Page 44 - Handbook of Civil Engineering Calculations, Second Edition

STATICS, STRESS AND STRAIN, AND FLEXURAL ANALYSIS 1.27

STRESS CAUSED BY AN AXIAL LOAD

A concentric load of 20,000 lb (88,960 N) is applied to a hanger having a cross-sectional
2
area of 1.6 sq.in. (1032.3 mm ). What is the axial stress in the hanger?

Calculation Procedure:
1. Compute the axial stress
Use the general stress relation s P/A 20,000/1.6 12,500 lb/sq.in. (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 BY AN
AXIAL LOAD

A member having a length of 16 ft (4.9 m) and a cross-sectional area of 2.4 sq.in. (1548.4
6
mm ) is subjected to a tensile force of 30,000 lb (133.4 kN). If E 15 10 lb/sq.in.
2
(103 GPa), how much does this member elongate?
Calculation Procedure:
1. Apply the general deformation equation
The general deformation equation is l PL/(AE) 30,000(16)(12)/[2.4(15 10 )1
6
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 OF A 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 ax-
ial loads of 30 kips (133 kN) and 10 kips (44 kN) are applied at A and B, respectively. If E
2000 kips/sq.in. (13,788 MPa), determine the horizontal displacement of A and B.

Calculation Procedure:
1. Express the axial force in terms of one reaction
Let R L and R R 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.