Simple Stress
September 3, 2022 | Author: Anonymous | Category: N/A
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MECHANICS OF DEFORMABLE BODIES
ENGR. JOHN PHILIP T. NADAL III Instructor 1
SIMPLE STRESS: Simple stress are expressed as the ratio of the applied force divided by the resisting area or: = It is the ex expr pres essio sion n of force orce per un unit it ar area ea to stru struct ctur ural al me memb mber erss th that at ar are e subjected to external forces and/or induced forces. Stress is the lead to accurately describe and predict the deformation of a body. Simple stress can be classified as normal stress, shear stress, and bearing stress. Normal stress develops when a force is applied perpendicular to the cross-sectional area of the material. If the force is going to pull the material, the stress is said to be tensile stress and compressive stress develops when the material is being stressed by two opposing forces. Shear stress is developed if the applied force is parallel to the resisting area. Example of which is the bolt that holds the tension rod in its anchor. Another condition of shearing is when we twist a bar along its longitudinal axis which is called torsion. Another type of simple stress is bearing stress, wherein it is the
contact pressure between two bodies.
Normal Stresses Stre tress ss
is defined defined as the streng strength th of a materi material al per unit area or unit strength strength.. It is the force on a member divided by area, which carries the force, formerly express in psi, now in N/mm2 or MPa. = where P is the applied normal load in Newton and A is the area in mm 2. The maximum stress in tension or compression occurs over a section normal to the load. Normal stress is either tensile stress or compressive stress. Members subject to pure tension (or tensile force) is under tensile stress, while compression members (members subject to compressive force) are under compressive stress.
Normal stress is either tensile stress or compressive stress. Members Members subject to pure tension (or tensile force) is under tensile stress, while compression members (members subject to compressive force) are under compressive stress.
force willthe tend to shorten the member. Tension force on the other handCompressive will tend to lengthen member. Example 1:
A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in the figure 1.1 below. Axial loads are applied at the positions indicated. Determine the stress in each section.
Figure 1.1
Solution:
To calculate the stresses, we must first determine the axial load in each section. The app appro ropr priat iate e fr freeee-bod body y dia diagr grams ams ar are e sho shown wn in figur figure e 1.2 bel below ow from wh which ich we determin dete rmine e Pbr=4000 lb (tension) and Pst=7000 lb (compr (compression). ession).
Figure 1.2 The stresses in each section are:
=
= =
= (tension)
. .
= (compression)
. . . .
= (compression)
Note that Note that neithe neitherr the length lengthss nor the materi materials als from from which which the the sect sectio ions ns ar are e made made afte afterr calculations of the stresses.
Example 2:
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is limited tto o 120 MN/m2. Solution: Given:
= 40 400 0 = 40000 400000 0 = 120 120
= 4 ( − )
= 400 40 0 000 000 = 120 120
4
− 10 0 00 00
Solving for outside diameter, D:
= .
Figure 1.3
Example 3:
A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.
Figure 1.4
Solution: Free-Body Diagram:
3924 =
By symmetry:
= =
1
(7848 )
2 = 39 3924
For bronze cable: =
= 3924
90
= . For steel cable:
= 3924 =
120
= .
PRACTICE PROBLEMS:
1. The homogeneous bar shown in Fig. 1.5 is supported by a smooth pin at C and a cable that runs from A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 inch and the bar weighs 6000 lb. (Answer: 10458.72 psi)
Figure 1.5
2.) A rod is composed of an aluminum section rigidly attached between steel and bronze sections, as shown in Fig. 1.6. Axial loads are applied at the positions indicated. If P = 3000 lb and the cross-sectional area of the rod is 0.5 in2, determine the stress in each section. (Answer: = ; ; = ; ; = )
Figure 1.6
3.) An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. 1.7. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa. (Ans (Answer: wer: 10000 N)
Figure 1.7
Shear Stress Forces parallel to the area resisting the force cause shearing stress. It differs to tensile and compressive stresses, which are caused by forces perpendicular to the area on which they act. Shearing stress is also known as tangential stress. Figure 1.8 shows the two types of shear stress namely single shear and double shear. = where V is the resultant shearing force which passes through the centroid of the area A being sheared.
Figure 1.8
Example 4: What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2
Solution:
The resisting area is the shaded area along the perimeter and the shear force V is equal to the punching force P.
=
;
= = 350
()(25 ()( 25 )( )(20 20) )
= . = .
Example 5: As in Fig. 1.9, a hole is to be punched out of a plate having a shearing strength of 40 ksi. The compressive compr essive stress stress in the punch is limited to 50 ksi. (a) Compute the t he maximum thickness thickness of plate in which a hole 2.5 inches in diameter can be punched. (b) If the plate is 0.25 inch thick, determine the diameter of the smallest hole that can be punched.
Solution:
a.) Maximum thickness of plate: Based on puncher strength: = = 50 50 00 000 0
2.5
4 = 245 436.93 (equivalent shear force of the plate) Based on shear strength of plate:
=
Figure 1.9
; 40 000
=
245 436.93 ; = .. (2.5 (2. 5 ) )() ()
b.) Diameter of smallest hole: Based on compression of puncher: = = 50 50 00 000 0
= 12 500 (equivalent shear force for plate)
Based on shearing of plate:
=
; 40 000
=
12 500 ()(0.25 )
; = .
Example 6:
Find the smallest diameter bolt that can be used in the clevis shown in Fig. 1.10 if P = 400 kN. The shearing strength strength of the bolt is 300 MPa. Solution:
The bolt is subject to double shear.
=
Figure 1.10
; =
400 1000 = 300 (2) 1 4 = .
Example 6: A 200-mm-diameter pulley is prevented from rotating relative to 60-mm-diameter shaft by a 70-mm-long key, as shown in Fig. 1.11 If a torque T = 2.2 kN·m is applied to the shaft, determine the width b if the allowable shearing stress stress in the key is 60 MPa.
Solution:
Figure 1.11
= 0.03 2.2 − = (0.0 (0.03 3 ) ) = 73.33 73.33 =
; = 1000 70 ( ()) 73.3 73 .33 3 = 60 1 = .
Bearing Stress Bearing stress is the contact pressure between the separa separate te bodies. It differ differss from compressive stress, as it is an internal stress caused by compressive forces.
=
Example 6: In Fig. 1.12, assume that a 20-mm-diameter rivet joins the plates that are each 110 mm wide. The allowable stresses are 120 MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine (a) the minimum thickness of each plate; and (b)
the largest average tensile stress in the plates.
Figure 1.12 Solution: a.) From shearing of rivet:
= = 60
= (20) = 18849 18849.56 .56 4
From bearing of plate material: =
1884 18 849.5 9.56 6 = 12 120 0 (2 (20) 0) = . .
a.) Largest average tensile stress:
1884 18 849.5 9.56 6 = σ(7. σ(7.85 85)(1 )(110 10 − 20 20)) = . .
Example 7: The lap joint shown in Fig. 1.13 is fastened by four ¾-in.-diameter rivets. Calculate the maximum safe load P that can be applied if the shearing stress in the rivets is limited to 14 ksi and the bearing stress in the plates is limited to 18 ksi. Assume the applied load
is uniformly distributed distributed among the four rivets.
Figure 1.12 Solution:
Based on shearing of rivet:
Based on bearing of plates:
=
=
= 14000
4
= . .
4
3 4
= 18000
.
3
4
4
=
∴ = ..
7 8
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