2 2 Classify the trusses in Figure P42 as stable or un hle I
Solution
a) Stable if (m+r>2j). Otherwise unstable.
m= number of members
r= number of reactions.
j= number of joints.
17+5>2*10 =22>20: Hence stable
Degree of indeterminacy = m+r-2j = 22-20 =2: Statically indeterminate to 2nd degree
b) Stable if (m+r>2j). Otherwise unstable.
m= number of members
r= number of reactions.
j= number of joints.
13+5=2*9 19=18: Hence stable
Degree of indeterminacy = m+r-2j = 19-18 =0: Statically determinate
c) Stable if (m+r>2j). Otherwise unstable.
m= number of members
r= number of reactions.
j= number of joints.
8+4=2*6 =12>12: Hence stable
Degree of indeterminacy = m+r-2j = 12-12=0: Statically determinate
d) Stable if (m+r>2j). Otherwise unstable.
m= number of members
r= number of reactions.
j= number of joints.
14+3>2*8 =17>16: Hence stable
Degree of indeterminacy = m+r-2j = 17-16 =1: Statically indeterminate to 1 degree
