Handbook of Civil Engineering Calculations

are zero. Resolve the reactions RA and Rc (Fig. 7) into their horizontal and vertical com-
ponents.
Considering the entire arch ABC as a free body, take moments with respect to A and C.
Thus SM 4 = 8(10) + 10(25) + 12(40) + 8(56) - 5(25.2) - 12RCV- IO.SRCH = O, or 12RCV

• \NoRCH= 1132, Eq. a. Also, 2MC = 12RAV- 10.8RAH- 8(62) 10(47) - 12(32) - 8(16)

• 5(14.4) = O, or 12RAV- IQ.SRAH= 1550, Eq. b.

2. Consider a segment of the arch and take moments
   Considering the segment BC as a free body, take moments with respect to B. Then SM 5 =
   8(16) + 5(4.8) - 32/?cr + \92RCH = O, or 32RCV- 19.2RCH= 152, Eq. c.


3. Consider another segment and take moments
   Considering segment AB as a free body, take moments with respect to B: ^MB 4QRAV-
   30RAH- 8(30) - 10(15) = O, or 40RAV- 30RAH = 300, Eq. d.


4. Solve the simultaneous moment equations
   Solve Eqs. b and d to determine RA. solve Eqs. a and c to determine Rc. Thus RAV =
   24.4 kips (108.5 kN); RAH = 19.6 kips (87.2 kN); Rcv = 13.6 kips (60.5 kN); RCH =
   14.6 kips (64.9 kN). Then RA = [(24A)
   2
   
   
   • (19.6)
     2
     ]
     0
     
     
     • 5
       = 31.3 kips (139.2 kN). Also Rc =
     
     
     
     
   
   
   
   

[(13.6)^2 + (14.6)^2 ]^05 = 20.0 kips (8.90 kN). And (^6) A = arctan (24.4/19.6) = 51°14';
Oc = arctan (13.6/14.6) - 42°58'.

LENGTH OF CABLE CARRYING

KNOWN LOADS

A cable is supported at points P and g (Fig. Sa) and carries two vertical loads, as shown.
If the tension in the cable is restricted to 1800 Ib (8006 N), determine the minimum length
of cable required to carry the loads.

FIGURE 7