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ES230 Strength of Materials
Lesson 5: Stress III
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LESSON 5: STRESS III
Wednesday, February 04, 2015
LESSON OBJECTIVES
1. Compute normal and shear stresses (the “state of stress”) on any arbitrary plane within a uniformly stressed body.
2. Identify brittle engineering materials that fail on planes of maximum tension and ductile engineering materials that
fail on planes that do not contain the maximum tensile stress.
3. Design engineering members that are subjected to pure shear (such as bolts in bolted connections) or pure tension
(such as tension rods in a truss), given the material’s allowable stress.
TEXTBOOK – PHILPOT: SECTIONS 1.1 TO 1.5
DEFINITIONS
 u = the ultimate tensile strength of a material. This is the maximum tensile stress that is acting on a tension test
specimen when it reaches its maximum tensile load
 u = the ultimate shear strength of a material. This is the shear stress that is acting on a shearing plane when the test
specimen reaches its maximum shearing load
 F.S. = factor of safety. This is a given level of safety. For many mechanical and civil applications, it ranges
between 1.5 and 2.
 allow = the allowable tensile strength of a material = u/F.S.
 allow = the allowable shear strength of a material = u/F.S.
 Ductile – the ability to be permanently deformed (e.g., steel, copper, gold)
 Brittle – the inability to be permanently deformed (e.g., glass, cast iron, ceramics)
HOMEWORK (DUE TOMORROW)
0. Do MecMovie M1.11, M1.12, and M1.13 (do not hand anything in). These are excellent practice. Also, see the
block shear video on the webpage.
1. A rectangular bar having a width w=6.00 inches and thickness t=1.50 inches is subjected to a tension load P, as
shown. The normal and shear stresses on plane AB must not exceed 16 ksi and 8 ksi, respectively. Determine the
maximum load P that can be applied without exceeding either stress limit.
2.
The hangers support the joists uniformly, so that it is assumed the four nails on each hanger carry an equal portion of
the load. Determine the smallest diameter nails at A and at B if the ultimate shear stress for the nails is  = 8ksi and
a factor of safety of 2 is needed. The hangers only support vertical loads.
2 nails on each side
(4 total, per hanger)
3.
The 3”x1/8” plate shown is connected to the larger plate with (4) 3/8” diameter bolts in 3/8” diameter holes in single
shear, as shown. Investigate the three following failure modes, then determine which failure mode is the weakest so
that you can report the maximum tension that the connection can withstand:
A. Tensile Fracture – See illustration, below, showing fracture path ab.
B. Block Shear – See illustration, below, showing fracture path cdef. The key to this problem is
understanding that some area has shear stress on it and some area has normal stress on it. However, if this
ES230 Strength of Materials
Lesson 5: Stress III
failure occurs, the shear areas will have reached the ultimate shear stress and the normal areas will have
reached the ultimate normal stress.
C. Bolt Shear
Given: The ultimate normal stress for all materials (bolts and plates) is  = 50 ksi. The ultimate shear stress for all
materials is = 30 ksi.
1”
P
Top View
1”
1”
1” 1”
P
1/8”
Side View
a
P
A). Fracture along
path ab
P
b
d
c
e
f
B). Block Shear
along path cdef
4.
The wooden joint is subjected to the axial member force of 5kN. Determine the average normal stress acting on
sections AB and BC. Assume that the connection is smooth (no FRICTION). The members are 50-mm thick (this
is the dimension into the page). Note: BC is horizontal
Horizontal