The tensile strength data provided in the table below is from Machinery's Handbook, and the fastener section of the McMaster-Carr catalog; it is consistent with data from other sources. Apologies in advance for the mixed units, but since most of my customers are from America, I'm guessing they're more comfortable with pounds, as I am.
In metric fasteners, the first number of the grade is approximately 1% of the minimum tensile strength in megapascals, and the second number is the ratio of the minimum yield stress to the minimum tensile strength. If you work through the conversion, you find that you just multiply the first part of the grade number by 14.5 to get the tensile strength in ksi (thousands of pounds per square inch).
The "tensile strength," also known as ultimate strength, is the stress
at which a material will fully fail. The material will retain permanent
deformation at the "yield strength." The shear stress for grades
10.9 and 12.9 is taken from the AIAA Design Handbook for Aerospace Engineers;
it indicates that the ultimate sheer stress is roughly 30% of the ultimate
tensile stress. I'm assuming that applies to the other grades until I find
The table below contains tap drill sizes for common metric threads,
along with the root and shank cross-sectional areas. The root areas are
estimated by taking the tap drill size to be the root diameter. In order
to determine the total load a fastener can withstand, multiply the cross-sectional
area by the ultimate strength for the fastener grade.
|Thread||Tap Drill||Root Area, sq in||Shank Area, sq in|
|M4 x 0.7||#30||0.0130||0.0195|
|M5 x 0.8||#19||0.0216||0.0304|
|M6 x 1.0||#10||0.0294||0.0438|
|M8 x 1.25||6.7 mm / 17/64||0.0546||0.0779|
|M10 x 1.5||8.4 mm / Q||0.0860||0.122|
For example, let's look at the KLR's upper subframe mount. The stock connection uses two, M8 10.9 screws loaded across the threads. The total shear load that the stock connection can support is then 2 x 0.0546 sq in x 44,000 psi = 4800 lbs. The Ultimate Blue upgrade uses an M10 12.9 screw, loaded across the shank on both sides. The maximum shear load for that connection is 2 x 0.122 sq in x 52,000 psi = 12,700 lbs.
These are only rough numbers for comparison, as they do not account for other factors that affect the strength of the joint. Stress concentrations (like the threads) and fatigue will affect joint strength. It is also important to consider the entire state of stress in the fastener when determining a yield point. In the subframe example, the screw is not only under shear stress; it is also under tensile stress from tightening (also known as preload). If the threads had a lot of drag, it could also be under torsional stress.
All of these stresses combine to create the total stress field in the
fastener. Given the axial, shear and torsional stresses, it is possible
to find the "principal axes" of the material in this loading. Stresses
are at a maximum along these axes (and a minimum, depending on what you
are looking at). In short, all of these stresses combine to put a higher
effective shear stress along a certain slice of the bolt, where it may
fail sooner than expected.
Ohm's law relates the voltage across a load to the current through it and its resistance:
Voltage = Current x Resistance
This can be arranged in different ways. For example, you're installing heated grips with a total resistance of 4 ohms. Neglecting the resistance of the wires, the current draw will be I=V/R=13.8 V / 4 ohms = 3.45 amps. Or, you have a load that draws two amps, and the wires leading to it have a resistance of 0.1 ohms. The voltage drop across the wires is V=IR= 2 A x 0.1 ohms = 0.2 V. The applications of Ohm's law are described below.
The power equation tells you the power a load draws given the voltage across it and the current through it. It is simply:
Power = Current x Voltage
Let's say you have a load that draws two amps. The power draw is then P=IV= 2 A x 13.8 V = 27.6 Watts. Another use is to determine resistive heating losses in wires. Using Ohm's law for voltage, the equation can be written as P=IxIxV, or P=I2V. If a wire has a resistance of 0.1 ohms and 5 amps of current running through it, the power loss due to heating of the wire is P=I2V=5*5*0.1=2.5 watts. Since the power loss goes up with the square of the current, it's important to keep the resistance low!
