1. (a) A glued lap splice is to be made in a 10 × 20 mm rectangular member at α =20°, as shown in Fig.1. Assuming that, the shear strength of the glued joint controls the design, what axial force P is to be applied to the member to get separated to 2 parts? Assume the shear strength of the glued joint to be 10MPa.

Fig: 1
(b) An exploded view of a bolted connection is as shown in Fig 2(b). The width of the plate is 60mm; their thicknesses are t1 =10mm. The snugly fitting bolt is 20mm in diameter. Calculate the maximum normal stress in the plates at the critical section due to an applied tensile force P=70 kN. For the same condition calculate the bearing and shear stresses in the bolt. t1 P/2

P
P/2
Fig:2(a) Fig: 2(b)
P.T.O
1
t1
d
t1
1
2. A 2‐mm‐ thick hollow circular tube of
30 mm outside diameter is subjected to a
constant shear of 10 Pa in the axial
direction, as shown in Fig 3. If the tube is
400 mm long what is the axial stress? Plot
the variation of axial stress along the tube.
Fig: 3
3. For the given state of stress determine the normal and shear stresses exerted on the oblique face AB of the element shown in Fig 4.
80MPa
A
40 MPa
B
Fig: 4
4. For the given state of stress as shown in Fig 5,
Find (a) principal planes and
(b) principal stresses.
48 MPa
16 MPa
60 MPa
Fig: 5
5. For the above problem (Fig 5) determine the following using Mohr’s Circle. (a) the orientation of the planes of maximum in‐plane shearing stress (b) the maximum in plane shearing stress,
(c) the corresponding normal stress.
2
55°

...graph. Different sands used produced different graphs representing different modes of failure. Failure modes were separated into 3 distinct modes, general shear failure, local shear failure and punching shear. By determining the given mode of failure, it can be assessed its ultimate limit load and how a foundation might react when heavily loaded.
Vesic further adapted Terzaghi’s equation to measure the effect of soil compressibility. From measuring this additional effect, factors were added to the initial equation to provide a more sufficient equation.
Effect of strength non-homogeneity on the shape of failure envelopes for combined loading of strip and circular foundations on clay, Gourvenec, S. & Randolph, M. (2003). Géotechnique 53 No.6, 575-586
The context of this journal is offshore skirt foundations on clay. The foundations are under combined loading from self-weight and lateral loading from waves and wind. Non-homogeneity is not taken into account in many of the traditional design calculations. The effect is analysed through the use of a finite element model. By assuming a linear gradient in strength, the shearstrength increases as depth increases, as shown from this equation:
This study has limitations, with the effect of non-homogeneity of a soil only applying to that with a linear strength gradient. With reference to...

...
EFFECT OF JOINT ROUGHNESS ON
SHEARSTRENGTH OF JOINT IN GRANITE
Introduction
Shearstrength is the extreme shearstrength where a material can support before rapture. Based on Lama and Vutukuri (1978)[1], it is one of the critical aspects used for designing purpose. There is no doubt that joint is a main factor that leads to failure in every rocks type including granite. In granite, shearstrength is one of the main characteristic that needs to be determined in order to classify its strength and stability The measurement of roughness of joint is known as Joint Roughness Coeffiecient (JRC). JRC is directly proportional towards the roughness of joint. Higher roughness of joint surface means higher value of JRC.
Project Objectives
The aim of this project is as followed:
* To determine the roughness of joint that affects its shearstrength.
* To undertake shear test on joint of known JRC
* To relate JRC with shearstrength of joint
Methodologies
In this research, the value of JRC will be determined by referring to the surface profile of granite based on a standard diagram introduced by Barton and Choubey (1977)[2]. Apart of using the observation method, JRC is also calculated empirically. As a result, the value of JRC obtained will be related to the...

...ANALYSIS
The vane shear apparatus makes use of the theory that the summation of the total shear stress is equal to the sum of the shear stress at failure along the cylindrical surface and the total shear stress on the top and the bottom of the surface of the cylinder. The various sizes of the vanes are accounted for by applying the multiplying factor. The value obtained from the medium sized vane is the value of the stress itself, and there is no need to multiply or divide the value. The values obtained from the larger vane is divided by two, since it is harder to rotate the vanes because there is more surfaces that create resistance. Lastly, the value obtained from the smallest vane six is multiplied by two since it is easier to rotate the vanes.
From Table 1.0, the values for the undrained shearstrength of the soil for the first trial using the medium vane, larger vane, and smaller vane are as follows: 18kN/m3, 17 kN/m3, and 19 kN/m3, respectively. It can be observed that the values are relatively large, this is because the soil sample is fully saturated. However, the results of each test using different vane sizes yielded different values. The differences in the values can be caused by 1) Personal Errors, 2) Overburden Stress 3) The moisture content of the soil and the 4) condition of the soil. Since the whole experiment is done by eye, it is very much susceptible to personal errors;...

...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.
where V is the resultant shearing force which passes through the centroid of the area A being sheared.
Problem 115
What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? Theshearstrength is 350 MN/m2.
The resisting area is the shaded area along the perimeter and the shear force is equal to the punching force .
answer
Problem 116
As in Fig. 1-11c, a hole is to be punched out of a plate having a shearing strength of 40 ksi. The compressive stress in the punch is limited to 50 ksi. (a) Compute the maximum 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.
(a) Maximum thickness of plate:
Based on puncher strength:
→ Equivalent shear force of the plate
Based on shearstrength of plate:
→
answer
(b) Diameter of smallest hole:
Based on compression of puncher:
→ Equivalent shear force for plate
Based on shearing of plate:
→ ...

