Find the same solution please

Figure $10\mathrm{.}31$ shows a $\frac{1}{2}$ in$.-13$ UNC grade $5$ steel bolt loaded in double shear $($i$.$e$.,$ the bolt has two shear

planes, as shown$).$ The clamped plates are made of steel and have clean and dry surfaces. The bolt is to

be tightened with a torque wrench to its full proof load; that is$,F_i=S_p{A}_t.$ What force $F$ is the joint capa$-$

ble of withstanding? $($Note: This double shear bolt loading is the same as that on the pin in Figure $2\mathrm{.}14.$

It is assumed that the bolt and plates have adequate strength to prevent the other failure modes discussed in

connection with Figures $2\mathrm{.}14$ and $2\mathrm{.}15..)$

SOLUTION

Known: A specified steel bolt clamps three steel plates and is loaded in double shear.

Find: Determine the force capacity of the joint.

Schematic and Given Data: Assumptions:

The bolt is tightened to its full proof load; that is$,F_i=S_p{A}_t.$

The bolt fails in double shear.

The bolt and plates have adequate strength to prevent other failure modes.

The wrench$-$torque variation is roughly $+-30\%.$

There is a $10\%$ initial loss in tension during the first few weeks of service $($see Section $10\mathrm{.}7).$

Analysis:

For the $\frac{1}{2}$ in$.-13$UNC grade $5$ steel bolt, Table $10\mathrm{.}1$ gives $A_t=0\mathrm{.}1419i{n}^2$ and Table $10\mathrm{.}4$ shows

that $S_p=85$ksi. Specified initial tension is $F_i=S_p{A}_t=85,0000\mathrm{.}1419i{n}^2=12,060lb.$ But

with a roughly estimated $+-30\%$ torque$-$wrench variation and $10\%$ initial$-$tension loss during the

first few weeks of service $($see Section $10\mathrm{.}7),$ a conservative assumption of working value of $F_i$ is

about $7600$ lb $.$

Reference $5$ gives a summary $($p$.78)$ of friction coefficients obtained with bolted plates. The coefficient

for semipolished steel is approximately $0\mathrm{.}3,$ and for sand or grit$-$blasted steel approximately $0\mathrm{.}5.$ Various

paints, platings, and other surface treatments can alter the coefficient markedly, usually downward. Here

a friction coefficient of $0\mathrm{.}4$ is assumed. This gives a force required to slip each of the two interfaces of

$7600lb0\mathrm{.}4=3040lb.$ Thus, the value of $F$ required to overcome friction is estimated to be in the

region of $6000$ lb $.$

Although it is often desirable to limit applied load $F$ to the value that can be transmitted by friction, we

should know the larger value of force that can be transmitted through the bolt itself. For the two shear

planes involved, this force is equal to $2S_{sy}A,$ where $A$ is the area of the bolt at the shear planes$-$in

this case, $\frac{(0\mathrm{.}5{)}^2}{4}=0\mathrm{.}196$ in$.{?}^2$ Taking advantage of the fact that the distortion energy theory gives a

good estimate of shear yield strength for ductile metals, we have $S_{sy}=0\mathrm{.}58S_y=0\mathrm{.}58(92$ksi$)=53$ksi.

Thus, for yielding of the two shear planes, $F=2(0\mathrm{.}196i{n}^2)(53,000)=21,000lb.$

The estimated $21,000-lb$ load would bring the shear stress to the yield strength over the entire cross

section of the shear planes, and the very small amount of yielding would probably result in losing most

or all of the clamping and friction forces. A further increase in load would cause total shear failure, as

indicated in Figure $10\mathrm{.}31b.$ This total failure load is calculated as in step $3,$ except for replacing $S_{sy}$ with

$S_{us}.$ From Eq$.10\mathrm{.}16,S_{us}$~~$74$ksi, the corresponding estimated load is $F=29,000lb.$

Comment: Note that in Figure $10\mathrm{.}31$ the threaded portion of the bolt does not extend to the shear plane. This

is important for a bolt loaded in shear. Extending the thread to the shear plane is conservatively considered to

reduce the she ar area to a circle equal to the thread root diameter; in this case, $A=\frac{(0\mathrm{.}4056{)}^2}{4}=0\mathrm{.}129i{n}^2,$

which is a reduction of $34\%.$