A (req) = required cross-sectional area of the axially loaded member being designed.
P = applied axial load
(all) = allowable axial stress
Example: Find what diameter steel rod is required to support a 2100# tension force.
P = 2100 lb
(all) = 24,000 psi (= lb/in 2 )
A (req) = P = 2,100 lb = .0875 in 2
(all) 24,000 lb/in 2
diameter = d 2100 lb 2100 lb
A (req) = .0875 in 2 = (r) 2 = (d req ) 2
(d req ) 2 = .0875 in 2 (4) = 0.1114 in 2
d req =(0.1114 in 2 ) 1/2 = .333” = 3/8”
Therefore a 3/8” diameter steel rod supports as much tensile force as 2” x 4”.
Note : App. A,B,C*,D* give cross-sectional areas for different steel shapes. E gives cross-sectional areas for lumber.
3.) Cautions concerning design and
a.) Some materials have a higher allowable compressive stress than tensile stress ( t(all) < c(all) )
Examples: Wood, concrete, cast iron
(refer to App. F, G, 722, 723)
b.) The length of a member effects how much compressive force it can sustain. Will discuss in depth in Week 14 (columns).
C all < T all
For now, we will assume compressive members are short enough to ignore effect of length.
c). Holes for bolts etc. must be subtracted from the gross cross-sectional area to obtain the net cross-sectional area (A net ).
d.) Design Manuals have been published by Steel (AISC), Wood (AITC, NDS), Concrete (ACI), and Aluminum associations.
e.) Assuming even force distribution over the cross-sectional area is usually valid, but not always .
C. STRESS ON NET AREA
Ex. 2 x 4 with bolt holes for 2-1/2” bolts
Determine allowable tensile load
9/16” dia. holes 3 1/2” 1 1/2” T T
T all = t,all A net
t(all) = 400 psi
A net = A - A holes
= (3.5” x 1.5”) - (2)(9/16”)(1.5”)
= 3.56 in 2
T all = (400 psi)(3.56 in 2 ) = 1425 lb
Note : 3.5” and 1.5” are “dressed dimensions” (see APP.E)
2 x 4 is actually 1.5” x 3.5”
2 x 8 is actually 1.5” x 7.25”
D.) Bearing Stress ( p )
-A contact pressure between separate bodies.
-A type of compressive stress
Example: Wood rafter in a building
2 X 4 2 X 8 2,000 lb
p = P A bear P = 2000 lb A bear = 1 1/2” x 71/4” = 10.88 in 2 p = 2,000 lb = 184 psi 10.88 in 2
E.) Shear Stress
Normal stress (tensile and compressive) acts in a direction perpendicular to the surface on which it acts. It is produced by a force whose line of action is perpendicular to the surface on which the stress is produced.
Shear stress acts in a direction parallel to the surface on which it acts. It is produced by a force whose line of action is parallel to the surface on which the stress is produced.
T T Tension -pulling apart V V Shear -sliding past
Example: Nail through boards
Shear Plane V/5 V/5 V V Shear Plane
v = V
A v = shear stress (psi) V = shear force (lb) A= cross-sectional area parallel to the force.
Example: Find the shear stress in the 1” diameter bolt shown.