M5-6 CORRECTION DUE TO SAG. A 50-m steel tape weighs 0.04 kg/m and is supported at its end points and at the 8-m and 25-m marks. If a pull of 6 kg is applied, determine the following: a) Correction due to sag between the 0-m and 8-m marks, 8-m and 25-m marks, and the 25-m and 50-m marks. b) Correction due to sag for one tape length. c) Correct distance between the ends of the tape. Steel Tape 50m-->P 8m 25m P4---Om Support at end Support at mid- point Support at end Intermediate support L3 M5-7 NORMAL TENSION. A steel tape having a cross-sectional area of 0.06 cm2 and a modulus of elasticity of 2.0x106 kg/cm², weighs 0.03 kg/m. Its length is 30.00 m when standardized at a pull of 5 kg and supported throughout its length. Determine the tension at which the effect of sag will be eliminated by the elongation of the tape due to increased tension. Fig. 10-6. Determining normal tension. -30-m -0-m mark -Steel Tape(Held above ground) mark - ? ! W = w L 2nd Support Ist Support 30.00 m -O-m mark -Steel Tope (Supported throughout on ground surtace) -30- m mark A W---P 5 kg M5-8 COMBINED CORRECTIONS. A 50-m steel tape is of standard length under a pull of 5.5 kg when supported throughout its entire length. The tape weighs 0.05 kg/m has a cross-sectional area of 0.04 sq.cm, and being made of steel, its modulus of elasticity is 2.10x106 kg/cm². This tape was used in the field to measure a distance that was determined to be 458.650 m. At the time the measurement was made, the constant pull applied was 8 ka with

Determine the 2nd moment of inertia, shearing force in each bolt, and average shearing stress in the bolts.

A short rectangular column 300 mm on one side and 400 mm on the other side is reinforced with 8-16-mm-diameter longitudinal bars equally distributed to the shorter sides of the column. Use fc = 21 MPa and fy = 420 Mpa.  Calculate the required spacing in mm of laterial ties 10-mm-diameter A. 300 B. 250 C. 450 D. None of the above