STRESS CAUSED BYAN AXIAL LOAD
A concentric load of 20,000 Ib (88,960 N) is applied to a hanger having a cross-sectional
area of 1.6 in
2
(1032.3 mm
2
). What is the axial stress in the hanger?
Calculation Procedure:
- Compute the axial stress
Use the general stress relation S = PIA = 20,000/1.6 = 12,500 lb/in
2
(86,187.5 kPa).
Related Calculations: Use this general stress relation for a member of any
cross-sectional shape, provided the area of the member can be computed and the member
is made of only one material.
DEFORMATION CAUSED BYAN
AXIALLOAD
A member having a length of 16 ft (4.9 m) and a cross-sectional area of 2.4 in^2 (1548.4
mm^2 ) is subjected to a tensile force of 30,000 Ib (133.4 kN). IfE = 15 x 106 lb/in^2 (103
GPa), how much does this member elongate?
Calculation Procedure:
- Apply the general deformation equation
The general deformation equation is A/ = PLI(AE) = 30,000(16)(12)/[2.4(15 x 106 )1 =
0.16 in (4.06 mm).
Related Calculations: Use this general deformation equation for any material
whose modulus of elasticity is known. For composite materials, this equation must be al-
tered before it can be used.
DEFORMATION OFA BUILT-UP MEMBER
A member is built up of three bars placed end to end, the bars having the lengths and
cross-sectional areas shown in Fig. 16. The member is placed between two rigid surfaces
and axial loads of 30 kips (133 kN) and 10 kips (44 kN) are applied at A and B, respec-
tively. If E = 2000 kips/in^2 (13,788 MPa), determine the horizontal displacement of A and
B.
Calculation Procedure:
- Express the axial force in terms of one reaction
Let RL and RR denote the reactions at the left and right ends, respectively. Assume that
both reactions are directed to the left. Consider a tensile force as positive and a compres-
sive force as negative. Consider a deformation positive if the body elongates and negative
if the body contracts.