...FundEng_Index.book Page 273 Wednesday, November 28, 2007 4:42 PM
9
CHAPTER
Strength
of Materials
James R. Hutchinson
OUTLINE
AXIALLY LOADED MEMBERS 274
Modulus of Elasticity s Poisson’s Ratio
Deformations s Variable Load
THIN-WALLED CYLINDER
s
Thermal
280
GENERAL STATE OF STRESS
282
PLANE STRESS 283
Mohr’s Circle—Stress
STRAIN 286
Plane Strain
HOOKE’S LAW
288
TORSION 289
Circular Shafts s Hollow, Thin-Walled Shafts
BEAMS 292Shear and Moment Diagrams
Stress s Deﬂection of Beams
Equation s Superposition
COMBINED STRESS
COLUMNS
s
s
Stresses in Beams s Shear
Fourth-Order Beam
307
309
SELECTED SYMBOLS AND ABBREVIATIONS
PROBLEMS
312
SOLUTIONS
311
319
Mechanics of materials deals with the determination of the internal forces
(stresses) and the deformation of solids such as metals, wood, concrete, plastics
and composites. In mechanics of materials there are three main considerations in
the solution of problems:
273
FundEng_Index.book Page 274 Wednesday, November 28, 2007 4:42 PM
274
Chapter 9
Strength of Materials
1. Equilibrium
2. Force-deformation relations
3. Compatibility
Equilibrium refers to the equilibrium of forces. The laws of statics must hold
for the body and all parts of the body. Force-deformation relations refer to the
relation of the applied forces to the deformation of the body. If certain forces are...

...BEAM DESIGN FORMULAS
WITH SHEAR AND MOMENT
DIAGRAMS
2005 EDITION
ANSI/AF&PA NDS-2005
Approval Date: JANUARY 6, 2005
ASD/LRFD
N DS
®
NATIONAL DESIGN SPECIFICATION®
FOR WOOD CONSTRUCTION
WITH COMMENTARY AND
SUPPLEMENT: DESIGN VALUES FOR WOOD CONSTRUCTION
American
Forest &
Paper
Association
x
w
Wood
American Wood Council
American Wood Council
R
R
2
2
V
Shear
V
Mmax
Moment
American
Forest &
DESIGN AID No. 6
DESIGN
Paper
Association
BEAM FORMULAS WITH
SHEAR AND MOMENT
DIAGRAMS
The American Wood Council (AWC) is part of the wood products group of the
American Forest & Paper Association (AF&PA). AF&PA is the national trade
association of the forest, paper, and wood products industry, representing member
companies engaged in growing, harvesting, and processing wood and wood fiber,
manufacturing pulp, paper, and paperboard products from both virgin and recycled
fiber, and producing engineered and traditional wood products. For more information
see www.afandpa.org.
While every effort has been made to insure the accuracy
of the information presented, and special effort has been
made to assure that the information reflects the state-ofthe-art, neither the American Forest & Paper Association
nor its members assume any responsibility for any
particular design prepared from this publication. Those
using this document assume all liability from its use....

...SELVAM COLLEGE OF TECHNOLOGY, NAMAKKAL – 637 003 DEPARTMENT OF MECHANICAL ENGINEERING YEAR/SEM: II/ IV
STRENGTH OF MATERIALS
Two marks questions UNIT-I 1. Define tensile stress and tensile strain. The stress induced in a body, when subjected to two equal and opposite pulls, as a result of which there is an increase in length, is known as tensile stress. The ratio of increase in length to th original length is known as tensile strain. p p
2. Define compressive stress and compressive strain. The stress induced in a body, when subjected to two equal and opposite pushes, as a result of which there is an decrease in length, is known as compressive stress. The ratio of increase in length to th original length is known as compressive strain. p p
3. Define shear stress and shear strain.
p p
The stress induced in a body, when subjected to two equal and opposite forces, which are acting tangentially across the resisting section as a result of which the body tends to shear off across the section is known as shear stress and corresponding strain is known as shear strain.
4. Sketch a composite bar
Material-1
Material-1 Material-2
A bar made up of two or more bars of equal length but of different materials rigidly fixed with each other behaving as one unit for extension or for compression when subjected to an axial tensile or compressive load is called as a composite bar. 5....

...College of Engineering
Civil Engineering Department
Experiment No. 1
Vane Shear Test in Cohesive Soil
Submitted by:
Dela Peña, Analyn A.
10948724
LBYCVG2
EJ
Submitted to:
Engr. Irene Ubay
Professor
Submitted on:
October 8, 2012
I. Introduction
Vane shear test is used to measure the shearstrength of a soil. It also estimated and measured the fully saturated clay’s undrained shearstrength without derangement in the specimen. This test can be conducted in field and laboratory however, in laboratory can only execute the experiment with low shearstrength (0.3 kg/cm2) for which unconfined test cannot be performed. The test apparatus are composed of 3 different diameters of 4-blade stainless vane that is attached in a steel rod that pushed vertically in the soil. The pocket value that can get in small vane should multiply by two however, the value can get in large vane should divide by two and the value that can get in medium vane is as it is. The test is performed by pushing the vane vertically in the soil and rotated it clockwise from the surface to determine the torsional force. The soil will resist the rotation of the vane and its resistance is the force of soil that causes the cylindrical area to be sheared by the vane. When the rotation of the vane is continues it means that the soil fails in shear and it is normal that...