Expressing the relationships among the
tensile stresses, we have AL = sALA/E =
sBLB/E = SCLCIE\ therefore, SA = sc, and
SA = sBLBILA = Q15sB for this arrangement
of rods. Since SB is the maximum stress,
the allowable stress first appears in rod B.- Evaluate the stresses at the
instant the load attains its
allowable value
Calculate the load carried by each rod, and
sum these loads to find /\now. Thus SB =
FIGURE 19 22,000 lb/in^2 (151,690.0 kPa); SB =
0.75(22,000) = 16,500 lb/in^2 (113,767.5
kPa); PA = PC= 16,500(1.2) = 19,800 Ib
(88,070.4 N); PB = 22,000(1.0) = 22,000 Ib
(97,856.0 N); />allow = 2(19,800) + 22,000
= 61,600 Ib (273,996.8 N).
Next, consider that the load is gradually increased from zero to its ultimate value.
When rod B attains its yield-point stress, its tendency to deform plastically is inhibited by
rods A and C because the rigidity of the bar constrains the three rods to elongate uniform-
ly. The structure therefore remains stable as the load is increased beyond the elastic range
until rods A and C also attain their yield-point stress. - Find the ultimate load
To find the ultimate load PU9 equate the stress in each rod to fy, calculate the load carried
by each rod, and sum these loads to find the ultimate load Pu. Thus, PA = PC = 36,000(1.2)
= 43,200 Ib (192,153.6 N); PB = 36,000(1.0) = 36,000 Ib (160,128.0 N); Pu = 2(43,200) +
36,000 = 122,400 Ib (544,435.2 N). - Apply the load factor to establish the allowable load
Thus, Pallow = PJLF = 122,400/1.85 = 66,200 Ib (294,457.6 N).
DETERMINATION OF SECTION
SHAPE FACTORS
Without applying the equations and numerical values of the plastic modulus given in the
AISC Manual, determine the shape factor associated with a rectangle, a circle, and a Wl6
x 40. Explain why the circle has the highest and the W section the lowest factor of the
three.
Calculation Procedure:- Calculate My for each section
Use the equation My = Sfy for each section. Thus, for a rectangle, My = bd^2 fy!6. For a cir-
cle, using the properties of a circle as given in the Manual, we find My = 7rd^3 fy/32. For a
W16 x 40, A = 11.77 in^2 (75.940 cm^2 ), S = 64.4 in^3 (1055.52 cm^3 ), and My = 64Afy. - Compute the resultant forces associated with plastification
In Fig. 20, the resultant forces are C and T. Once these forces are known, their action lines
and Mp should be computed.