Here's below shows one of the simple shear forces calculation of bolts connection joints when received Mxx = 46.223 kN/M of maximum bending moment.
Mxx = M dead + M live
= 16.147 + 30.076
= 46.223 kN/M
Sunday, March 29, 2020
How To Calculate Shear Forces On Bolts Connection Joints
How to determine shear forces on bolts?
Here's below shows one of the simple shear forces calculation of bolts connection joints when received Mxx = 46.223 kN/M of maximum bending moment.
Mxx = M dead + M live
= 16.147 + 30.076
= 46.223 kN/M
Here's below shows one of the simple shear forces calculation of bolts connection joints when received Mxx = 46.223 kN/M of maximum bending moment.
Mxx = M dead + M live
= 16.147 + 30.076
= 46.223 kN/M
Wednesday, March 18, 2020
Flexible Pavement Design
Flexible Pavement Design
Design Procedure of IRC
Design Approach and Criteria - The following three types of pavement distress resulting from repeated application of traffic are considered:
(i) Horizontal tensile strain at the bottom of the bituminous layer
(ii) Vertical compressive strain at the top of the subgrade
(iii) Pavement deformation with in the bituminous layer
The deformation with in the bituminous layer is assumed to be controlled by meeting the mix design requirements
Failure Criteria
1. Fatigue Cracking
Design Procedure of IRC
Design Approach and Criteria - The following three types of pavement distress resulting from repeated application of traffic are considered:
(i) Horizontal tensile strain at the bottom of the bituminous layer
(ii) Vertical compressive strain at the top of the subgrade
(iii) Pavement deformation with in the bituminous layer
The deformation with in the bituminous layer is assumed to be controlled by meeting the mix design requirements
Failure Criteria
1. Fatigue Cracking
- Is due to the build up of tensile strain at the bottom of Asphaltic Concrete Layer
- Pavement is considered failed if 20% of the surface has cracked
- Is due to the build up of excessive compressive strain at the top of subgrade layer
- Pavement is considered failed if it exhibits a rut depth of 20 mm.
IRC Design Approach and Criteria
The pavement has been modeled as a three layer structure and stresses and strains at
critical locations have been computed using the linear elastic structural model FPAVE.
Friday, October 27, 2017
Nondestructive Testing with Bruker Handheld XRF Analyzers
Non-destructive testing or NDT – also called non destructive examination (NDE), nondestructive inspection (NDI), and non-destructive evaluation (NDE) – with Bruker handheld XRF analyzers can be achieved in seconds or minutes for a broad variety of applications. By contrast, other elemental analysis techniques, such as OES (Optical Emission Spectrometry) leave a spark mark on the alloys being analyzed; and ICP (Inductively Coupled Plasma) analysis or AA (Atomic Absorption) analysis are destructive to the sample. Handheld XRF allows you to perform completely nondestructive testing on practically any material.
When it is crucial that materials not be marred in any way yet elemental analysis is required, nondestructive testing with handheld XRF guns make it possible for many manufacturers, retailers, distributors and consultants to remain compliant with regulations:
- Art Conservation and Restoration
- Consumer Products/H.R.4040 with the (S1 TITAN) XRF Gun
- Lead in Apparel with the S1 TITAN
- TPCH- Toxics in Packaging Clearing House Laws
- ASTM F-963 - The Mandatory Toy Standard
- RoHS- Restriction of Hazardous Substances
- Prop 65- Lead restricted from all products sold in CA
- Jewelry Evaluation - Precious Metals, Gold, Silver
- PMI Analysis
- Positive Materials Identification in Aerospace Alloys
- QA/QC for Alloys
There are also applications in industries where nondestructive testing is not required yet it makes life a lot easier if the sample is not destroyed and can be confirmed with back up analysis in a laboratory:
- Soil Remediation
- Mining Exploration
Whether you are striving to comply with regulations or conducting elemental analysis where nondestructive testing is necessary for optimum efficiency, handheld XRF analyzers from Bruker can get the job done, quickly, effectively and nondestructively.
