DA + DB Solution: DA = 1 ÷ tan 𝜙/2 DA = 1 ÷ tan 00025’30”/2 DA = 269.62 meters Answer DB = 1 ÷ tan 00023’15”/2 DB = 295.72 meters Answer DA-B = 269.62 m + 295.72 m DA-B = 565.34 m Answer A slope distance is measured with a steel tape and found to be 1240.32 ft. If the vertical angle is measured with a theodolite and found to be 3027’, what is the horizontal distance? Given: 𝜙 = 3027’ s= 1240.32 ft D = unknown Solution: d= s x cos𝜙 d = 1240.32 ft x cos 3027’ d= 1238.07 ft Answer A 50 m. steel tape is used to measure a line which was found to be 532.28 meters long when the temperature was 350 centigrade. Determine the following: a). Temperature correction per tape length. b). Temperature correction for the measured line c). Correct length of the line. Note: When MEASURING, the correction is ADDED when tape is TOO LONG When LAYING OUT, the correction is SUBTRACTED when tape is TOO LONG. Given: T= 350C Ts = 200C C= 0.0000116 /0C L = 50 m. L =532.28m Solution:
Ct = CL ( T –Ts) Ct = (0.0000116 /0C) ( 50m) (350C – 200C) Ct = +0.0087 m (Correction per tape length. Positive sign indicates that tape is too long) Answer b.) Ct = (0.0000116 /0C) (532.28m) (350C – 200C) Ct’ = + 0.0926m. (Correction for the measured line) Answer c.) L’= L Ct’ L’ = 532.28 + 0.0926 L’ = 532.37 m (Correct length of the measured line. The correction is added since the tape is too long) Answer A 30-m tape is supported only at its end and under a steady pull of 8 kg. If the tape weighs 0.91 kg, determine the sag correction and the correct distance between the ends of the tape. Given: L=30 m. P= 8 kg.
W= 0.91 kg. Required: Sag Correction (cs) Solution: Cs= W2 L / 24p2 Cs= (0.91)2 (30) / 24 (8)2 Cs= 0.0162 m. (correction due to sag between the end supports) L’= L - Cs L’= 30 m. – 0.0162 m
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