Series and Parallel total resistance
When resistors/loads are connected in series, the total resistance is the sum of the individual resistances: Rt = R1 + R2 + R3... When resistors are connected in parallel, the reciporocal of the sum is equal to the sum of the reciporocals. Or, 1/Rt = 1/R1 + 1/R2 + 1/R3... For example, Kimpex grip heaters have a resistance of about 7 ohms each. When you hook them up in parallel, as shown in the instructions, the total resistance becomes 1/Rt = 1/7 + 1/7 = 2/7, so Rt = 7/2 = 3.5 ohms.
What size wire do I need for my accessory?
Wires have resistance, and this causes power loss in a circuit as shown above. To maintain maximum circuit effectiveness, the voltage drop across the wires running to a load should be no more than 2%.
For example, let's say you're installing a pair of 35 W off-road lights on your bike. Each light has a 3-foot long power wire, and a two-foot long ground wire, for a total of five feet. From the power equation, the current in the wire is I = P/ V= 35/13.8 = 2.5 A. We want the voltage drop across the wire to be less than 2% of 13.8 volts, or 0.28 V. From Ohm's law, the voltage drop across the wire is the product of the current and the resistance, or 0.28 V = 2.5 A x R, or R = 0.28/2.5 = 0.112 ohms. That is a maximum resistance of 0.112 ohms per five feet of wire, or 22.4 ohms per thousand feet. Looking in the table below, the minimum wire size is 22 gage.
Resistance of copper wire. Values are for 77 degrees Farenheit.
Resistance goes up with temperature; for instance, the resistance
of 10 gage wire is 1.18 ohms per thousand feet at 149 degrees Farenheit.
Light Emitting Diodes are great for use as indicators on motorcycles. Compared to a regular incandescent light, they use much less current, aren't suceptible to vibration, and have a service life that is probably greater than the bike's. It is important to note that an LED only passes current in one direction, and typically has a maximum voltage around 2-3 volts.
For a typical example, we'll look at a yellow, 5mm LED from Radio Shack (part number 276-301). It requires 30 mA (miliamps, or thousandths of an amp) at 2.1-2.8 volts. I usually add a little safety factor in case the bike's charging system gets over 13.8 volts, and assume that the LED will run at 2.6 V with the supply voltage at 13.8 V. To keep the voltage across the LED at 2.6 V, we need to drop the remaining voltage across a resistor wired in series with the LED. In this example, the resistor needs to drop 13.8 - 2.6 = 11.2 V, at a current of 30 mA. Using Ohm's law, V = IR, or R = V/I = 11.2/0.030 = 373 ohms. Resistors generally come in standard increments, and it's best to go to the next higher one, which is 390 ohms in this case. You can solder the resistor directly to either lead of the LED, but take care to solder quicky, as the devices can be suceptible to heat. It is also a good idea to coat the leads and resistor in epoxy, as the vibration of the KLR can cause the leads to snap off of the resistor. If you are merely seeking an indicator light in red, amber or green, it is best to just use a pre-wired LED assembly, with the LED and resistor encased in a housing. This protects the components from vibration, and the wires are color-coded for easy hook-up.
Because of the voltage range of LEDs, the LED will still function at 12V (engine off) if it is wired with a resistor designed for 13.8 V operation.
If you are making something big, like a board full of LEDs to put in your tail light or turn signal, it is best to do away with the resistors and wire the LEDs in series in groups of six. If you need an odd number to fill out your display, you can wire them in series and then use an apropriate dropping resistor. Let's say that you have one circuit of four LED's like the one mentioned above. The LEDs would accept a voltage of 4 x 2.6, or 10.4 V. The resistor must then drop 13.8 - 10.4 = 3.4 V. The same current flows through all of the components in a series circuit, so the current is still 30 mA. The dropping resistor value is then 3.4/0.030 = 113 ohms (120 ohms in standard values).
A note on relays
When a DC relay coil is powered, it stores energy. When the switch controlling the coil is turned off, this energy must be dissipated, and it will generate a high enough voltage to arc across the switch contacts. Over time, this will burn them out. The problem can be avoided by placing a diode across the coil terminals, known as an "inductive flyback" diode. An article describing this procedure can be found here.
If you're into building electronic circuits, check out Discover Circuits and Electronic Circuit Diagrams.