Saturday, July 11, 2015
Analysis and Design of the Fastener
Analysis and retrofitting design of the fastener group
for the spreader
bar of
SM3 disassembly
Background:
Try to re-use (or to modify) the existing Mino
project spreader bar for SM3
disassembly project. the capacity of the spreader bar for SM3 project
is:
Overall load Pt = 9.4
tons with span distance Ls = 201 in
Applicable codes:
ASME B30.20; “Below – the – Hook Lifting Devices”
ASD, AISC 9th edition
References:
ME – 397459
ME – 397426
MD – 397452
“Steel Structures Design and Behavior” by C. Salmon & J. Johnson, 3rd
edition
Assumptions:
Slip critical connection, single
shear
Figure 1 of
page 1 is showing that when the applying load P is eccentric to the
centroid of the bolt group, this physical configuration is the actual design of
the spread bar using for sm3 project.
Where:
P = 9,400 lbs, applying load
L = 47.625 in
Location A is the geometrical centroid of the bolt group, 6 bolts are located as
showing in Figure 1. currently, it is assuming: A325, ¾ - 10, UNC
n = 6
per Table I –D, part 4 of ASD, Rav
= 7.51 kip (allowable shear load)
Find out the local properties of the
fastener group:
∑x2 = 4 (4)2
= 64 in2
∑y2 = 6 (2)2 = 24 in2
∑x2 + ∑y2 = 88 in2
The primary shear load Rv of
each bolt subject to the applying load P:
Rv = P/n = 9,400 lbs / 6
= 1,567 lbs ↓
The secondary torsional shear load of the
bolt subject to the moment PL:
To
pick the bolt of the most right top one as showing in figure 1,
Where: Rx = PLy ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x
(2 in) ÷ 88 in2
= 10,174 lbs. →
Ry = PLx ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x (4 in) ÷ 88 in2
= 20,349 lbs ↓
The resultant force applying to the most
right top bolt:
R = [(Rv + Ry)2 + Rx2
]1/2
= [(1,567 +
20,349)2 + (10,174)2 ]1/2 lbs
= 23,259 lbs > Rav
= 7.51 kip
It is necessary to look for:
a. Different specifications of the fastener with the same fastener
group.
b. Another pattern of the fastener group
A.Different specification of the fastener
with the same fastener group:
A1.
If using 6 bolts with A325, 1 3/8”- 6 UNC,*
then Rav = 25.2 ksi > R
= 23.26 ksi,
*: The
hole ctr. to hole ctr. distance Le = 4.0 in < 3d = 4.13 in
A2.
If using same fastener group with bolt of A490, 1 ¼”- 7, UNC,
then Rav = 25.8 ksi > R =
23.26 ksi (per Table I-D, part 4 of ASD, 9th edition)
where: Le = 4.0 in > 3d = 3.75
in
B. Modify the current pattern of the fastener
group:
Figure 2 on page 3 is the new fastener group
with adding additional 10 fasteners to
the original group, it can be found the new properties of the fastener
group:
∑x2 = (4 (4)2
+ 4 (2)2 + 6 (6)2) in2
= (64 + 16 + 216) in2
= 296 in2
∑y2 = (6(2)2
+ 8(3.375)2) in2
= 115 in2
∑x2 + ∑y2
= (296 + 115) in2
= 411 in2
Also
n = 16
The primary shear load Rv of
each bolt subject to the applying load P:
Rv = P/n = 9,400 lbs / 16
= 588 lbs ↓
The secondary torsional shear load of the
bolt subject to the moment PL,
To pick the bolt of the most right top one
as denoted as bolt A of Figure 2:
Where: Rx = PLy ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x
(3.375 in) ÷ 411 in2
= 3,677 lbs. →
Ry = PLx ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x (6 in) ÷ 411 in2
= 6,536 lbs ↓
The resultant force R applying to the most
right top bolt A:
R = [(Rv + Ry)2 + Rx2
]1/2
= [(588 + 6,536)2
+ (3,677)2 ]1/2 lbs
= 8,017 lbs > Rav
= 7.51 kip
If the bolt material change to ASTM A490,
then Rav = 9.28 kip
R = 8.017 kip < Rav = 9.28 kip
Since only 2 bolts of the fastener group
will experience shear load of R ~ 8.017 kip, all the rest bolt shear load is
less than 7.51 kip, so there are two choices:
1.
The most top right and bottom
right bolts use A490, the rest bolts use A325. (3/4 – 10, UNC.)
2.
Or all of them use A490 bolts
(3/4 – 10, UNC)
The
conclusions:
There are two ways to modify the current
fastener group to meet the new design criteria of spreader bar for SM3
disassembly:
1.
Using (6) A490 high strength
structural bolts with spec. of 1¼ - 7, UNC (original is ¾ -10, UNC), or
2.
Using (16) A490 high strength
structural bolts with spec. of ¾ -10, UNC.
Saturday, May 3, 2014
Determine the Operating Group of the Hoist
General Comparison
Summarizing
To select correct crane duty, crane structure and mechanical components, the user must identify and pass on the following information to the supplier:
- Average lifts and trolley and bridge movements made in an hour.
- Average length of each movement.
- Estimate the load lifted each time.
- Total operating hour per day.
Wednesday, April 23, 2014
FEM SERVICE CLASS
To determine your crane duty group (according to FEM, Fédération Européene de la Manutention) you need following factors:
1) Load spectrum (Indicates the frequency of maximum and smaller loadings during examined time
period).
2) Class of utilization (This is determined according to number of hoisting cycles during lifetime of crane)
3) Combining these factors is how a duty group is selected.
Example of different load spectrums:
Calculate the Average Daily Operating Time
t = (2 x H x N x T) / (V x 60)
Where:
H = average hoisting height (m or feet)
N = number of work cycles per hour (cycle/hour)
T = daily working time (h)
V = hoisting speed (m/min or feet/min)
1) Load spectrum (Indicates the frequency of maximum and smaller loadings during examined time
period).
2) Class of utilization (This is determined according to number of hoisting cycles during lifetime of crane)
3) Combining these factors is how a duty group is selected.
Example of different load spectrums:
Calculate the Average Daily Operating Time
t = (2 x H x N x T) / (V x 60)
Where:
H = average hoisting height (m or feet)
N = number of work cycles per hour (cycle/hour)
T = daily working time (h)
V = hoisting speed (m/min or feet/min)
Tuesday, April 8, 2014
AISE SERVICE CLASS
AISE also provides for different service classes for cranes covered under AISE Technical Report No. 6, "Specifications for Electric Overhead Traveling Cranes for Steel Mill Service". Like CMAA, AISE also provides a numerical method for determining crane class based on the expected load spectrum. Without getting into the specifics of this method, AISE does generally describe the different service classes (load cycles) as follows:
1. Service Class 1 (N1): Less than 100,000 cycles
2. Service Class 2 (N2): 100,000 to 500,000 cycles
3. Service Class 3 (N3): 500,000 to 2,000,000 cycles
4. Service Class 4 (N4): Over 2,000,000 cycles
Further AISE describe the different Load Classes as
1. L1= Cranes which hoist the rated load exceptionally, and normally hoist very light loads
2. L2= Cranes which rarely hoist the rated load, and normally hoist loads about 1/3 the rated capacity
3. L3= Cranes which hoist the rated load fairly frequently, and normally hoist loads between 1/2 and 2/3 or the rated capacity
4. L4= Cranes which are regularly loaded close to the rated capacity
Based on the load classes and load cycles, the CMMA chart below helps determine the class of the crane.
1. Service Class 1 (N1): Less than 100,000 cycles
2. Service Class 2 (N2): 100,000 to 500,000 cycles
3. Service Class 3 (N3): 500,000 to 2,000,000 cycles
4. Service Class 4 (N4): Over 2,000,000 cycles
Further AISE describe the different Load Classes as
1. L1= Cranes which hoist the rated load exceptionally, and normally hoist very light loads
2. L2= Cranes which rarely hoist the rated load, and normally hoist loads about 1/3 the rated capacity
3. L3= Cranes which hoist the rated load fairly frequently, and normally hoist loads between 1/2 and 2/3 or the rated capacity
4. L4= Cranes which are regularly loaded close to the rated capacity
Based on the load classes and load cycles, the CMMA chart below helps determine the class of the crane.
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